suggestion

# Results about Sets of Desirable Gamble Sets

Catrin Campbell-Moore111University of Bristol. Thanks to Arthur van Camp,
Jasper de Bock, Arne Decadt Gert de Cooman, Kevin Blackwell and Jason Konek.
This paper was primarily written in 2021, overlapping with my finishing up
campbellmoore2021isipta. There is some overlap between this paper and
decooman2023filterdesirable; the results of this paper were shown
independently. There was a paper by Jasper de Bock, a very early version of
debock2023desirablethings, available at the time of writing, which also had an
axiomatisation allowing the gamble sets to be infinite and a representation
result in line with sec:repinf, although he adopted axiom
$\mathrm{K}_{\mathrm{AddInf}}$ which we reject. Both papers also omit axiom
$\mathrm{K}_{\mathrm{Dom}}$, which I needed; it follows from the remaining
axioms when we adopt axiom $\mathrm{K}_{\mathrm{AddInf}}$ (thm:dom from
infadd) or when $\Omega$ is finite [Proposition 13]campbellmoore2021isipta.

###### Abstract

Coherent sets of desirable gamble sets is used as a model for representing an
agents opinions and choice preferences under uncertainty. In this paper we
provide some results about the axioms required for coherence and the natural
extension of a given set of desirable gamble sets. We also show that coherent
sets of desirable gamble sets can be represented by a proper filter of
coherent sets of desirable gambles.

###### Contents

  1. 1 Setup - Definition of Coherence
    1. 1.1 Gambles
    2. 1.2 Desirable gambles - usual results
    3. 1.3 Axioms for coherence for sets of desirable gamble sets
  2. 2 Natural Extension for sets of desirable gamble sets
    1. 2.1 The key result about coherence for sets of desirable gamble sets
    2. 2.2 Understanding this
    3. 2.3 Defining the Natural Extension
    4. 2.4 Comments on the Formulation of it
      1. 2.4.1 Quick notes
      2. 2.4.2 On the case when $\mathcal{A}=\varnothing$
      3. 2.4.3 Removing the use of $\mathcal{G}_{\gneq 0}$ in the $\mathrm{posi}$:
    5. 2.5 Relationship to De Bock and de Cooman (2018)
    6. 2.6 Does DBdC get infinite addition axiom?
    7. 2.7 Extending
      1. 2.7.1 Regularity
      2. 2.7.2 Extending to infinite addition
  3. 3 Axioms when gamble sets must be finite
    1. 3.1 Finite addition axiom - axiom
    2. 3.2 Dominators axiom - axiom $\mathrm{K}_{\mathrm{Dom}}$
  4. 4 Representation with desirable gambles
    1. 4.1 Sets of coherent desirable gamble sets
    2. 4.2 Representation of all coherent $\mathcal{K}$ using ${D}$s
  5. A Finite addition proof without supersets

##  1 Setup - Definition of Coherence

###  1.1 Gambles

######  Setup 1.1.

$\mathcal{G}$ is the set of all gambles, which are the bounded functions from
$\Omega$ to $\mathbb{R}$.

When $f(\omega)\geqslant g(\omega)$ for all $\omega\in\Omega$, we will say
$f\geqslant g$.

When $f(\omega)>g(\omega)$ for all $\omega\in\Omega$, we will say $f>g$, or
that $f$ _strictly dominates_ $g$.

When $f\geqslant g$ and $g\not\geqslant f$ we say $f\gneq g$, or that $f$
_weakly dominates_ $g$. That is, when $f(\omega)\geqslant g(\omega)$ for all
$\omega\in\Omega$ and $f(\omega)>g(\omega)$ for some $\omega\in\Omega$.

$\mathcal{G}_{\geqslant 0}$ is the set of gambles where $f\geqslant 0$.

$\mathcal{G}_{>0}$ is the set of gambles which strictly dominate $0$. I.e.,
where $f>0$.

$\mathcal{G}_{\gneq 0}$ is the set of gambles which weakly dominate $0$. I.e.,
where $f\gneq 0$.

We will also make use of the positive linear hull of a set:
$\mathrm{posi}(B):=\\{\sum_{i=1}^{n}\lambda_{i}g_{i}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
n\in\mathbb{N},\lambda_{i}>0,g_{i}\in B\\}$. ∎

###  1.2 Desirable gambles - usual results

######  Definition 1.2.

${D}\subseteq\mathcal{G}$ is coherent if:

  1. ($\mathrm{D}_{0}$)

$0\notin{D}$

  2. ($\mathrm{D}_{\gneq 0}$)

If $g\in\mathcal{G}_{\gneq 0}$, then $g\in{D}$

  3. ($\mathrm{D}_{\lambda}$)

If $g\in{D}$ and $\lambda>0$, then $\lambda g\in{D}$

  4. ($\mathrm{D}_{+}$)

If $f,g\in{D}$, then $f+g\in{D}$ ∎

######  Proposition 1.3.

For $E\subseteq\mathcal{G}$,

  1. (i)

If $0\notin\mathrm{posi}(E\cup\mathcal{G}_{\gneq 0})$ then there is some
coherent ${D}$ extending $E$; and the minimal such coherent ${D}$ is
$\mathrm{posi}(E\cup\mathcal{G}_{\gneq 0})$.

  2. (ii)

If $0\in\mathrm{posi}(E\cup\mathcal{G}_{\gneq 0})$ then there is no coherent
${D}$ extending $E$.

######  Proposition 1.4.

${D}$ is coherent iff ${D}=\mathrm{posi}(E\cup\mathcal{G}_{\gneq 0})$ for some
$E\subseteq\mathcal{G}$, and $0\notin{D}$.

###  1.3 Axioms for coherence for sets of desirable gamble sets

I will give definitions of coherence as in Campbell-Moore (2021), which are
based on those of De Bock and de Cooman (2018) but diverge from them in order
to accommodate the fact that I am not restricting to finite sets of gambles.
In section 3 we’ll show they’re equivalent in the case where gamble sets must
be finite.

######  Definition 1.5.

$\mathcal{K}\subseteq\wp(\mathcal{G})$ is _coherent_ if it satisfies

  1. ($\mathrm{K}_{\varnothing}$)

$\varnothing\not\in\mathcal{K}$

  2. ($\mathrm{K}_{0}$)

If $A\in\mathcal{K}$ then $A\setminus\\{0\\}\in\mathcal{K}$.

  3. ($\mathrm{K}_{\gneq 0}$)

If $g\in\mathcal{G}_{\gneq 0}$, then $\\{g\\}\in\mathcal{K}$

  4. ($\mathrm{K}_{\supseteq}$)

If $A\in\mathcal{K}$ and $B\supseteq A$, then $B\in\mathcal{K}$

  5. ($\mathrm{K}_{\mathrm{Dom}}$)

If $A\in\mathcal{K}$ and for each $g\in A$, $f_{g}$ is some gamble where
$f_{g}\geqslant g$, then
$\\{f_{g}\nonscript\>|\nonscript\>\mathopen{}\allowbreak g\in
A\\}\in\mathcal{K}$.

  6. ($\mathrm{K}_{\mathrm{Add}}$)

If $A_{1},\ldots,A_{n}\in\mathcal{K}$ and for each sequence $\langle
g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times A_{n}$, $f_{\langle
g_{i}\rangle}$ is some member of $\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$,
then

$\\{f_{\langle
g_{i}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times A_{n}\\}\in\mathcal{K}$

∎

Recall
$\mathrm{posi}(B):=\\{\sum_{i=1}^{n}\lambda_{i}g_{i}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
n\in\mathbb{N},\lambda_{i}>0,g_{i}\in B\\}$; so axiom
$\mathrm{K}_{\mathrm{Add}}$ could also be written as:

  * If $A_{1},\ldots,A_{n}\in\mathcal{K}$ then

$\\{\sum_{i}\lambda^{\langle
g_{1},\ldots,g_{n}\rangle}_{i}g_{i}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times A_{n}\\}\in\mathcal{K}$

where $\lambda^{\langle g_{1},\ldots,g_{n}\rangle}_{1},\ldots,\lambda^{\langle
g_{1},\ldots,g_{n}\rangle}_{n}$ are reals all $\geqslant 0$ with at least one
$>0$.

Campbell-Moore (2021) contains some comments on the formulation of axiom
$\mathrm{K}_{\mathrm{Dom}}$.

##  2 Natural Extension for sets of desirable gamble sets

###  2.1 The key result about coherence for sets of desirable gamble sets

This section gives a result which turns out to be the key important property
of coherent sets of desirable gamble sets. We will afterwards show that this
is is a characteristic result of coherence, i.e., it gives us the natural
extension.

######  Theorem 2.1.

Suppose $\mathcal{K}$ is coherent and $A_{1},\ldots,A_{n}\in\mathcal{K}$.

If $B$ is such that for each sequence $\langle g_{1},\ldots,g_{n}\rangle\in
A_{1}\times\ldots\times A_{n}$, either
$0\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$ or
there is some $f_{\langle g_{1},\ldots,g_{n}\rangle}\in B$ with $f_{\langle
g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})$;
Then $B\in\mathcal{K}$.

###### Proof.

We start with a (not terribly interesting) lemma:

######  Sublemma 2.1.1.

For $f\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$,
whenever $f\notin\mathcal{G}_{\gneq 0}$, there is some
$h\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$ with $f\geqslant h$.

###### Proof.

$f\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$ so
$f=\sum_{i}\lambda_{i}g_{i}+\sum_{j}\mu_{j}p_{j}$ for some
$p_{j}\in\mathcal{G}_{\gneq 0}$ and $\lambda_{i},\mu_{j}\geqslant 0$, with at
least one $>0$.

Suppose $f\notin\mathcal{G}_{\gneq 0}$. Then some $\lambda_{i}>0$ because
otherwise $f=\sum_{j}\mu_{j}p_{j}$ with each $p_{j}\in\mathcal{G}_{\gneq 0}$,
and so we’d also have that $f\in\mathcal{G}_{\gneq 0}$.

So, let $h=\sum_{i}\lambda_{i}g_{i}$, and we then know that
$h\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$. Since
$f=h+\sum_{j}\mu_{j}p_{j}$ with each $p_{j}\in\mathcal{G}_{\gneq 0}$, we know
that $f\geqslant h$. ∎

Assume we have $\mathcal{K}$, $A_{1},\ldots,A_{n}$ and $B$ satisfying the
assumptions of the theorem, and we want to show that $B\in\mathcal{K}$.

We know that for any sequence $\langle g_{i}\rangle\in\bigtimes_{i}A_{i}$ with
$0\notin\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$,
there is some $f_{\langle
g_{i}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})$ with $f_{\langle g_{i}\rangle}\in B$. When
$0\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$, we can
let $f_{\langle g_{i}\rangle}$ denote $0$; so we have that for all $\langle
g_{i}\rangle\in\bigtimes_{i}A_{i}$, $f_{\langle g_{i}\rangle}\in B\cup\\{0\\}$
with $f_{\langle
g_{i}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})$.

