# A class of genuinely high-dimensionally entangled states with a positive
partial transpose

Károly F. Pál  Institute of Nuclear Research of the Hungarian Academy of
Sciences H-4001 Debrecen, P.O. Box 51, Hungary  Tamás Vértesi  Institute of
Nuclear Research of the Hungarian Academy of Sciences H-4001 Debrecen, P.O.
Box 51, Hungary

###### Abstract

Entangled states with a positive partial transpose (so-called PPT states) are
central to many interesting problems in quantum theory. On one hand, they are
considered to be weakly entangled, since no pure state entanglement can be
distilled from them. On the other hand, it has been shown recently that some
of these PPT states exhibit genuinely high-dimensional entanglement, i.e. they
have a high Schmidt number. Here we investigate $d\times d$ dimensional PPT
states for $d\geq 4$ discussed recently by Sindici and Piani, and by
generalizing their methods to the calculation of Schmidt numbers we show that
a linear $d/2$ scaling of its Schmidt number in the local dimension $d$ can be
attained.

##  I Introduction

Entanglement is at the heart of quantum theory and is also a key resource in
quantum information applications horo_review ; GT_review . Bipartite systems
which are entangled across many degrees of freedom are especially interesting
in this respect. Firstly, they are usually more robust to noise than systems
with less degrees of freedom grob ; dada . Secondly, they allow us to devise
protocols which are genuinely high-dimensional. In particular, low dimensional
systems are not enough for these protocols to work lanyon .

Experimentally, high-dimensional entanglement also became feasible in the
recent years. Higher and higher dimensional systems can be prepared and
controlled in optics experiments stein ; dada . Therefore it is an important
question to decide whether an experiment managed to create genuinely high-
dimensional entanglement or the experimental data can alternatively be
explained by assuming low dimensional entanglement. A measure which detects
states with genuinely entangled degrees of freedom is based on the Schmidt
number schmidt_number . Schmidt number $r$ of a bipartite system certifies
that the state is entangled in at least $r$ degrees of freedom.

As an illustration, let us consider the $d\times d$ isotropic state defined by

$\displaystyle\hat{\rho}_{d}^{\rm
iso}(F)=F|\Psi_{d}^{+}\rangle\langle\Psi_{d}^{+}|+(1-F)\frac{\leavevmode\hbox{\small
1\normalsize\kern-3.30002pt1}-|\Psi_{d}^{+}\rangle\langle\Psi_{d}^{+}|}{d^{2}-1},$
(1)

where

$\displaystyle|\Psi_{d}^{+}\rangle=\frac{1}{\sqrt{d}}\sum_{k=1}^{d}|k\rangle|k\rangle$
(2)

is the $d\times d$ maximally entangled state and $F$ is the entanglement
fraction hh  of the state $\hat{\rho}_{d}^{\rm iso}$. Based on the results of
Ref. schmidt_number  the state (1) can be shown to have Schmidt number at
least $r$ for the entanglement fraction parameter

$\displaystyle F\geq\frac{r-1}{d},$ (3)

where $r$ can take $r=(2,\ldots,d)$ in the above formula. Loosely speaking,
$r$ is the minimum Schmidt rank of the pure states needed to construct it
(later we will also give a formal definition of the Schmidt number). Hence
tuning the parameter $F$ in a $d\times d$ isotropic state (1), we can change
the Schmidt number $r$ of the state between two (i.e. the case of standard
entanglement) and $d$ (the maximum possible Schmidt number of a $d\times d$
state).

In this paper, our focus is on entangled states with a positive partial
transpose (PPT). These are states which cannot be transformed into pure
singlet states using local operations and classical communication. This
procedure is called entanglement distillation ED . Such entangled states which
cannot be distilled are called bound entangled BE , and they are not useful in
protocols which are based on distillation. Nevertheless, PPT bound entangled
states turn out to be useful in a couple of other quantum information tasks
such as quantum key distribution hllo ; ozols , superactivation of capacity of
quantum channels smithyard , quantum metrology Toth18 , EPR steering tobias ,
Bell-nonlocal correlations VB ; yuoh ; PV17 , and channel discrimination PW09
.

Several interesting constructions of PPT states have been given in the
literature (see e.g. Refs. BDM ; breuer ; PM ; hhho ; sentis ; hllo ), however
the question whether PPT states can be genuinely high-dimensionally entangled
have been investigated only recently szarek ; chen ; huber . In particular,
Huber et al. huber  found that for a special class of $(d\times
d)$-dimensional PPT entangled states the Schmidt number scales as $d/4$. This
entails that one can generate PPT state with any number of genuinely entangled
dimension provided the dimension $d$ is high enough. In the present work we
strengthen this result by presenting a family of PPT states in dimensions
$d\times d$ for which the Schmidt number scales as $d/2$. The proof relies on
the special properties of the projections on the two-qudit symmetric and
antisymmetric subspaces.

The structure of the paper is as follows. In Sec. II we give the necessary
definitions including the definition of the Schmidt number and we define the
class of PPT states investigated by Sindici and Piani SinPia . We recall the
two methods of Ref. SinPia  to prove that the PPT states in question are
entangled, that is, their Schmidt number is greater than one. In Sec. III we
generalize the above methods of Sindici and Piani to lowerbound the Schmidt
number of PPT states. In Sec. III.1 we show how to construct PPT bound
entangled states with Schmidt number $r\geq d/2$ for any even $d\geq 4$
dimension. In Sec. III.2 a technical result is given concerning the normal
form of an antisymmetric pure state. Then in Sec. III.3 we show that starting
from an arbitrary entangled antisymmetric state $\hat{\rho}_{A}$, a
semidefinite program sdp  provides a lower bound on its Schmidt number. From
this lower bound we in turn find PPT states with certain Schmidt number whose
antisymmetric projections are proportional to $\hat{\rho}_{A}$. In Sec. IV we
conclude our study and raise open questions.

##  II Preliminaries and overview of the two methods of Sindici and Piani

Any bipartite pure state can be written in the form of Schmidt decomposition:

$|\varphi\rangle=\sum_{i=1}^{r}\sqrt{\pi_{i}}|a_{i}\rangle\otimes|b_{i}\rangle,$
(4)

where $\pi_{i}>0$, $\sum_{i}\pi_{i}=1$, $|a_{i}\rangle$ and $|b_{i}\rangle$
are orthonormal vectors in the component spaces and $r$, the Schmidt rank of
$|\varphi\rangle$, is not larger than the dimensionality of either of the
component spaces. A general state is represented by a positive semidefinite
operator of trace one in the composite space, and it can be expressed as a
convex combination of pure states (represented now by projections onto their
state vectors) as:

$\hat{\rho}=\sum_{k}q_{k}|\varphi_{k}\rangle\langle\varphi_{k}|.$ (5)

A mixed state can usually be written in many ways in forms like above. A state
has Schmidt number $r$ if all possible such expressions contain at least one
pure state whose Schmidt rank is at least $r$, and there is at least one
expression in which neither of the Schmidt ranks exceeds $r$. From the
definition it follows that for pure states the Schmidt number is the same as
the Schmidt rank. A general state is separable if its Schmidt number is one,
otherwise it is entangled.

In a recent work Sindici and Piani SinPia  have given simple methods to
construct entangled states with a positive partial transpose (PPT). The
construction relied on properties of the antisymmetric and the symmetric
subspaces of the joint Hilbert space of two systems of the same
dimensionality. The projectors $\hat{P}_{\cal S}$ and $\hat{P}_{\cal A}$
defining the symmetric and the antisymmetric subspace, respectively, can be
written as

$\displaystyle\hat{P}_{\cal S}=$ $\displaystyle\frac{\hat{I}+\hat{V}}{2}$ (6)
$\displaystyle\hat{P}_{\cal A}=$ $\displaystyle\frac{\hat{I}-\hat{V}}{2},$ (7)

where $\hat{I}$ and $\hat{V}$ are the identity and the swap operators,
respectively. The swap operator can be written as