If there is some $f\in\mathcal{G}_{\gneq 0}$ with $f\in B$ then by axioms
items $\mathrm{K}_{\gneq 0}$ and $\mathrm{K}_{\supseteq}$, $B\in\mathcal{K}$.
So we can assume that each $f_{\langle g_{i}\rangle}\notin\mathcal{G}_{\gneq
0}$ (noting that $0\notin\mathcal{G}_{\gneq 0}$), so by 2.1.1, we can find
$h_{\langle g_{i}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$ with
$f_{\langle g_{i}\rangle}\geqslant h_{\langle g_{i}\rangle}$. By axiom
$\mathrm{K}_{\mathrm{Add}}$,

$\\{h_{\langle
g_{i}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes_{i}A_{i}\\}\in\mathcal{K}.$ (1)

and then by axiom $\mathrm{K}_{\mathrm{Dom}}$,

$\\{f_{\langle
g_{i}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes_{i}A_{i}\\}\in\mathcal{K}.$ (2)

Since each $f_{\langle g_{i}\rangle}\in B\cup\\{0\\}$, by axiom
$\mathrm{K}_{\supseteq}$, $B\cup\\{0\\}\in\mathcal{K}$. And thus
$B\in\mathcal{K}$ by axiom $\mathrm{K}_{0}$. ∎

Note that we have restricted this to finitely many $A_{1},\ldots,A_{n}$ in
$\mathcal{K}$. This is because axiom $\mathrm{K}_{\mathrm{Add}}$ is restricted
to finitely many members.

###  2.2 Understanding this

This will in fact give us maximal information about coherence.

But first we give an idea of why this might be the right thing to do and to
help understand the criterion.

Suppose $A_{1},A_{2}\in\mathcal{K}$. We can then consider what else should be
in $\mathcal{K}$.

You think that $A_{1}$ contains some desirable gamble, and so does $A_{2}$.
But you leave it open which member of $A_{1}$ is desirable and which member of
$A_{2}$.

We can go through each member of $A_{1}$ and consider what are the
consequences of that gamble being the one in virtue of which you think $A_{1}$
contains a desirable gamble. Similarly for $A_{2}$. Consider the possibility
that $A_{1}$ is desirable in virtue of $g_{1}$, and $A_{2}$ is desirable in
virtue of $g_{2}$. In that case, also $g_{1}+g_{2}$ is desirable, and more
generally any $f\in\mathrm{posi}(\\{g_{1},g_{2}\\}\cup\mathcal{G}_{\gneq 0})$
is desirable; so any $B$ which contains some member of
$\mathrm{posi}(\\{g_{1},g_{2}\\}\cup\mathcal{G}_{\gneq 0})$, i.e., when
$\varnothing\neq B\cap\mathrm{posi}(\\{g_{1},g_{2}\\}\cup\mathcal{G}_{\gneq
0})$, contains some desirable gamble.

$\mathrm{posi}(\\{g_{1},g_{2}\\}\cup\mathcal{G}_{\gneq
0})$$g_{1}$$g_{2}$$g_{1}+g_{2}$

But this is all conditional on $g_{1}$ and $g_{2}$ being the desirable
members. If $B$ is a set such that _whichever_ members of $A_{1}$ and $A_{2}$
are considered to be the relevant ones in virtue of which $A_{i}$ desirable,
we can have that some member of $B$ is desirable, then we can outright
conclude that $B$ contains a desirable gamble. We needn’t be able to point to
any particular member of $B$ which we evaluate as desirable. But whenever a
member of $A_{1}$ and a member of $A_{2}$ are desirable, then so is a member
of $B$.

This motivates the general constraint:

  * If for each pair $\langle g_{1},g_{2}\rangle\in A_{1}\times A_{2}$, there is some $f_{\langle g_{1},g_{2}\rangle}\in\mathrm{posi}(\\{g_{1},g_{2}\\}\cup\mathcal{G}_{\gneq 0})$ with $f_{\langle g_{1},g_{2}\rangle}\in B$. Then $B\in\mathcal{K}$. 3

To see how this works in practice, consider fig. 1.

Figure 1:  Suppose $A_{1}=\\{g_{1}^{1},g_{1}^{2}\\}\in\mathcal{K}$ and
$A_{2}=\\{g_{2}^{1},g_{2}^{2}\\}\in\mathcal{K}$. Then this reasoning
(described in section 2.2) gets us that $\\{f_{\langle
g_{1}^{1},g_{2}^{1}\rangle},f_{\langle g_{1}^{1},g_{2}^{2}\rangle},f_{\langle
g_{1}^{2},g_{2}^{1}\rangle},f_{\langle
g_{1}^{2},g_{2}^{2}\rangle}\\}\in\mathcal{K}$.

A proof of the legitimacy of this, given the axioms on coherence, is in
theorem 2.1 since it is a weaker condition than that used in that result.

This story can easily be extended to multiple sets, if $A_{1},\ldots,A_{n}$
are all taken to contain at least one desirable gamble, and whenever we pick
some particular members of them and consider those to be desirable, a member
of $B$ is also desirable, then we conclude that $B$ contains a desirable
gamble. That is:

  * If for each sequence $\langle g_{i}\rangle\in\bigtimes A_{i}$, there is some $f_{\langle g_{i}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$ with $f_{\langle g_{i}\rangle}\in B$. Then $B\in\mathcal{K}$.

Can it be extended to infinitely many sets? Well, this will go beyond the
axioms that have been imposed for coherence. A thought behind this: I am
unable to perform infinite reasoning with desirability. Or alternatively,
maybe I am unable to simultaneously consider the infinitely many sets and pick
the relevant members. Or perhaps one can only make finitely many conditional
assumptions.

So, we have motivated something quite close to that of theorem 2.1. However,
it misses the clause that says that when
$0\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$ we
don’t need to find a $f_{\langle g_{i}\rangle}\in B$. This is because when
$0\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$,
$g_{1},\ldots,g_{n}$ are mutually incompatible. Suppose we have $g_{1}\in
A_{1}$ and $g_{2}\in A_{2}$ with
$0\in\mathrm{posi}(\\{g_{1},g_{2}\\}\cup\mathcal{G}_{\gneq 0})$. It cannot be
that, simultaneously, $A_{1}$ is desirable in virtue of $g_{1}$ and $A_{2}$ is
desirable in virtue of $g_{2}$. The desirability of $g_{1}$ rules out the
desirability of $g_{2}$. So we can ignore this choice of $\langle
g_{1},g_{2}\rangle$. We only need to find the relevant $f_{\langle
g_{i}\rangle}$ when the choice of these $g_{i}$ are mutually compatible.

###  2.3 Defining the Natural Extension

If we start with a given set of desirable gamble sets, $\mathcal{A}$, we
define $\mathrm{Ext}(\mathcal{A})$ by closing it under the construction of
theorem 2.1 (except when we start with $\varnothing$ in which case we give a
separate definition because that construction then doesn’t add anything).

######  Definition 2.2.

For $\mathcal{A}\subseteq\wp(\mathcal{G})$, let $\mathrm{Ext}(\mathcal{A})$ be
given by:
When $\mathcal{A}\neq\varnothing$:

  * •

$B\in\mathrm{Ext}(\mathcal{A})$ iff there are some
$A_{1},\ldots,A_{n}\in\mathcal{A}$ such that for each sequence $\langle
g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times A_{n}$, whenever
$0\notin\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$,
there is some $f_{\langle g_{1},\ldots,g_{n}\rangle}\in B$ where also
$f_{\langle
g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})$.

When $\mathcal{A}=\varnothing$,

  * •

$B\in\mathrm{Ext}(\varnothing)$ iff there is some $f\in B$ where also
$f\in\mathcal{G}_{\gneq 0}$. ∎

This will give us the natural extension notion: the minimal coherent set of
desirable gamble sets extending the given set, $\mathcal{A}$. The formulation
is a bit different to that of De Bock and de Cooman (2018), but they’re
ultimately equivalent. We’ll discuss this in section 2.5.

We now see that this gives us the natural extension:

######  Theorem 2.3.

  1. (i)

If $\varnothing\notin\mathrm{Ext}(\mathcal{A})$ then
$\mathrm{Ext}(\mathcal{A})$ is coherent and it is the minimal coherent
$\mathcal{K}$ extending $\mathcal{A}$.

  2. (ii)

If $\varnothing\in\mathrm{Ext}(\mathcal{A})$ then $\mathrm{Ext}(\mathcal{A})$
is incoherent and there is no coherent $\mathcal{K}$ extending $\mathcal{A}$.

###### Proof.

We first deal with the easy case, when $\mathcal{A}=\varnothing$: If
$\mathcal{K}$ is coherent then by $\mathrm{K}_{\gneq 0}$ and
$\mathrm{K}_{\supseteq}$, any $B\in\mathrm{Ext}(\varnothing)$ has
$B\in\mathcal{K}$. Also observe that $\mathrm{Ext}(\varnothing)$ is coherent.
So it is the minimal coherent extension of $\varnothing$.

So we now just consider when $\mathcal{A}\neq\varnothing$. As a corollary of
theorem 2.1, when $\mathcal{K}$ is coherent extending $\mathcal{A}$, then
$\mathcal{K}\supseteq\mathrm{Ext}(\mathcal{A})$. This is because
$\mathrm{Ext}(\mathcal{A})$ was just defined to take the closure in accordance
with theorem 2.1. We then need to check that $\mathrm{Ext}(\mathcal{A})$ is
coherent (so long as it doesn’t contain $\varnothing$).

######  Sublemma 2.3.1.

Suppose $\mathcal{A}\neq\varnothing$. If
$\varnothing\notin\mathrm{Ext}(\mathcal{A})$, then $\mathrm{Ext}(\mathcal{A})$
is coherent.

###### Proof.

We check each of the axioms.

  * •

$\mathrm{K}_{\varnothing}$: $\varnothing\notin\mathrm{Ext}(\mathcal{A})$ by
assumption.

  * •

$\mathrm{K}_{0}$: Suppose $A\in\mathrm{Ext}(\mathcal{A})$. We need to show
that $A\setminus\\{0\\}\in\mathrm{Ext}(\mathcal{A})$. Since
$A\in\mathrm{Ext}(\mathcal{A})$, there are $B_{1},\ldots,B_{n}\in\mathcal{A}$
where for each $\langle e_{i}\rangle\in\bigtimes B_{i}$, whenever
$0\notin\mathrm{posi}(\\{e_{1},\ldots,e_{n}\\}\cup\mathcal{G}_{\gneq 0})$
there is some $h_{\langle e_{i}\rangle}\in A$ with $h_{\langle
e_{i}\rangle}\in\mathrm{posi}(\\{e_{1},\ldots,e_{n}\\}\cup\mathcal{G}_{\gneq
0})$. For such $\langle e_{i}\rangle$, $h_{\langle e_{i}\rangle}\neq 0$ so
also $h_{\langle e_{i}\rangle}\in(A\setminus\\{0\\})$. Thus,
$A\setminus\\{0\\}\in\mathrm{Ext}(\mathcal{A})$.

  * •

$\mathrm{K}_{\gneq 0}$: Let $g\in\mathcal{G}_{\gneq 0}$. Since
$\mathcal{A}\neq\varnothing$ there is some $B\in\mathcal{A}$. For every $e\in
B$, $g\in\mathrm{posi}(\\{e\\}\cup\mathcal{G}_{\gneq 0})$. So by definition of
$\mathrm{Ext}$, $\\{g\\}\in\mathrm{Ext}(\mathcal{A})$.

  * •

$\mathrm{K}_{\supseteq}$: Suppose $A\in\mathrm{Ext}(\mathcal{A})$ and
$C\supseteq A$. We need to show that $C\in\mathrm{Ext}(\mathcal{A})$. There
are some $B_{i}\in\mathcal{A}$ where for each $\langle
e_{i}\rangle\in\bigtimes B_{i}$, whenever
$0\notin\mathrm{posi}(\\{e_{1},\ldots,e_{n}\\}\cup\mathcal{G}_{\gneq 0})$, we
have some $h_{\langle e_{i}\rangle}\in A$ with $h_{\langle
e_{i}\rangle}\in\mathrm{posi}(\\{e_{1},\ldots,e_{n}\\}\cup\mathcal{G}_{\gneq
0})$. Since $C\supseteq A$, also $h_{\langle e_{i}\rangle}\in C$. So
$C\in\mathrm{Ext}(\mathcal{A})$.