$\hat{V}=\sum_{i=1}^{d}\sum_{j=1}^{d}|j,i\rangle\langle i,j|,$ (8)

where $d$ is the dimensionality of the Hilbert spaces of the component
systems. We use the shorthand notation of
$|i,j\rangle\equiv|i\rangle\otimes|j\rangle$. The effect of $\hat{V}$ on a
product state is
$\hat{V}|\varphi_{1}\rangle\otimes|\varphi_{2}\rangle=|\varphi_{2}\rangle\otimes|\varphi_{1}\rangle$.
The swap operator depends on the choice of local bases, states in the
component spaces are regarded the same if their vector components are the same
in the bases chosen. Therefore, the symmetric and the antisymmetric subspaces
also depend on this choice. For example, any pure antisymmetric state becomes
symmetric in the bases corresponding to its Schmidt decomposition. However,
the symmetricity or antisymmetricity of states are preserved if both parties
perform the same local transformation werner . In Ref. SinPia  the authors
consider only identical systems, where the choice of the bases is not
completely arbitrary, but their methods are more general. As $\hat{P}_{\cal
S}+\hat{P}_{\cal A}=\hat{I}$, the full space is the direct sum of the
symmetric and antisymmetric subspaces. It is easy to check that the symmetric
and the antisymmetric subspaces have dimensions $d_{\cal S}=d(d+1)/2$ and
$d_{\cal A}=d(d-1)/2$, respectively.

The first method given in Ref. SinPia  is based on semidefinite programming
(SDP) sdp . One starts with an antisymmetric state $\hat{\rho}_{\cal A}$. Then
the task is to find the PPT state $\hat{\sigma}$ (a positive semidefinite
matrix of trace one, whose partial transpose is also positive semidefinite)
whose projection to the antisymmetric subspace is proportional to the starting
state (that is $\hat{P}_{\cal A}\hat{\sigma}\hat{P}_{\cal A}={\rm
Tr}(\hat{P}_{\cal A}\hat{\sigma})\hat{\rho}_{\cal A}$) and whose overlap ${\rm
Tr}(\hat{P}_{\cal A}\hat{\sigma})$ with the starting state is maximal. The
procedure is successful if this maximal overlap denoted by
$p^{PPT}(\hat{\rho}_{\cal A})$ is smaller than $1/2$, as it is proven in the
paper SinPia  that in this case the state $\hat{\sigma}$ is entangled. If the
result is $1/2$, which is the upper bound for this quantity, one can start
from another $\hat{\rho}_{\cal A}$. It has also been proven SinPia  that if
$\hat{\sigma}$ is an optimal solution of the problem above, so is
$\hat{P}_{\cal A}\hat{\sigma}\hat{P}_{\cal A}+\hat{P}_{\cal
S}\hat{\sigma}\hat{P}_{\cal S}=p^{PPT}(\hat{\rho}_{\cal A})\hat{\rho}_{\cal
A}+(1-p^{PPT}(\hat{\rho}_{\cal A}))\hat{\rho}_{\cal S}^{opt}$, which is a
convex mixture of the starting antisymmetric state and a symmetric state
($\hat{P}_{\cal S}\hat{\sigma}\hat{P}_{\cal S}$ properly normalized).
Therefore, an alternative way to get $p^{PPT}(\hat{\rho}_{\cal A})$ is to find
the optimal symmetric state whose convex mixture with $\hat{\rho}_{\cal A}$ is
PPT, and the weight of $\hat{\rho}_{\cal A}$ in this mixed state is maximal.
Then if this weight is less than $1/2$, the mixture is entangled.

The second method given in Ref. SinPia  to construct a PPT entangled state is
based on the fact that the Schmidt number of the projection of a separable
state onto the antisymmetric subspace is always two. Therefore, any PPT state
whose antisymmetric projection is proportional to a state whose Schmidt number
is larger than two is necessarily entangled. For example, if we start from any
antisymmetric state with a Schmidt number larger than two, and mix it with a
symmetric state such that the mixture is PPT, the result will be a PPT
entangled state. The symmetric state need not be the optimal one.