  * •

$\mathrm{K}_{\mathrm{Dom}}$: Suppose $A\in\mathrm{Ext}(\mathcal{A})$ and for
each $g\in A$, $f_{g}$ is a gamble with $f_{g}\geqslant g$. We need to show
that $C:=\\{f_{g}\nonscript\>|\nonscript\>\mathopen{}\allowbreak g\in
A\\}\in\mathrm{Ext}(\mathcal{A})$. There are some $B_{i}\in\mathcal{A}$ where
for each $\langle e_{i}\rangle\in\bigtimes B_{i}$, whenever
$0\notin\mathrm{posi}(\\{e_{1},\ldots,e_{n}\\}\cup\mathcal{G}_{\gneq 0})$,
there is some $h_{\langle e_{i}\rangle}\in A$ with $h_{\langle
e_{i}\rangle}\in\mathrm{posi}(\\{e_{1},\ldots,e_{n}\\}\cup\mathcal{G}_{\gneq
0})$. For each such $\langle e_{i}\rangle$, let $f_{\langle e_{i}\rangle}$ be
the relevant $f_{h_{\langle e_{i}\rangle}}\geqslant h_{\langle e_{i}\rangle}$.
So since $h_{\langle
e_{i}\rangle}\in\mathrm{posi}(\\{e_{1},\ldots,e_{n}\\}\cup\mathcal{G}_{\gneq
0})$ with $f_{\langle e_{i}\rangle}\geqslant h_{\langle e_{i}\rangle}$, also
$f_{\langle
e_{i}\rangle}\in\mathrm{posi}(\\{e_{1},\ldots,e_{n}\\}\cup\mathcal{G}_{\gneq
0})$. And $f_{\langle e_{i}\rangle}\in C$; as required.

  * •

$\mathrm{K}_{\mathrm{Add}}$: Suppose
$A_{1},\ldots,A_{n}\in\mathrm{Ext}(\mathcal{A})$, and $C:=\\{f_{\langle
g_{i}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes A_{i}\\}$, with $f_{\langle
g_{i}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$ (as in
$\mathrm{K}_{\mathrm{Add}}$). We need to show that
$C\in\mathrm{Ext}(\mathcal{A})$.

For each $i$, there are $B^{i}_{1},\ldots,B^{i}_{m_{i}}$ where for any
$\langle e_{j}\rangle\in\bigtimes_{j}B^{i}_{j}$, whenever
$0\notin\mathrm{posi}(\\{e_{1},\ldots,e_{m_{i}}\\}\cup\mathcal{G}_{\gneq 0})$,
there is some $g_{\langle e_{j}\rangle}\in A_{i}$ and $g_{\langle
e_{j}\rangle}\in\mathrm{posi}(\\{e_{1},\ldots,e_{m_{i}}\\}\cup\mathcal{G}_{\gneq
0})$.

Now, consider all of $B^{i}_{j}$ for $i,j$. This is a finite collection. Any
sequence of members of it has the form $\langle
e^{i}_{j}\rangle_{i,j}\in\bigtimes_{i,j}B^{i}_{j}$. Fix such a sequence with
$0\notin\mathrm{posi}(\\{e^{i}_{j}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
i,j\\}\cup\mathcal{G}_{\gneq 0})$. We need to show there is some $f\in C$ with
$f\in\mathrm{posi}(\\{e^{i}_{j}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
i,j\\}\cup\mathcal{G}_{\gneq 0})$.

We also have
$0\notin\mathrm{posi}(\\{e^{i}_{j}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
j\\}\cup\mathcal{G}_{\gneq 0})$ for each $i$; so we have some
$g_{i}:=g_{\langle e^{i}_{j}\rangle_{j}}\in A_{i}$ with
$g_{i}\in\mathrm{posi}(\\{e^{i}_{j}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
j\\}\cup\mathcal{G}_{\gneq 0})$. Since $\langle g_{i}\rangle\in\bigtimes
A_{i}$, by our choice of $C$ there is some $f\in C$ with
$f\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$.

We need to show that
$f\in\mathrm{posi}(\\{e^{i}_{j}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
i,j\\}\cup\mathcal{G}_{\gneq 0})$. This follows from
$f\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$ and each
$g_{i}\in\mathrm{posi}(\\{e^{i}_{j}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
i,j\\}\cup\mathcal{G}_{\gneq 0})$ (since it is in
$\mathrm{posi}(\\{e^{i}_{j}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
j\\}\cup\mathcal{G}_{\gneq 0})$).

So $C\in\mathrm{Ext}(\mathcal{A})$. ∎

So we have shown that when $\varnothing\notin\mathrm{Ext}(\mathcal{A})$,
$\mathrm{Ext}(\mathcal{A})$ is the minimal coherent extension of
$\mathcal{A}$.

We now just need to check that when $\varnothing\in\mathrm{Ext}(\mathcal{A})$,
there is no coherent $\mathcal{K}$ extending $\mathcal{A}$. This follows from
theorem 2.1 because then $\varnothing$ is in any coherent $\mathcal{K}$
extending $\mathcal{A}$, contradicting the supposed coherence of
$\mathcal{K}$. ∎

###  2.4 Comments on the Formulation of it

####  2.4.1 Quick notes

One component of our natural extension formulation is:

> whenever
> $0\notin\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$,
> there is some $f_{\langle g_{1},\ldots,g_{n}\rangle}\in B$ where also
> $f_{\langle
> g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
> 0})$.

This can equivalently be phrased:

> there is some $f_{\langle g_{1},\ldots,g_{n}\rangle}\in B\cup\\{0\\}$ where
> also $f_{\langle
> g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
> 0})$.

We can also phrase things with intersections. I.e.,

> there is some $f_{\langle
> g_{1},\ldots,g_{n}\rangle}\in(B\cup\\{0\\})\cap\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
> 0})$.

Or without the reference to $f_{\langle g_{1},\ldots,g_{n}\rangle}$ at all:

>
> $(B\cup\\{0\\})\cap\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
> 0})\neq\varnothing$.

Just to see what this would look like, we’d then have the definition of
$\mathrm{Ext}$ for $\mathcal{A}\neq\varnothing$ as:

  * •

$B\in\mathrm{Ext}(\mathcal{A})$ iff there are some
$A_{1},\ldots,A_{n}\in\mathcal{A}$ such that for each sequence $\langle
g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times A_{n}$, such that
$(B\cup\\{0\\})\cap\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})\neq\varnothing.$

And we might then want to write the case for $\mathcal{A}=\varnothing$ as:

  * •

$B\in\mathrm{Ext}(\varnothing)$ iff $B\cap\mathcal{G}_{\gneq
0}\neq\varnothing.$

I quite like the reference to the particular member, rather than talking about
the intersection being non-empty, because I think it helps thinking about it
and is useful to have for proofs, which is why I’ve opted for that.

####  2.4.2 On the case when $\mathcal{A}=\varnothing$

I have chosen to define the case of $\mathcal{A}\neq\varnothing$ differently
from when $\mathcal{A}=\varnothing$.

To give a single definition we could include an additional separate clause for
the case where $B\cap\mathcal{G}_{\gneq 0}\neq\varnothing$ (Decadt, 2021). So,
we would offer:

> For any $\mathcal{A}$ (including $\varnothing$),
> $B\in\mathrm{Ext}(\mathcal{A})$ iff either:
>
>   * •
>
> There is some $f\in B$ where also $f\in\mathcal{G}_{\gneq 0}$. Or
>
>   * •
>
> There are some $A_{1},\ldots,A_{n}\in\mathcal{A}$ such that for each
> sequence $\langle g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times
> A_{n}$, whenever
> $0\notin\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$,
> there is some $f_{\langle g_{1},\ldots,g_{n}\rangle}\in B$ where also
> $f_{\langle
> g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
> 0})$.
>
>

An alternative would be to first add something that we know to be in all
coherent $\mathcal{K}$ so that we essentially ensure we’re working with non-
empty $\mathcal{A}$, for example we could add $\mathcal{G}_{\gneq 0}$ itself
or each singleton $\\{g\\}$ for $g\in\mathcal{G}_{\gneq 0}$. This would then
allow us to use the theorem 2.1 characterisation. But I want to keep the non-
empty case as simple as possible as it’s really the interesting one, so I keep
them separate.

Both of these have the disadvantage of introducing additional clauses to be
checked. I have thus opted to keep the definition simpler and simply deal with
$\mathcal{A}=\varnothing$ manually.

####  2.4.3 Removing the use of $\mathcal{G}_{\gneq 0}$ in the
$\mathrm{posi}$:

Decant doesn’t work with
$\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$ but instead
simply with $\mathrm{posi}$ and $\geqslant$. His definition is (with different
formulation):

> For any $\mathcal{A}$ (including $\varnothing$),
> $B\in\mathrm{Ext}(\mathcal{A})$ iff either:
>
>   * •
>
> There is some there are some $f\in B$ where also $f\in\mathcal{G}_{\gneq
> 0}$. Or
>
>   * •
>
> There are some $A_{1},\ldots,A_{n}\in\mathcal{A}$ such that for each
> sequence $\langle g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times
> A_{n}$, whenever
> $0\notin\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$,
> there is some $f_{\langle g_{1},\ldots,g_{n}\rangle}\in B$ and $h_{\langle
> g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$ where
> $f_{\langle g_{1},\ldots,g_{n}\rangle}\geqslant h_{\langle
> g_{1},\ldots,g_{n}\rangle}$.
>
>

######  Proposition 2.4.

This is equivalent to 2.2

###### Proof.

######  Sublemma 2.4.1.

$f\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$ iff
$f\in\mathcal{G}_{\gneq 0}$ or there is some
$h\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$ with $f\geqslant h$.

###### Proof.

The left-to-right direction is 2.1.1.

The right-to-left direction is simple: When $f\in\mathcal{G}_{\gneq 0}$
clearly $f\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})$. When $f\geqslant h$ with $h\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$,
then $f-h\geqslant 0$. If $f=h\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$,
then also $f\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})$. When $f\neq h$, then $f-h\in\mathcal{G}_{\gneq 0}$, so
$f=h+(f-h)\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})$. ∎

Thus, 2.2 is equivalent to:

> When $\mathcal{A}\neq\varnothing$:
>
>   * •
>
> $B\in\mathrm{Ext}(\mathcal{A})$ iff there are some
> $A_{1},\ldots,A_{n}\in\mathcal{A}$ such that for each sequence $\langle
> g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times A_{n}$, whenever
> $0\notin\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$,
> there is some $f_{\langle g_{1},\ldots,g_{n}\rangle}\in B$ where
>
>     * –
>
> either $f_{\langle g_{1},\ldots,g_{n}\rangle}\in\mathcal{G}_{\gneq 0}$
>
>     * –
>
> or there is some $h_{\langle
> g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$ with
> $f_{\langle g_{1},\ldots,g_{n}\rangle}\geqslant h_{\langle
> g_{1},\ldots,g_{n}\rangle}$.
>
>

>
> When $\mathcal{A}=\varnothing$,
>
>   * •
>
> $B\in\mathrm{Ext}(\varnothing)$ iff there is some $f\in B$ where also
> $f\in\mathcal{G}_{\gneq 0}$.
>
>

The first condition of the $\mathcal{A}\neq\varnothing$ case and the
$\mathcal{A}=\varnothing$ case can be combined into the single condition that
$B\cap\mathcal{G}_{\gneq 0}\neq\varnothing$. We are then left with the
alternative condition that

> There are some $A_{1},\ldots,A_{n}\in\mathcal{A}$ such that for each
> sequence $\langle g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times
> A_{n}$, whenever
> $0\notin\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$,
> there is some $f_{\langle g_{1},\ldots,g_{n}\rangle}\in B$ and $h_{\langle
> g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$ where
> $f_{\langle g_{1},\ldots,g_{n}\rangle}\geqslant h_{\langle
> g_{1},\ldots,g_{n}\rangle}$.

So we see that the two formulations are equivalent. ∎

Note that it is important in doing this that the additional clause about
$f\in\mathcal{G}_{\gneq 0}$ is added. That is, we cannot define
$\mathrm{Ext}(\mathcal{A})$ by simply taking our definition (2.2) and
replacing the existence of $f\in
B\cap\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$ with
the existence of $f\in B$ and $h\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$
with $f\geqslant h$. [However, interestingly if we move to strict dominance in
our criteria that would be possible.]

###  2.5 Relationship to De Bock and de Cooman (2018)

De Bock and de Cooman (2018) also give a formulation of the natural extension.
Since they both characterise natural extensions, they’ll be equivalent, but in
this section we consider the difference in more detail.

They:

  1. (i)

Add the singletons from $\mathcal{G}_{\gneq 0}$ to $\mathcal{A}$. This gives
us $\mathcal{A}\cup\\{\\{g\\}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
g\in\mathcal{G}_{\gneq 0}\\}$.

  2. (ii)

We add sets using $\mathrm{posi}$ from these sets. More carefully: For any
$C_{1},\ldots,C_{n}$ in
$\mathcal{A}\cup\\{\\{g\\}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
g\in\mathcal{G}_{\gneq 0}\\}$ (from (i)), if for each sequence $\langle
g_{i}\rangle\in\bigtimes C_{i}$, $f_{\langle
g_{i}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$, then we add
$\\{f_{\langle
g_{i}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes C_{i}\\}$. Our resultant set is called
$\mathrm{Posi}(\mathcal{A}\cup\\{\\{g\\}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
g\in\mathcal{G}_{\gneq 0}\\})$.

  3. (iii)

Add any sets obtained by removing some gambles which are $\leqslant 0$.

  4. (iv)

Add any supersets thereof.

This gives us the final set, called:
$\mathrm{Rs}(\mathrm{Posi}(\mathcal{A}\cup\\{\\{g\\}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
g\in\mathcal{G}_{\gneq 0}\\}))$. They show that this is the natural extension
in the finite setting. So in this setting, we know that it will be equivalent
to ours, however it’s open whether it gives the natural extension in the
infinite setting.

It’s got the same sort of moving parts as our definition: there’s addition of
some $\mathcal{G}_{\gneq 0}$ (in (i)), there’s taking posi’s (in (ii)),
there’s removing some obviously-bad stuff (in (iii)) and taking supersets (in
(iv)). These are the same sorts of moving parts that I have.

Let’s try to rework their formulation to present it in a way that is more
similar to our presentation so we can evaluate how/if they are different.
Firstly, we can summarise the construction to:

  * •

$B\in\mathrm{Ext}_{\mathrm{DBdC}}(\mathcal{A})$ iff there are some
$C_{1},\ldots,C_{n}\in\mathcal{A}\cup\\{\\{g\\}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
g\in\mathcal{G}_{\gneq 0}\\}$ with $\langle g_{1},\ldots,g_{n}\rangle\in
C_{1}\times\ldots\times C_{n}$, there is some $f_{\langle
g_{1},\ldots,g_{n}\rangle}\in B\cup\mathcal{G}_{\leqslant 0}$ where also
$f_{\langle
g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$.

We can then split the choice of $C_{1},\ldots,C_{n}$ from
$\mathcal{A}\cup\\{\\{g\\}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
g\in\mathcal{G}_{\gneq 0}\\}$ into a choice separately of
$A_{1},\ldots,A_{n}\in\mathcal{A}$, and
$p_{1},\ldots,p_{m}\in\mathcal{G}_{\gneq 0}$. Sometimes no members of
$\mathcal{A}$ are chosen, which should be presented as a separate clause. So
we see that:

  * •

$B\in\mathrm{Ext}_{\mathrm{DBdC}}(\mathcal{A})$ iff either:

    * –

There are some $A_{1},\ldots,A_{n}\in\mathcal{A}$ and
$p_{1},\ldots,p_{m}\in\mathcal{G}_{\gneq 0}$ such that for each sequence
$\langle g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times A_{n}$, there is
some $f_{\langle g_{1},\ldots,g_{n}\rangle}\in B\cup\mathcal{G}_{\leqslant 0}$
where also $f_{\langle
g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n},p_{1},\ldots,p_{m}\\})$.

    * –

There are some $p_{1},\ldots,p_{m}\in\mathcal{G}_{\gneq 0}$ such that there is
some $f\in B\cup\mathcal{G}_{\leqslant 0}$ where also
$f\in\mathrm{posi}(\\{p_{1},\ldots,p_{m}\\})$.

Note that this is different to our:

  * •

$B\in\mathrm{Ext}(\mathcal{A})$ iff there are some
$A_{1},\ldots,A_{n}\in\mathcal{A}$ and for each sequence $\langle
g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times A_{n}$, there is some
$f_{\langle g_{1},\ldots,g_{n}\rangle}\in B\cup\\{0\\}$ where also $f_{\langle
g_{1},\ldots,g_{n}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})$.

Some of the differences between these are simply choices of presentation.

However, substantial differences are:

  * •

DBdC allow the choice of the sequence $C_{1},\ldots,C_{n}$ to contain no
members of $\mathcal{A}$ whatsoever. This allows their definition to apply
properly to the case of $\mathcal{A}=\varnothing$, whereas we need a separate
definition for that case.

  * •

Their $\mathsf{Posi}$ is defined working with finitely many
$C_{1},\ldots,C_{n}$, which means in any use of it we can only use finitely
many members of $\mathcal{G}_{\gneq 0}$. Is that enough? It’s closely related
to my addition of item $\mathrm{K}_{\mathrm{Dom}}$.

For example, suppose $A\in\mathcal{A}$ and we have for each $g$ in $A$ a
distinct $p_{g}\in\mathcal{G}_{\gneq 0}$, we want to ensure that
$\\{g+p_{g}\nonscript\>|\nonscript\>\mathopen{}\allowbreak g\in
A\\}\in\mathrm{Ext}(\mathcal{A})$.

This doesn’t immediately follow from the construction of $\mathsf{Posi}$
because it involves infinitely many sets: $A$ plus the infinitely many
singletons $\\{p_{g}\\}$.

In fact, like the dominance axiom, when $\Omega$ is finite it’ll be fine
because
$g+p_{g}\in\mathrm{posi}(\\{g,I_{\omega_{1}},\ldots,I_{\omega_{n}}\\})$ so we
can just work with the finitely many sets
$A,\\{I_{\omega_{1}}\\},\ldots,\\{I_{\omega_{n}}\\}$. But perhaps when
$\Omega$ is infinite it’ll be different.

###  2.6 Does DBdC get infinite addition axiom?

Should DBdC extend the construction to define $\mathsf{Posi}$ that it directly
works with infinitely many sets?

###  2.7 Extending

####  2.7.1 Regularity

We can replace $\mathcal{G}_{\gneq 0}$ by $\mathcal{G}_{>0}$ and everything
still works. Theorem 2.1 then becomes simpler because any $f_{\langle
g_{i}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{>0})$,
has some $h_{\langle g_{i}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\})$
with $f_{\langle g_{i}\rangle}\geqslant h_{\langle g_{i}\rangle}$ so we don’t
need a separate clause.

In fact this shows us that in the strict-dominance setting, axiom
$\mathrm{K}_{\gneq 0}$ is not required as a separate axiom, though one should
then include a non-trivial axiom that $\mathcal{K}$ is non-empty.

####  2.7.2 Extending to infinite addition

If we extend to:

  1. ($\mathrm{K}_{\mathrm{AddInf}}$)

If $A_{i}\in\mathcal{K}$ for each $i\in I$ (possibly infinite) and for each
sequence $\langle g_{i}\rangle_{i\in I}$ with each $g_{i}\in A_{i}$,
$f_{\langle g_{i}\rangle}$ is some member of
$\mathrm{posi}(\\{g_{i}\nonscript\>|\nonscript\>\mathopen{}\allowbreak i\in
I\\})$, then

$\\{f_{\langle
g_{i}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes_{i\in I}A_{i}\\}\in\mathcal{K}$

we obtain a strictly stronger system (Campbell-Moore, 2021, §4.4.1). This
resultant axiom system is an instance of the axioms in De Bock (2023).

The results immediately carry through now instead working with:

  * •

$B\in\mathrm{Ext}(\mathcal{A})$ iff there are some $A_{i}\in\mathcal{K}$ (a
possible infinite collection) where for each sequence $\langle
g_{i}\rangle\in\bigtimes_{i}A_{i}$, there is some $f_{\langle g_{i}\rangle}\in
B\cup\\{0\\}$ with $f_{\langle
g_{i}\rangle}\in\mathrm{posi}(\\{g_{i}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
i\in I\\}\cup\mathcal{G}_{\gneq 0})$.

##  3 Axioms when gamble sets must be finite

This section shows that in the special case when gamble sets are finite, we
can use the axioms from De Bock and de Cooman (2018)

######  Definition 3.1.

$\mathcal{K}\subseteq\wp^{\mathrm{finite}}(\mathcal{G})$ is _coherent_ if it
satisfies $\mathrm{K}_{\varnothing}$ $\mathrm{K}_{0}$ $\mathrm{K}_{\gneq 0}$
$\mathrm{K}_{\supseteq}$ and:

  1. ($\mathrm{K}_{\mathrm{AddPair}}$)

If $A_{1},A_{2}\in\mathcal{K}$ and for each pair $g_{1}\in A_{1}$ and
$g_{2}\in A_{2}$, $f_{\langle g_{1},g_{2}\rangle}$ is some member of
$\mathrm{posi}(\\{g_{1},g_{2}\\})$, then

$\\{f_{\langle
g_{1},g_{2}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak g_{1}\in
A_{1},g_{2}\in A_{2}\\}\in\mathcal{K}$

∎

These are equivalent to our earlier axioms when we are restricted to finite
sets. The proof of this is contained in the next two sections.

###  3.1 Finite addition axiom - axiom $\mathrm{K}_{\mathrm{Add}}$

######  Proposition 3.2.

222Thanks to Arthur van Camp for a key insight, and Jasper De Bock and Arne
Decadt for discussion.

Assume we are restricted to finite sets. Axiom $\mathrm{K}_{\mathrm{Add}}$
follows from axiom $\mathrm{K}_{\mathrm{AddPair}}$ and axiom
$\mathrm{K}_{\supseteq}$.

###### Proof.

We in fact prove:

  * •

If $A_{1},\ldots,A_{n}\in\mathcal{K}$ and for each sequence $\langle
g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times A_{n}$, there is some
$f_{\langle g_{1},\ldots,g_{n}\rangle}\in
B\cap\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}$;
Then $B\in\mathcal{K}$.

which entails axiom $\mathrm{K}_{\mathrm{Add}}$. (It essentially just adds any
supersets. This allows us to avoid some fiddliness in the proof.)