##  III Our results

###  III.1 Construction of PPT states with any Schmidt number

As we will show in this subsection, the second method of Sindici and Piani
reviewed in Sec. I can be used to construct PPT bound entangled states with
any Schmidt number. For that we will prove that the Schmidt number of the
projection of a state onto the antisymmetric subspace $\hat{P}_{\cal
A}\hat{\sigma}\hat{P}_{\cal A}$ can not be more than twice the Schmidt number
of the state $\hat{\sigma}$. First let us consider a pure state
$|\varphi\rangle$, whose Schmidt decomposition is defined by Eq. (4). Then

$\displaystyle\hat{P}_{\cal
A}|\varphi\rangle=\frac{\sum_{i=1}^{r}\sqrt{\pi_{i}}(|a_{i}\rangle\otimes|b_{i}\rangle-|b_{i}\rangle\otimes|a_{i}\rangle)}{2},$
(9)

where $r$ is the Schmidt rank of the state (4). Vectors $|a_{i}\rangle$ and
$|b_{i}\rangle$ together can not span more than $2r$-dimensional subspaces of
the component spaces, and the projected vector fully resides within the tensor
product of those subspaces. Therefore, its Schmidt rank can not be more than
$2r$. Now let us consider a mixed state. If it has Schmidt number $r$ it can
be expressed as a convex mixture of pure states with Schmidt ranks at most
$r$. If one applies the projection onto this form of the mixed state one gets
an expansion in terms of vectors such that neither of them has Schmidt rank
more than $2r$. Consequently, the Schmidt number of the projection can not be
more than $2r$. Therefore, if one starts from an antisymmetric state whose
Schmidt number is at least $r$ (where $r$ is even), and construct any PPT
state whose antisymmetric projection is proportional to this state, then the
Schmidt number of this PPT state has to be at least $r/2$. Again, we can do
that by mixing the antisymmetric state with a sufficient amount of suitable
symmetric state.

When choosing the antisymmetric state for the construction, it may be a
problem that the Schmidt number of a mixed state is usually very hard to
determine. However, as suggested in Ref. SinPia , one may start from a pure
state. In case of $d$-dimensional component spaces, if $d$ is even, a generic
antisymmetric pure state, which one can get by antisymmetrizing a random pure
state has Schmidt rank $d$. By starting the construction from such a state
with $d=2r$, one gets a PPT state with a Schmidt number at least $r$. In Ref.
SinPia  the authors considered the following states:

$|\psi_{\cal A}\rangle=\sum_{\mu=1}^{d-1,{\rm
odd}}c_{\mu}(|\mu,\mu+1\rangle-|\mu+1,\mu\rangle),$ (10)

where

$\sum_{\mu=1}^{d-1,{\rm odd}}c^{2}_{\mu}=\frac{1}{2},\quad\quad c_{\mu}\geq 0$
(11)

If all $c_{\mu}>0$, the Schmidt rank of the state is $d$ (the Schmidt-
coefficients are $c_{\mu}$, and each of them occurs twice). A special case is
when all coefficients are equal:

$|\psi_{0\cal A}\rangle=\frac{1}{\sqrt{d}}\sum_{\mu=1}^{d-1,{\rm
odd}}(|\mu,\mu+1\rangle-|\mu+1,\mu\rangle).$ (12)

We will prove in the Appendix A that $p|\psi_{0\cal
A}\rangle\langle\psi_{0\cal A}|+(1-p)\hat{P}_{\cal S}/d_{\cal S}$ with
$p=1/(d+2)$ is a PPT bound entangled state with Schmidt number at least $d/2$,
where $d$ is even.

###  III.2 Normal form of antisymmetric pure states

Now we will prove that any antisymmetric pure state may be transformed into
the simple form given in Eq. (10) (for odd $d$ the summation goes from $1$ to
$d-2$) with local transformations preserving the symmetric and antisymmetric
subspaces. Let $|\varphi_{\cal A}\rangle$ be such an antisymmetric state in a
$(d\times d)$-dimensional space:

$|\varphi_{\cal A}\rangle=\sum_{i=1}^{d}\sum_{j=1}^{d}a_{ij}|i,j\rangle,$ (13)

where $A=(a_{ij})$ is a $d\times d$ skew symmetric matrix, that is
$a_{ij}=-a_{ji}$. Then there exists a unitary matrix $U$ Zum62  such that