We work by induction on $n$.

Base case: For $n=1$ use axiom $\mathrm{K}_{\mathrm{AddPair}}$ and axiom
$\mathrm{K}_{\supseteq}$, letting $A_{1}=A_{2}$.

Inductive step: Assume it holds for $n$ and consider $n+1$.

So, suppose $A_{1},\ldots,A_{n},A_{n+1}\in\mathcal{K}$ and for each sequence
$\langle g_{i}\rangle\in\bigtimes_{i}A_{i}$ and $a\in A_{n+1}$, there is some
$f_{\langle g_{1},\ldots,g_{n},a\rangle}\in
B\cap\mathrm{posi}(\\{g_{1},\ldots,g_{n},a\\})$. We need to show that
$B\in\mathcal{K}$.

We will abuse notation and write $f_{\langle g_{i},a\rangle}$ instead of
$f_{\langle g_{1},\ldots,g_{n},a\rangle}$.

$A_{n+1}$ is finite, so enumerate it, $A_{n+1}=\\{a_{1},\ldots,a_{K}\\}$.
Define:

$B_{k}:=\left\\{f_{\langle
g_{i},a\rangle}\nonscript\>\middle|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes_{i\leqslant
n}A_{i},\;a\in\\{a_{1},\ldots,a_{k}\\}\\}\cup\\{a_{k+1},\ldots,a_{K}\\}Wherewehavealreadyreplacedk-
manyofA_{n+1}withtherelevantmembersofB.WeshowbyainductiononkthatB_{k}\in\mathcal{K}.SinceB_{K}\subseteq
B,thiswillsuffice.\par Basecase:B_{0}=A_{n+1}\in\mathcal{K}.\par
Inductivestep:wecanassumethatB_{k}\in\mathcal{K}andneedtoshowB_{k+1}\in\mathcal{K}.\par\par
WewillconstructasetCwhichweknowtobein\mathcal{K},whichwecanthencombinewithB_{k}usingaxiom\ref{ax:sets:addpair}(andaxiom\ref{ax:sets:supersets})togetB_{k+1}\in\mathcal{K}.\par\par
Byassumption,f_{\langle
g_{i},a_{k+1}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n},a_{k+1}\\}).Sothereissomeh_{\langle
g_{i},a_{k+1}\rangle}where:\begin{aligned} &h_{\langle
g_{i},a_{k+1}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}),\text{ and
}\\\ &f_{\langle g_{i},a_{k+1}\rangle}\in\mathrm{posi}(\\{a_{k+1},h_{\langle
g_{i},a_{k+1}\rangle}\\}).\end{aligned}Toseethis:f_{\langle
g_{i},a_{k+1}\rangle}hastheform\sum_{i\leqslant n}\lambda_{i}g_{i}+\mu
a_{k+1}.If\lambda_{i}>0forsomei,thenweputh_{\langle
g_{i},a_{k+1}\rangle}:=\sum_{i\leqslant
n}\lambda_{i}g_{i}.Ifnot,thenwesimplyletitbe,forexample,g_{1}.Thisisasrequired.\par
Welet\begin{equation}C:=\\{h_{\langle
g_{i},a_{k+1}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes_{i\leqslant
n}A_{i}\\}\end{equation}C\in\mathcal{K}bytheinductionhypothesis(onn),using\lx@cref{creftype~refnum}{eq:finaddn:h}.\par\par
Wewillshowthatforevery\langle c,b\rangle\in B_{k}\times
C,thereissomed\in\mathrm{posi}(\\{c,b\\})\cap
B_{k+1};whichwillthenallowustouseaxiom\ref{ax:sets:addpair}(and\ref{ax:sets:supersets})togetthatB_{k+1}\in\mathcal{K}.\par
Ifc\in B_{k}\cap
B_{k+1},wejustputd=c.TheonlyremainingmemberofB_{k}toconsiderisc=a_{k+1}.Foranyb\in
C,b=h_{\langle g_{i},a_{k+1}\rangle}forsome\langle
g_{i}\rangle\in\bigtimes_{i}A_{i}.Bydefinitionofthis(\lx@cref{creftype~refnum}{eq:finaddn:f}),f_{\langle
g_{i},a_{k+1}\rangle}\in\mathrm{posi}(\\{a_{k+1},b\\}).AndbylookingatthedefinitionofB_{k+1}(\lx@cref{creftype~refnum}{eq:finaddn:B}),alsof_{\langle
g_{i},a_{k+1}\rangle}\in B_{k+1}.Sof_{\langle
g_{i},a_{k+1}\rangle}istherequiredd\in\mathrm{posi}(\\{c,b\\})\cap
B_{k+1}.\par
So,byaxiom\ref{ax:sets:addpair}andaxiom\ref{ax:sets:supersets},B_{k+1}\in\mathcal{K},completingtheinductivestep.\par
ThishasshownbyinductionthatB_{K}\in\mathcal{K}.B_{K}\subseteq
B,soB\in\mathcal{K}(using\ref{ax:sets:supersets}).\par
Thiscompletestheinductivestepforourinitialinduction(onn).Thuswehaveshownthatourinitialstatementholdsforalln.Andsinceitentailsaxiom\ref{ax:sets:add},wehaveshownthatthisfollows.\qed\end@proof\par
Ithinkthatonecandoitwithoutaxiom\ref{ax:sets:supersets};thoughit^{\prime}llbefiddlier.\leavevmode\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}See\lx@cref{creftype~refnum}{sec:addpair}foranattemptedproof.\par\par\par\par\par$
(5)

###  3.2 Dominators axiom - axiom $\mathrm{K}_{\mathrm{Dom}}$

######  Proposition 3.3.

Assume we are restricted to finite sets. Axiom $\mathrm{K}_{\mathrm{Dom}}$
follows from axiom $\mathrm{K}_{\mathrm{Add}}$ and $\mathrm{K}_{\gneq 0}$.

The proof is closely related to De Bock and de Cooman (2019, Lemma 34). There
they show that when one replaces a single member of a desirable gamble set by
a dominator it remains desirable. We could directly use this and iterate it
(finitely many times) to show our result (to write this down carefully, we’d
have to do a proof by induction). But it is slightly cleaner to directly rely
on our already proved axiom $\mathrm{K}_{\mathrm{Add}}$ and do the
replacements simultaneously.

###### Proof.

Suppose $B=\\{f_{g}\nonscript\>|\nonscript\>\mathopen{}\allowbreak g\in A\\}$,
where $f_{g}\geqslant g$.

Now, $h_{g}:=f_{g}-g\in\mathcal{G}_{\geqslant 0}$.

If $h_{g}\neq 0$, the singleton $\\{h_{g}\\}\in\mathcal{K}$.

So we have $A\in\mathcal{K}$ and each singleton $\\{h_{g}\\}\in\mathcal{K}$
whenever $f_{g}\gneq g$.

Since $A$ is finite, this is finitely many sets.

For any $g^{*}\in A$,
$f_{g^{*}}\in\mathrm{posi}(\\{g^{*}\\}\cup\\{h_{g}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
g\in A\\})$.

So by axiom $\mathrm{K}_{\mathrm{Add}}$, $B\in\mathcal{K}$. ∎

We can similarly see:

######  Proposition 3.4.

Axiom $\mathrm{K}_{\mathrm{Dom}}$ follows from axiom
$\mathrm{K}_{\mathrm{AddInf}}$ and axiom $\mathrm{K}_{\gneq 0}$ even when sets
can be infinite.

###### Proof.

The argument from proposition 3.3. ∎

##  4 Representation with desirable gambles

There is a representation result in the finite setting which says that every
coherent $\mathcal{K}$ can be captured by a set of coherent desirable gambles:
De Bock and de Cooman (2019, Theorem 9). We have to generalise this result in
the infinite setting if we want to avoid item $\mathrm{K}_{\mathrm{AddInf}}$
and merely impose item $\mathrm{K}_{\mathrm{Add}}$.

###  4.1 Sets of coherent desirable gamble sets

If we have a set of desirable gambles, ${D}$, we can extract a set of
desirable gamble sets, $\mathcal{K}_{D}$: ${D}$ evaluates a gamble set $B$ as
desirable when it thinks that some member of the set is desirable

$\displaystyle B\in\mathcal{K}_{D}$ iff there is some $g\in B$ with $g\in{D}$
(6) $\displaystyle\text{ iff }B\cap{D}\neq\varnothing$ (7)

This results in very special kinds of sets of desirable gamble sets. We can
get more if we instead look at generating a set of desirable gamble sets by
using a _set_ ($\mathbb{D}$) of coherent sets of desirable gambles (${D}$s).

If we have a set of coherent ${D}$s, $\mathbb{D}$, we can consider evaluations
of a set of gambles, $B$, as desirable if every ${D}\in\mathbb{D}$ evaluates
it as desirable. That is:

$\displaystyle B\in\mathcal{K}_{\mathbb{D}}$ iff $B\in\mathcal{K}_{D}$ for
each ${D}\in\mathbb{D}$ (8) iff for each ${D}\in\mathbb{D}$ there is some
$g\in B$ with $g\in{D}$ (9) iff for each ${D}\in\mathbb{D}$ there is some
$g\in B\cap{D}$ (10)

More concisely,
$\mathcal{K}_{\mathbb{D}}:=\bigcap_{{D}\in\mathbb{D}}\mathcal{K}_{D}$.

When $\mathbb{D}$ is a nonempty set of coherent ${D}$,
$\mathcal{K}_{\mathbb{D}}$ is coherent. But it also satisfies axiom
$\mathrm{K}_{\mathrm{AddInf}}$.

######  Proposition 4.1.

Let $\mathbb{D}$ be a nonempty set of coherent ${D}$s.
$\mathcal{K}_{\mathbb{D}}$ satisfies K${}_{\text{infadd}}$.444This result is
also De Bock (2023, Theorem 9, Example 9)

###### Proof.

Let $A_{i}\in\mathcal{K}_{\mathbb{D}}$ with $f_{\langle
g_{i}\rangle}\in\mathrm{posi}(\\{g_{i}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
i\in I\\})$ with $B=\\{f_{\langle
g_{i}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes A_{i}\\}$.

So for each ${D}\in\mathbb{D}$, ${D}\cap A_{i}\neq\varnothing$.

Fix any ${D}\in\mathbb{D}$. Take $g_{i}$ to be some member of $A_{i}\cap{D}$.
Now, $f_{\langle
g_{i}\rangle}\in\mathrm{posi}(\\{g_{i}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
i\in I\\})$, so since each $g_{i}\in{D}$, also $f_{\langle
g_{i}\rangle}\in{D}$ by coherence of ${D}$. Thus $B\cap{D}\neq\varnothing$.

So, for any ${D}\in\mathbb{D}$, $B\cap{D}\neq\varnothing$. Thus,
$B\in\mathcal{K}_{\mathbb{D}}$. ∎

Note that, as we show in Campbell-Moore (2021, §4.4.1), some coherent
$\mathcal{K}$ fail axiom $\mathrm{K}_{\mathrm{AddInf}}$ So these are still
special kinds of coherent sets of desirable gamble sets.

In fact, $\mathcal{K}_{\mathbb{D}}$ are exactly the coherent $\mathcal{K}$
that satisfy axiom item $\mathrm{K}_{\mathrm{AddInf}}$. See De Bock (2023,
Theorem 9, Example 9) for this result and generalisations. Instead of giving a
direct proof here, we move to consider the more general question: What about
when we allow for mere finite addition? Can we generate _all_ coherent
$\mathcal{K}$ in an analogous way? Answer: yes!