$U^{T}AU=\left(\begin{array}[]{ccccc}0&c_{1}&0&0&\dots\\\
-c_{1}&0&0&0&\dots\\\ 0&0&0&c_{3}&\dots\\\ 0&0&-c_{3}&0&\dots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\\\ \end{array}\right),$ (14)

where $c_{\mu}\geq 0$ ($\mu$ odd) are real numbers, and the last row and
column contains zeros if $d$ is odd (a skew symmetric matrix of odd $d$ can
not have full rank). The matrix on the right-hand side is called the normal
form of $A$. The state characterized by such a matrix is just the one of Eq.
(10), therefore we will also refer to it as the normal form. It is easy to see
that this matrix transformation corresponds to both parties performing the
same local basis transformation $|i^{\prime}\rangle=\hat{U}^{*}|i\rangle$,
where $\hat{U}^{*}=\sum_{kl}U^{*}_{kl}|k\rangle\langle l|$. Such simultaneous
local transformations preserve the antisymmetricity and the symmetricity of a
state ($\hat{U}^{*}\otimes\hat{U}^{*}$ commutes with both $P_{\cal A}$ and
$P_{\cal S}$). Starting from $\hat{\rho}_{\cal A}$, the constructions
considered above will give the same results in the transformed local basis as
in the original one. Therefore, we may calculate $p^{PPT}$ for any pure
antisymmetric state after transforming it into its normal form. From the
normal form it is also clear that all antisymmetric pure states have an even
Schmidt rank in any number of dimensions, and from this it follows that all
antisymmetric mixed states have an even Schmidt number. We will also use later
that if the Schmidt rank $r$ of a $(d\times d)$-dimensional pure state in its
normal form is less than $d$, then the subspace necessary to accommodate the
state is spanned by $r$ basis vectors in each component space.

###  III.3 Schmidt number from the value of $p^{PPT}(\hat{\rho}_{\cal A})$

The important result of the first method of Ref. SinPia  is based on the
observation that $p^{PPT}(\hat{\rho}_{\cal A})<1/2$ proves that the PPT state
resulting from the construction is entangled. We will show that from the value
of $p^{PPT}(\hat{\rho}_{\cal A})$ we may be able to tell that the Schmidt
number of $\hat{\rho}_{\cal A}$ is larger than a certain value, consequently
any PPT state constructed from it with the methods given above has Schmidt
number at least half that value. It has been proven SinPia  that there exists
a lower bound for $p^{PPT}(\hat{\rho}_{\cal A})$ depending on the dimension of
the component spaces $d$, which decreases monotonously as a function of $d$.
The bound derived is $2/[d(d+1)+2]$, which is not tight, as we will show it
later by deriving a much better one. Let us denote the $d$-dependent
monotonously decreasing lower bound by $L(d)$. What we will show is that
$L(d)$ is actually $L(r)$, where $r$ is the Schmidt number of
$\hat{\rho}_{\cal A}$, that is in any space dimensions
$p^{PPT}(\hat{\rho}_{\cal A})$ can not be smaller than $L(r)$. First let us
consider a pure state in a $d\times d$-dimensional space with Schmidt rank
$r<d$. We are allowed to transform it into its normal form. Let us reduce the
space to the $r\times r$-dimensional subspace that accommodates the state by
dropping the basis vectors orthogonal to this subspace. Determined in this
space $p^{PPT}$ can not be smaller than $L(r)$. If the dropped basis vectors
are added back, $p^{PPT}$ can not decrease, because it is the maximum of the
overlaps between the state and all PPT states $\sigma$ whose antisymmetric
projection is proportional to the state, and the extended space means an even
larger variety of appropriate $\sigma$ states. Now let us consider mixed
states. First we will show that $p^{PPT}$ belonging to a convex mixture of two
antisymmetric states can not be smaller than $p^{PPT}$ belonging to any of the
two states. Let $p_{1}\equiv p^{PPT}(\hat{\rho}_{1\cal A})$, $p_{2}\equiv
p^{PPT}(\hat{\rho}_{2\cal A})$ and $p_{1}<p_{2}$. Then there exist symmetric
states $\hat{\rho}_{1\cal S}$ and $\hat{\rho}_{2\cal S}$ such that
$p_{1}\hat{\rho}_{1\cal A}+(1-p_{1})\hat{\rho}_{1\cal S}$ and
$p_{2}\hat{\rho}_{2\cal A}+(1-p_{2})\hat{\rho}_{2\cal S}$ are PPT. Let us mix
the second state with any symmetric PPT state such that the weight of
$\hat{\rho}_{2\cal A}$ in the mixture is reduced to $p_{1}$. We get
$p_{1}\hat{\rho}_{2\cal A}+(1-p_{1})\hat{\rho}^{\prime}_{2\cal S}$, which is
also a PPT state. Then $p_{1}[\lambda\hat{\rho}_{1\cal
A}+(1-\lambda)\hat{\rho}_{2\cal A}]+(1-p_{1})[\lambda\hat{\rho}_{1\cal
S}+(1-\lambda)\hat{\rho}_{2\cal S}]$, where $0\leq\lambda\leq 1$ is also PPT,
which proves that $p^{PPT}$ belonging to the convex mixture is not smaller
than $p_{1}$. Then it follows that $p^{PPT}$ belonging to a convex mixture of
any number of states is not smaller than the smallest of $p^{PPT}$ belonging
to the components: mixing can not make $p^{PPT}$ smaller. A mixed state of
Schmidt number $r$ can be written as a convex mixture of pure states with a
maximum Schmidt number of $r$. As $p^{PPT}$ of any of the components can not
be smaller than the lower bound belonging to $r$, $p^{PPT}$ belonging to the
mixture can not be smaller either, which proves the statement for mixed states
as well. The result means that if $p^{PPT}(\hat{\rho}_{\cal A})<L(d)$, its
Schmidt number is larger than $d$ (actually, due to the non-existence of
antisymmetric states with odd Schmidt numbers, it has to be at least $d+2$).

In the Appendix A we will show that $L(d)=1/(d+2)$ for even $d\geq 4$ and
$L(2)=1/2$ are lower bounds for $p^{PPT}(\hat{\rho}_{\cal A})$. For $d=2$ any
smaller value would imply the existence of a bound entangled state. For odd
$d$ the bound has to be the same as for $d-1$. There are no antisymmetric
states with an odd Schmidt number, so the maximum Schmidt number of
$\hat{\rho}_{\cal A}$ is $d-1$. However, as we have shown earlier,
$p^{PPT}(\hat{\rho}_{\cal A})$ can not be smaller than the lower bound
corresponding to its Schmidt number. Therefore, $L(3)=1/2$, and for odd $d\geq
5$ it is $L(d)=1/(d+1)$.

To prove the lower bound for even $d$ we have given above, it is enough to
consider pure states, because mixing can not make $p^{PPT}$ smaller, as we
have shown earlier. We may also confine ourselves to their normal form given
by Eq. 10.

The first step of the proof given in the Appendix A is to take the state
$|\psi_{0\cal A}\rangle$ with equal amplitudes given by Eq. (12), and
calculate how much admixture of $\hat{P}_{\cal S}/d_{\cal S}$ is necessary to
make it a PPT state. This is the same symmetric state considered in Theorem 1
of Ref. SinPia  calculating the lower bound for any state. Their bound is not
tight because for the antisymmetric state for which this choice is optimal
less admixture is enough (that is $p^{PPT}$ is larger), while for all other
states the choice is not optimal. From numerical calculations in smaller
dimensions we believe that for $|\psi_{0\cal A}\rangle$ this choice is
actually optimal, therefore the lower bound $1/(d+2)$ we get is the true value
of $p^{PPT}(|\psi_{0\cal A}\rangle\langle\psi_{0\cal A}|)$. Then we show that
for states of normal form with non-equal amplitudes $p^{PPT}$ is larger,
therefore, $1/(d+2)$ is a lower bound for them, too. Our conjecture is that
this bound is tight.

As $L(2)=1/2$, any smaller value for $p^{PPT}(\hat{\rho}_{\cal A})$ proves
that the Schmidt number of $\hat{\rho}_{\cal A}$ is at least $4$, which also
means that any PPT state whose antisymmetric projection is proportional to
$\hat{\rho}_{\cal A}$ is entangled (that is its Schmidt number is at least 2).
This is true for any such PPT state, not only for the one that comes out of
the construction providing the value of $p^{PPT}(\hat{\rho}_{\cal A})$. For
$d\geq 4$, $p^{PPT}(\hat{\rho}_{\cal A})<L(d)$ with $L(d)=1/(d+2)$ means that
the Schmidt number of $\hat{\rho}_{\cal A}$ is at least $d+2$, and all PPT
states whose antisymmetric projections are proportional to $\hat{\rho}_{\cal
A}$ have Schmidt numbers at least $(d/2)+1$.