###  4.2 Representation of all coherent $\mathcal{K}$ using ${D}$s

A gamble set $B$ was evaluated by $\mathbb{D}$ by seeing if every ${D}$ in
$\mathbb{D}$ evaluated $B$ as desirable. Now, we consider different possible
evaluations, as given by a collection of $\mathbb{D}$, and simply ask that one
of them evaluates $B$ as desirable.

That is, we put:

$\displaystyle B\in\mathcal{K}_{\mathfrak{D}}$ iff there is some
$\mathbb{D}\in\mathfrak{D}$ with $B\in\mathcal{K}_{\mathbb{D}}$ (11) iff there
is some $\mathbb{D}\in\mathfrak{D}$ st for all ${D}\in\mathbb{D}$,
$B\in\mathcal{K}_{D}$ (12) iff there is some $\mathbb{D}\in\mathfrak{D}$ st
for all ${D}\in\mathbb{D}$ there is some $g\in B\cap{D}$ (13)

Or, more concisely,

$\mathcal{K}_{\mathfrak{D}}:=\bigcup_{\mathbb{D}\in\mathfrak{D}}\bigcap_{{D}\in\mathbb{D}}\mathcal{K}_{D}$
(14)

Whereas we went from ${D}$ to $\mathbb{D}$ using universal quantification, we
go from $\mathbb{D}$ to $\mathfrak{D}$ using existential quantification.
$\mathfrak{D}$ contains the $\mathbb{D}$ that one is happy to evaluate gamble
sets with respect to. This is different from a $\mathbb{D}$ which contains the
${D}$ that one thinks are still open. One is not sure about which
${D}\in\mathbb{D}$ is “right” so will only make judgements when agreed on by
all ${D}\in\mathbb{D}$. But each $\mathbb{D}\in\mathfrak{D}$ is in some sense
good and can be trusted to make decisions.

In general, this will not generate something coherent. In particular, there is
no guarantee that axiom $\mathrm{K}_{\mathrm{Add}}$ will hold, as we can
select $A_{i}$s evaluated as desirable by different $\mathbb{D}_{i}$, and
there’s no guarantee that the relevant $\\{f_{\langle
g_{i}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes A_{i}\\}$ is evaluated as desirable by any
$\mathbb{D}$ in $\mathfrak{D}$. To ensure that axiom
$\mathrm{K}_{\mathrm{Add}}$ holds, we will add a requirement on
$\mathfrak{D}$: we will require it to be downwards closed:555I had originally
required that it is actually closed under finite intersection. Thanks to
Jasper De Bock for noticing that my proof didn’t then work.  when
$\mathbb{D}_{1},\mathbb{D}_{2}\in\mathfrak{D}$, there is some
$\mathbb{D}\in\mathfrak{D}$ with
$\mathbb{D}\subseteq\mathbb{D}_{1}\cap\mathbb{D}_{2}$. This will ensure that
by taking some $A_{1}$ evaluated as desirable by $\mathbb{D}_{1}$ and $A_{2}$
by $\mathbb{D}_{2}$, then the
$\mathbb{D}\subseteq\mathbb{D}_{1}\cap\mathbb{D}_{2}$ will evaluate the
required $\\{f_{\langle
g_{1},g_{2}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{1},g_{2}\rangle\in A_{1}\times A_{2}\\}$ as desirable; and it can be
extended to finitely many members to get axiom $\mathrm{K}_{\mathrm{Add}}$.

inline]Change this to filters not downward closed sets This then allows us to
represent all coherent $\mathcal{K}$, where we impose axiom
$\mathrm{K}_{\mathrm{Add}}$ but not necessarily axiom
$\mathrm{K}_{\mathrm{AddInf}}$.

######  Theorem 4.2.

$\mathcal{K}$ is coherent iff there is some $\mathfrak{D}$ a non-empty set of
$\mathbb{D}$s, which are themselves non-empty sets of coherent ${D}$s, and
where $\mathfrak{D}$ is downwards closed, with
$\mathcal{K}=\mathcal{K}_{\mathfrak{D}}$, i.e.,:

$\displaystyle B\in\mathcal{K}$ iff there is some $\mathbb{D}\in\mathfrak{D}$
st for all ${D}\in\mathbb{D}$ there is some $g_{D}\in B\cap{D}$ (15)

###### Proof.

Suppose $\mathcal{K}$ is coherent.

Let
$\mathfrak{D}_{\mathcal{K}}:=\\{\mathbb{D}_{\\{A_{1},\ldots,A_{n}\\}}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
A_{1},\ldots,A_{n}\text{ are finitely many members of $\mathcal{K}$}\\}$ where

$\mathbb{D}_{\\{A_{1},\ldots,A_{n}\\}}:=\left\\{D\text{
coherent}\nonscript\>\middle|\nonscript\>\mathopen{}\allowbreak\begin{split}&\text{There
is $\langle g_{1},\ldots,g_{n}\rangle\in A_{1}\times\ldots\times A_{n}$ with
}\\\ &0\notin\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})\text{ and }\\\
&D\supseteq\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})\end{split}\right\\}$ (16)

######  Sublemma 4.2.1.

$\mathfrak{D}_{\mathcal{K}}$ is a non-empty set of $\mathbb{D}$s, which are
themselves non-empty sets of coherent ${D}$s, and $\mathfrak{D}_{\mathcal{K}}$
is downwards closed

###### Proof.

$\mathfrak{D}_{\mathcal{K}}$ is non-empty because $\mathcal{K}$ is non-empty
and for any $A\in\mathcal{K}$,
$\mathbb{D}_{\\{A\\}}\in\mathfrak{D}_{\mathcal{K}}$.

Each $\mathbb{D}_{\\{A_{1},\ldots,A_{n}\\}}$ is a non-empty set of coherent
${D}$s: It is a set of coherent ${D}$s by definition; we just need to show
that it is non-empty. If there is some $\langle g_{i}\rangle\in\bigtimes
A_{i}$ with
$0\notin\mathrm{posi}(\\{g_{1},\dots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$, then
$\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$ is coherent
by proposition 1.3. So then $\mathbb{D}_{\\{A_{1},\ldots,A_{n}\\}}$ is non-
empty. Otherwise, every $\langle g_{i}\rangle\in\bigtimes A_{i}$ has
$0\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$; but in
this case, $\varnothing\in\mathcal{K}$ by theorem 2.1, contradicting the
coherence of $\mathcal{K}$.

It remains to show that $\mathfrak{D}_{\mathcal{K}}$ is downwards closed. We
will show that

$\mathbb{D}_{A_{1}^{1},\ldots,A_{n_{1}}^{1}}\cap\mathbb{D}_{A_{1}^{2},\ldots,A_{n_{2}}^{2}}\supseteq\mathbb{D}_{A_{1}^{1},\ldots,A_{n_{1}}^{1},A_{1}^{2},\ldots,A_{n_{2}}^{2}}$
(17)

Suppose
${D}\in\mathbb{D}_{A_{1}^{1},\ldots,A_{n_{1}}^{1},A_{1}^{2},\ldots,A_{n_{2}}^{2}}$.
So there is $\langle
g_{1}^{1},\ldots,g_{n_{1}}^{1},g_{1}^{2},\ldots,g_{n_{2}}^{2}\rangle\in
A_{1}^{1}\times\ldots\times A_{n_{1}}^{1}\times A_{1}^{2}\times\ldots\times
A_{n_{2}}^{2}$ with
$0\notin\mathrm{posi}(\\{g_{1}^{1},\ldots,g_{n_{1}}^{1},g_{1}^{2},\ldots,g_{n_{2}}^{2}\\}\cup\mathcal{G}_{\gneq
0})$ and
$D\supseteq\mathrm{posi}(\\{g_{1}^{1},\ldots,g_{n_{1}}^{1},g_{1}^{2},\ldots,g_{n_{2}}^{2}\\}\cup\mathcal{G}_{\gneq
0})$.

Since
$\mathrm{posi}(\\{g_{1}^{1},\ldots,g_{n_{1}}^{1},g_{1}^{2},\ldots,g_{n_{2}}^{2}\\}\cup\mathcal{G}_{\gneq
0})\supseteq\mathrm{posi}(\\{g_{1}^{1},\ldots,g_{n_{1}}^{1}\\}\cup\mathcal{G}_{\gneq
0})$, also
${D}\supseteq\mathrm{posi}(\\{g_{1}^{1},\ldots,g_{n_{1}}^{1}\\}\cup\mathcal{G}_{\gneq
0})$ and
$0\notin\mathrm{posi}(\\{g_{1}^{1},\ldots,g_{n_{1}}^{1}\\}\cup\mathcal{G}_{\gneq
0})$. So ${D}\in\mathbb{D}_{A_{1}^{1},\ldots,A_{n_{1}}^{1}}$. Similarly
${D}\in\mathbb{D}_{A_{1}^{2},\ldots,A_{n_{2}}^{2}}$. Thus
${D}\in\mathbb{D}_{A_{1}^{1},\ldots,A_{n_{1}}^{1}}\cap\mathbb{D}_{A_{1}^{2},\ldots,A_{n_{2}}^{2}}$,
as required. ∎

######  Sublemma 4.2.2.

$\displaystyle B\in\mathcal{K}$ iff there is some $\mathbb{D}\in\mathfrak{D}$
st for all ${D}\in\mathbb{D}$ there is some $g\in B\cap{D}$ (18)

###### Proof.

$\displaystyle B\in\mathcal{K}_{\mathfrak{D}_{\mathcal{K}}}$ (19) iff there is
some $\mathbb{D}\in\mathfrak{D}_{\mathcal{K}}$ st for all ${D}\in\mathbb{D}$,
$B\cap{D}\neq\varnothing$ (20) iff there is some
$A_{1},\ldots,A_{n}\in\mathcal{K}$ st
$\mathbb{D}\supseteq\mathbb{D}_{\\{A_{1},\ldots,A_{n}\\}}$, and for all
${D}\in\mathbb{D}$, $B\cap{D}\neq\varnothing$ (21) iff there is some
$A_{1},\ldots,A_{n}\in\mathcal{K}$ st for all
${D}\in\mathbb{D}_{\\{A_{1},\ldots,A_{n}\\}}$, $B\cap{D}\neq\varnothing$ (22)
$\displaystyle\text{ iff }\begin{aligned} &\text{there is some
$A_{1},\ldots,A_{n}\in\mathcal{K}$ st for all $\langle
g_{i}\rangle\in\bigtimes A_{i}$, }\\\ &\quad\text{
$B\cap\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})\neq\varnothing$ or
$0\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq
0})$}\end{aligned}$ (23)

By theorem 2.1, we know that any $B$ satisfying eq. 23 is in $\mathcal{K}$. It
is also easy to see that any $B\in\mathcal{K}$ satisfies eq. 23 (by $n=1$,
$A_{1}=B$). So we have that
$\mathcal{K}_{\mathfrak{D}_{\mathcal{K}}}=\mathcal{K}$. ∎

######  Sublemma 4.2.3.

When $\mathfrak{D}$ is a non-empty set of $\mathbb{D}$s, which are themselves
non-empty sets of coherent ${D}$s, and $\mathfrak{D}$ is downwards closed,
then $\mathcal{K}_{\mathfrak{D}}$ is coherent, where
$B\in\mathcal{K}_{\mathfrak{D}}$ iff there is some $\mathbb{D}\in\mathfrak{D}$
st for all ${D}\in\mathbb{D}$, $B\cap{D}\neq\varnothing$.

###### Proof.

We check each of the axioms. It is only axiom $\mathrm{K}_{\mathrm{Add}}$
which requires the structure of $\mathfrak{D}$, the rest simply hold because
they hold of each $\mathcal{K}_{\mathbb{D}}$.

Axiom $\mathrm{K}_{\varnothing}$: Every $\mathbb{D}\in\mathfrak{D}$ is non-
empty by assumption, so has some ${D}\in\mathbb{D}$. And
$\varnothing\cap{D}=\varnothing$, so for every $\mathbb{D}\in\mathfrak{D}$,
there’s some ${D}\in\mathbb{D}$ with $\varnothing\cap{D}=\varnothing$; so
$\varnothing\notin\mathcal{K}_{\mathfrak{D}}$.

Axiom $\mathrm{K}_{0}$: Suppose $A\in\mathcal{K}_{\mathfrak{D}}$. Then there
is some $\mathbb{D}$ with $A\in\mathcal{K}_{\mathbb{D}}$. And thus
$A\in\mathcal{K}_{D}$ for each ${D}\in\mathbb{D}$. That is, there is some
$g\in A\cap{D}$. Since $0\notin{D}$, also
$(A\setminus\\{0\\})\cap{D}\neq\varnothing$. So
$(A\setminus\\{0\\})\in\mathcal{K}_{\mathbb{D}}$. Thus
$(A\setminus\\{0\\})\in\mathcal{K}_{\mathfrak{D}}$.

Axiom $\mathrm{K}_{\gneq 0}$: Suppose $g\in\mathcal{G}_{\gneq 0}$. Then
$g\in{D}$ for each coherent ${D}$. Consider any $\mathbb{D}\in\mathfrak{D}$.
Then $g\in{D}$ for each ${D}\in\mathbb{D}$. Thus
$g\in\mathcal{K}_{\mathbb{D}}$; thus $g\in\mathcal{K}_{\mathfrak{D}}$.

Axiom $\mathrm{K}_{\supseteq}$: Suppose $A\in\mathcal{K}_{\mathfrak{D}}$ and
$B\supseteq A$. Then there is some $\mathbb{D}$ with
$A\in\mathcal{K}_{\mathbb{D}}$. And thus $A\cap{D}\neq\varnothing$ for each
${D}\in\mathbb{D}$. So also $B\cap{D}\neq\varnothing$ for each
${D}\in\mathbb{D}$. So $B\in\mathcal{K}_{\mathbb{D}}$. So
$B\in\mathcal{K}_{\mathfrak{D}}$.

Axiom $\mathrm{K}_{\mathrm{Dom}}$: Suppose $A\in\mathcal{K}_{\mathfrak{D}}$
and we have $f_{g}\geqslant g$ for each $g\in A$. There is some $\mathbb{D}$
with $A\in\mathcal{K}_{\mathbb{D}}$. And thus $A\in\mathcal{K}_{D}$ for each
${D}\in\mathbb{D}$. That is, there is some $g\in A\cap{D}$. But therefore
$f_{g}\in{D}$ by properties of coherent ${D}$, thus $B\in\mathcal{K}_{D}$ for
each ${D}\in\mathbb{D}$. Thus $B\in\mathcal{K}_{\mathbb{D}}$, and so
$B\in\mathcal{K}_{\mathfrak{D}}$.

For axiom $\mathrm{K}_{\mathrm{Add}}$: Let
$A_{1},\ldots,A_{n}\in\mathcal{K}_{\mathfrak{D}}$ and $B=\\{f_{\langle
g_{i}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes A_{i}\\}$ where $f_{\langle
g_{i}\rangle}\in\mathrm{posi}({\\{g_{i}\\}})$. Since
$A_{i}\in\mathcal{K}_{\mathfrak{D}}$, we have some
$\mathbb{D}_{i}\in\mathfrak{D}$ with $A_{i}\in\mathcal{K}_{\mathbb{D}_{i}}$.
By downwards closure, there is some $\mathbb{D}^{*}\in\mathfrak{D}$ with
$\mathbb{D}^{*}\subseteq\mathbb{D}_{1}\cap\ldots\cap\mathbb{D}_{n}$. We will
show that $B\in\mathcal{K}_{\mathbb{D}^{*}}$. For any ${D}\in\mathbb{D}^{*}$,
also ${D}\in\mathbb{D}_{i}$ for each $i$, and thus there is some
$g_{i}\in{D}\cap A_{i}$. Since ${D}$ is coherent containing
$g_{1},\ldots,g_{n}$ and $f_{\langle
g_{i}\rangle}\in\mathrm{posi}\\{g_{1},\ldots,g_{n}\\}$, also $f_{\langle
g_{i}\rangle}\in{D}$. So ${D}\cap B\neq\varnothing$. Since this holds for all
${D}\in\mathbb{D}^{*}$, $B\in\mathcal{K}_{\mathbb{D}^{*}}$. So
$B\in\mathcal{K}_{\mathfrak{D}}$, as required. ∎

This proves our result. ∎

It may be more intuitive to think about $\mathfrak{D}$ not simply as a
downwards closed set of $\mathbb{D}$s, but as a filter of $\mathbb{D}$s. That
is, we might also close it under supersets. This won’t make a difference to
the representation, because adding supersets doesn’t add any new members to
$\mathcal{K}_{\mathfrak{D}}$.

We can then think of $\mathfrak{D}$ as containing all one’s judgements about
what the “correct” ${D}$ is like. This would allow us to give a different
formulation.

$\displaystyle B\in\mathcal{K}_{\mathfrak{D}}$ iff there is some
$\mathbb{D}\in\mathfrak{D}$ st for all ${D}\in\mathbb{D}$ there is some $g\in
B\cap{D}$ (24) iff there is some $\mathbb{D}\in\mathfrak{D}$ st
$\mathbb{D}\subseteq\\{{D}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\text{there
is some $g\in B\cap{D}$}\\}$ (25) $\displaystyle\text{ iff
}\\{{D}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\text{there is some
$g\in B\cap{D}$}\\}\in\mathfrak{D}$ (26)

######  Theorem 4.3.

$\mathcal{K}$ is coherent iff there is some $\mathfrak{D}$ a proper filter of
coherent ${D}$s, that is:
$\mathfrak{D}\subseteq\wp(\\{D\subseteq\mathcal{G}\nonscript\>|\nonscript\>\mathopen{}\allowbreak
D\text{ is coherent}\\})$ where:

  * •

$\mathfrak{D}\neq\varnothing$, $\varnothing\notin\mathfrak{D}$,

  * •

If $\mathbb{D}_{1},\mathbb{D}_{2}\in\mathfrak{D}$ then
$\mathbb{D}_{1}\cap\mathbb{D}_{2}\in\mathfrak{D}$.

  * •

If $\mathbb{D}_{1}\in\mathfrak{D}$ and $\mathbb{D}_{2}\supseteq\mathbb{D}_{1}$
then $\mathbb{D}_{2}\in\mathfrak{D}$.

with $\mathcal{K}=\mathcal{K}_{\mathfrak{D}}$, i.e.,:

$\displaystyle B\in\mathcal{K}$ iff there is some $\mathbb{D}\in\mathfrak{D}$
st for all ${D}\in\mathbb{D}$ there is some $g_{D}\in B\cap{D}$ (27)

or equivalently:

$\displaystyle B\in\mathcal{K}$ $\displaystyle\text{ iff
}\\{{D}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\text{there is some
$g_{D}\in B\cap{D}$}\\}\in\mathfrak{D}$ (28)

###### Proof.

When $\mathfrak{D}$ is a non-empty set of $\mathbb{D}$s, which are themselves
non-empty sets of coherent ${D}$s, and $\mathfrak{D}$ is downwards closed,
then by closing it under supersets gets a proper filter of coherent ${D}$s.
Also observe that adding supersets does not add any new gamble sets satisfying
eq. 27. So the representation of coherent $\mathcal{K}$ by such a
$\mathfrak{D}$ from theorem 4.2 can be closed under supersets to get
$\mathcal{K}$ represented by a proper filter of coherent ${D}$s.

The right-to-left holds because a filter of coherent ${D}$s is itself a is a
non-empty set of $\mathbb{D}$s, which are themselves non-empty sets of
coherent ${D}$s, and $\mathfrak{D}$ is downwards closed.

For the equivalence between eqs. 27 and 28, note that if there is some such
$\mathbb{D}\in\mathfrak{D}$ then
$\\{{D}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\text{there is some
$g_{D}\in B\cap{D}$}\\}$ is a superset of it, so is also in $\mathfrak{D}$. ∎

Note that one thing that this result does not show is an equivalence between
$\mathfrak{D}$ and $\mathcal{K}$s. Multiple $\mathfrak{D}$ will correspond to
the same $\mathcal{K}$.666See also Campbell-Moore (2021, section 5) for the
related observation in that more restrictive setting. See de Cooman et al.
(2023).

######  Proposition 4.4.

The minimal filter $\mathfrak{D}$ corresponding to coherent $\mathcal{K}$ is
$\mathfrak{D}_{\mathcal{K}}$ where $\mathbb{D}\in\mathfrak{D}_{\mathcal{K}}$
iff there are finitely many members $A_{1},\ldots,A_{n}\in\mathcal{K}$ with
$\mathbb{D}\supseteq\mathbb{D}_{\\{A_{1},\ldots,A_{n}\\}}$ (in the sense of
eq. 16).

###### Proof.

Should follow very quickly. ∎

######  Proposition 4.5 (Gert and Arthur).

There is an isomorphism between coherent $\mathcal{K}$s and $\mathfrak{D}$
that satisfy:

  * •

$\mathfrak{D}$ is a non-empty set of non-empty sets of coherent ${D}$ closed
under finite intersection and superset.

  * •

$\mathbb{D}\in\mathfrak{D}$ iff $\uparrow\mathbb{D}\in\mathfrak{D}$, where
$\uparrow\mathbb{D}=\\{{D}^{\prime}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\exists{D}\in\mathbb{D},\;{D}^{\prime}\subseteq{D}\\}.$

A version of Arthur and Gert’s result:

######  Proposition 4.6.

There is an isomorphism between coherent $\mathcal{K}$s and $\mathfrak{D}$
that satisfy:

  * •

$\mathfrak{D}$ is a non-empty set of non-empty sets of coherent ${D}$ closed
under finite intersection and superset.

  * •

If $\mathbb{D}\in\mathfrak{D}$ then there are sets of gambles
$A_{1},\ldots,A_{n}$ such that
$\mathbb{D}\supseteq\mathbb{D}_{\\{A_{1},\ldots,A_{n}\\}}$; i.e., where for
every coherent ${D}$ such that there are $\langle g_{1},\ldots,g_{n}\rangle\in
A_{1}\times\ldots\times A_{n}$ with
$0\notin\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$ and
${D}\supseteq\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}\cup\mathcal{G}_{\gneq 0})$
is itself $\in\mathbb{D}$.

  * •

If $\mathbb{D}\in\mathfrak{D}$ and $\mathbb{D}^{\prime}$ is st
$\uparrow\mathbb{D}=\uparrow\mathbb{D}^{\prime}$ then
$\mathbb{D}^{\prime}\in\mathfrak{D}$.

There is a further alternative we might do which removes requirements on
$\mathfrak{D}$: we can drop the requirement that $\mathfrak{D}$ be downwards
closed, and instead bake that into the characterisation of
$\mathcal{K}_{\mathfrak{D}}$. We could say:

$\displaystyle B\in\mathcal{K}_{\mathfrak{D}}$ $\displaystyle\text{ iff
}\begin{split}&\text{there are some
$\mathbb{D}_{1},\ldots,\mathbb{D}_{n}\in\mathfrak{D}$}\\\ &\text{ st for all
${D}\in\mathbb{D}_{1}\cap\ldots\cap\mathbb{D}_{n}$, there is some $g_{D}\in
B\cap{D}$}\end{split}$ (29)

Now we get axiom $\mathrm{K}_{\mathrm{Add}}$ for free, however axiom
$\mathrm{K}_{0}$ might now fail. We still need a “coherence” requirement on
this. The relevant requirement is that $\mathfrak{D}$ has the finite
intersection property: that we never have any
$\mathbb{D}_{1}\cap\ldots\cap\mathbb{D}_{n}=\varnothing$. It is then
equivalent.

######  Corollary 4.7.

$\mathcal{K}$ is coherent iff there is some $\mathfrak{D}$ a non-empty set of
$\mathbb{D}$s, which are themselves non-empty sets of coherent ${D}$s, where
$\mathfrak{D}$ has the finite intersection property (i.e., no
$\mathbb{D}_{1}\cap\ldots\cap\mathbb{D}_{n}=\varnothing$), and

$\displaystyle B\in\mathcal{K}$ $\displaystyle\text{ iff
}\begin{split}&\text{there are some
$\mathbb{D}_{1},\ldots,\mathbb{D}_{n}\in\mathfrak{D}$}\\\ &\text{ st for all
${D}\in\mathbb{D}_{1}\cap\ldots\cap\mathbb{D}_{n}$, there is some $g_{D}\in
B\cap{D}$}\end{split}$ (30)

###### Proof.

If $\mathfrak{D}$ is such a set which has the finite intersection property,
consider $\mathfrak{D}^{\prime}$ which closes it under finite intersections,
i.e.,
$\mathfrak{D}^{\prime}:=\\{\mathbb{D}_{1}\cap\ldots\cap\mathbb{D}_{n}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\text{$\mathbb{D}_{1},\ldots,\mathbb{D}_{n}$
are finitely many members of $\mathfrak{D}$}\\}$. By assumption that
$\mathfrak{D}$ has the finite intersection property, $\mathfrak{D}^{\prime}$
doesn’t contain the emptyset. It immediately satisfies all the other
assumptions of theorem 4.2.

Also observe that:

There is some $\mathbb{D}_{1},\ldots,\mathbb{D}_{n}\in\mathfrak{D}$ st for all
${D}\in\mathbb{D}_{1}\cap\ldots\cap\mathbb{D}_{n}$, $\varnothing\neq B\cap{D}$
(31) iff There is some $\mathbb{D}^{\prime}$ st st for all ${D}\in\mathbb{D}$,
$\varnothing\neq B\cap{D}$ (32)

So we see that for any such $\mathfrak{D}$ with the finite intersection
property, we can simply close it under finite intersections and use the
representation of theorem 4.2 ∎

## References

  * Campbell-Moore (2021) Catrin Campbell-Moore.  Probability filters as a model of belief; comparisons to the framework of desirable gambles.  In Andrés Cano, Jasper De Bock, Enrique Miranda, and Serafín Moral, editors, _Proceedings of the Twelveth International Symposium on Imprecise Probability: Theories and Applications_ , volume 147 of _Proceedings of Machine Learning Research_ , pages 42–50–42–50. PMLR, 2021.
  * De Bock (2023) Jasper De Bock.  A theory of desirable things.  In _International Symposium on Imprecise Probability: Theories and Applications_ , pages 141–152. PMLR, 2023.
  * De Bock and de Cooman (2018) Jasper De Bock and Gert de Cooman.  A desirability-based axiomatisation for coherent choice functions.  In S. Destercke, T. Denoeux, M.A. Gil, P. Grzergorzewski, and O. Hryniewicz, editors, _Uncertainty Modelling in Data Science: Proceedings of SMPS_ , pages 46–53. Springer, 2018.
  * De Bock and de Cooman (2019) Jasper De Bock and Gert de Cooman.  Interpreting, axiomatising and representing coherent choice functions in terms of desirability.  In _International Symposium on Imprecise Probabilities: Theories and Applications_ , pages 125–134. PMLR, 2019.
  * de Cooman et al. (2023) Gert de Cooman, Arthur Van Camp, and Jasper De Bock.  Desirable sets of things and their logic.  In Enrique Miranda, Ignacio Montes, Erik Quaeghebeur, and Barbara Vantaggi, editors, _Proceedings of the Thirteenth International Symposium on Imprecise Probability: Theories and Applications_ , volume 215 of _Proceedings of Machine Learning Research_ , pages 153–164. PMLR, 11–14 Jul 2023.  URL https://proceedings.mlr.press/v215/cooman23a.html.
  * Decadt (2021) Arne Decadt.  Decide quicker with total choice functions.  In _ISIPTA 2021_ , 2021.

##  Appendix A Finite addition proof without supersets

######  Proposition A.1.

Assume we are restricted to finite sets. Axiom $\mathrm{K}_{\mathrm{Add}}$
follows from axiom $\mathrm{K}_{\mathrm{AddPair}}$.

inline] Jasper: the pf doesn’t work: what if multiple of the f’s are all 0?

######  Proof Attempt.

The argument gets slightly fiddly because members may coincide and repetitions
get “deleted” by sets. To avoid this issue, we need to include constants
throughout to ensure distinctness.

We work by induction on $n$.

Base case: $n=1$ follows directly from axiom $\mathrm{K}_{\mathrm{AddPair}}$
by putting $A_{1}=A_{2}$.

Inductive step: Assume it holds for $n$ and consider $n+1$.

So, suppose $A_{1},\ldots,A_{n},A_{n+1}\in\mathcal{K}$ and for each sequence
$\langle g_{i}\rangle\in\bigtimes_{i}A_{i}$ and $a\in A_{n+1}$, we have
$f_{\langle
g_{1},\ldots,g_{n},a\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n},a\\})$ with

$B=\\{f_{\langle
g_{1},\ldots,g_{n},a\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes_{i\leqslant n}A_{i},\;a\in A_{n+1}\\}$ (33)

We need to show that $B\in\mathcal{K}$.

We will abuse notation and write $f_{\langle g_{i},a\rangle}$ instead of
$f_{\langle g_{1},\ldots,g_{n},a\rangle}$.

Consider adding some scalar constants: let

$f_{\langle g_{i},a\rangle}^{*}:=\delta_{\langle g_{i},a\rangle}\times
f_{\langle g_{i},a\rangle}$ (34)

where $\delta_{\langle g_{i},a\rangle}$ are chosen to ensure that these are
all distinct, and distinct from any $a^{\prime}$. We will show that

$B^{*}=\\{f^{*}_{\langle
g_{i},a\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes_{i}A_{i}\\}$ (35)

$A_{n+1}$ is finite, so enumerate it, $A_{n+1}=\\{a_{1},\ldots,a_{K}\\}$.
Define:

$B^{*}_{k}:=\\{f^{*}_{\langle
g_{i},a_{j}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes_{i\leqslant
n}A_{i},\;a_{j}\in\\{a_{1},\ldots,a_{k}\\}\\}\cup\\{a_{k+1},\ldots,a_{K}\\}$
(36)

Where we have already replaced $k$-many of $A_{n+1}$ with the relevant members
of $B^{*}$. We show by a induction on $k$ that
$B^{*}_{k}\in\mathcal{K}$.777Arthur van Camp’s insight that this can be done
in such a step-by-step way to just use axiom $\mathrm{K}_{\mathrm{AddPair}}$,
i.e., it can be shown by a sub-induction.

Base case: $B_{0}=A_{n+1}\in\mathcal{K}$.

Inductive step: we can assume that $B_{k}\in\mathcal{K}$ and need to show
$B_{k+1}\in\mathcal{K}$.

We will construct a set $C$ which we know to be in $\mathcal{K}$, which we can
then combine with $B_{k}$ using axiom $\mathrm{K}_{\mathrm{AddPair}}$ (and
axiom $\mathrm{K}_{\supseteq}$) to get $B^{*}_{k+1}\in\mathcal{K}$.

######  Sublemma A.1.1.

We can find distinct $h^{*}_{\langle g_{i},a_{k+1}\rangle}$ where:

$\displaystyle h^{*}_{\langle
g_{i},a_{k+1}\rangle}\in\mathrm{posi}(\\{g_{1},\ldots,g_{n}\\}),\text{ and }$
(37) $\displaystyle f^{*}_{\langle
g_{i},a_{k+1}\rangle}\in\mathrm{posi}(\\{h^{*}_{\langle
g_{i},a_{k+1}\rangle},a_{k+1}\\}).$ (38)

###### Proof.

$f_{\langle g_{i},a_{k+1}\rangle}$ has the form $\sum_{i\leqslant
n}\lambda_{i}g_{i}+\mu a_{k+1}$. If $\lambda_{i}>0$ for some $i$, then we put
$h_{\langle g_{i},a_{k+1}\rangle}:=\sum_{i\leqslant n}\lambda_{i}g_{i}$. If
not, then we simply let it be, for example, $g_{1}$.

To ensure they are distinct, we simply multiply the $h$ by appropriate
constants to find $h^{*}$. ∎

We let

$C:=\\{h^{*}_{\langle
g_{i},a_{k+1}\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak\langle
g_{i}\rangle\in\bigtimes_{i}A_{i}\\}$ (39)

$C\in\mathcal{K}$ by the induction hypothesis (on $n$), using eq. 37.

Now, for each $\langle c,b\rangle$, let

$\displaystyle e_{\langle c,b\rangle}:=\begin{cases}c&b\in B^{*}_{k}\cap
B^{*}_{k+1}\\\ f^{*}_{\langle g_{i},a_{k+1}\rangle}&c=a_{k+1}\text{ and
}b=h^{*}_{\langle g_{i},a_{k+1}\rangle}\end{cases}$ (40)

Since each $e_{\langle c,b\rangle}\in\mathrm{posi}(\\{c,b\\})$, by axiom
$\mathrm{K}_{\mathrm{AddPair}}$,

$\\{e_{\langle c,b\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak c\in
B_{k},\,b\in C\\}\in\mathcal{K}$ (41)

######  Sublemma A.1.2.

$B^{*}_{k+1}=\\{e_{\langle
c,b\rangle}\nonscript\>|\nonscript\>\mathopen{}\allowbreak c\in B_{k},\,b\in
C\\}$ (42)

###### Proof.

Can easily be checked. ∎

So, by axiom $\mathrm{K}_{\mathrm{AddPair}}$ and axiom
$\mathrm{K}_{\supseteq}$, $B^{*}_{k+1}\in\mathcal{K}$, completing the
inductive step.

This has shown by induction that $B^{*}_{K}\in\mathcal{K}$. $B^{*}_{K}=B^{*}$,
so $B^{*}\in\mathcal{K}$.

We need to show that $B\in\mathcal{K}$. This works because it is obtained from
$B^{*}$ simply by multiplying everything by a scalar, so follows from the
$n=1$ version of axiom $\mathrm{K}_{\mathrm{Add}}$, which itself directly
follows from axiom $\mathrm{K}_{\mathrm{AddPair}}$.

This shows that the statement holds for $n+1$, completing the inductive step.
Thus we have shown that it holds for all $n$; as required. ∎