##  IV Conclusions

In this paper, we proved that a class of $d\times d$ states with a positive
partial transposition gives rise to Schmidt number $d/2$. This family is the
one investigated by Sindici and Piani SinPia  with respect to their
entanglement properties. Here we generalized their methods to explore the
dimensionality of the entanglement of such states. In particular, the Schmidt
number $d/2$ of these states improves the Schmidt number of the states
investigated by Huber et al. huber  which latter scales as $d/4$.

There are a couple of open questions left. We first ask whether such highly
entangled states are useful for communication tasks beyond known protocols.
Secondly, is it possible to construct PPT states, which have Schmidt number
higher than $d/2$? In particular, is it possible to approach the value of
$(d-1)$? In the recent paper YLT , it has been shown that no ($3\times
3$)-dimensional PPT state has Schmidt number three. In this respect, one may
raise the question: what is the dimensionality of the smallest PPT state which
has Schmidt number at least three? Our present result provides such a PPT
state for dimension $d=6$. But is there possibly a smaller dimensional
example? Finally, we believe that our investigations are relevant from an
experimental point of view as well given the recent advances in the
experimental implementation of bound entangled states exp1 ; exp2 ; exp3 .

##  V Acknowledgements

This work was supported by the National Research, Development and Innovation
Office NKFIH (Grant No. KH125096).

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##  Appendix A Lower bound for $\boldsymbol{p^{PPT}(\hat{\rho}_{\cal A})}$

As we have shown in the main text, to calculate the $d$-dependent lower bound
$L(d)$ for $p^{PPT}(\hat{\rho}_{\cal A})$ it is enough to consider pure states
in their normal form in $(d\times d)$-dimensional spaces, where $d$ is even.

First we determine a lower bound of $p^{PPT}$ for the special state
$|\psi_{0\cal A}\rangle$ given by Eq. (12). The problem we are going to solve
is to find the largest value of $p$ such that the partial transpose of

$\hat{\sigma}_{0}=p|\psi_{0\cal A}\rangle\langle\psi_{0\cal
A}|+(1-p)\frac{\hat{P}_{\cal S}}{d_{\cal S}}.$ (15)

is positive semidefinite. In particular, we show that this value is $1/(d+2)$.
If $\hat{P}_{\cal S}/d_{\cal S}$ is the optimal symmetric operator, then this
way we get $p^{PPT}(|\psi_{0{\cal A}}\rangle\langle\psi_{0{\cal A}}|)$ itself,
if not, we get a lower bound. For smaller $d$ values we have numerical
evidence that this is the optimal choice, and our conjecture is that this is
true for any dimensions. The partial transposes (denoted by upper index
$\Gamma$) of the operators appearing in the right-hand side of Eq. (15) are:

$\displaystyle|\psi_{0\cal A}\rangle\langle\psi_{0\cal
A}|^{\Gamma}=\frac{1}{d}\sum_{\mu=1}^{d-1,{\rm odd}}\sum_{\nu=1}^{d-1,{\rm
odd}}$
$\displaystyle(|\mu,\nu+1\rangle\langle\nu,\mu+1|+|\mu+1,\nu\rangle\langle\nu+1,\mu|-$
$\displaystyle|\mu,\nu\rangle\langle\nu+1,\mu+1|-|\nu+1,\mu+1)\rangle\langle\mu,\nu|),$
(16)

and

$\displaystyle\frac{\hat{P}^{\Gamma}_{\cal S}}{d_{\cal
S}}=\frac{\sum_{i=1}^{d}\sum_{j=1}^{d}|i,j\rangle\langle
i,j|+\sum_{i=1}^{d}\sum_{j=1}^{d}|i,i\rangle\langle j,j|}{d(d+1)}.$ (17)

Eq. (16) can be derived from Eq. (12), while Eq. (17) from Eqs. (6) and (8),
and $d_{\cal S}=d(d+1)/2$.

Now we will solve the eigenvalue problem, and determine the largest $p$ value
such that none of the eigenvalues are negative. The calculation can be
simplified by realizing that the matrix representing operator
$\hat{\sigma}^{\Gamma}_{0}$ has a block-diagonal structure. This structure is
not apparent, as the rows and columns belonging to each block do not follow
each other. Nevertheless, if there exists a subset of indices such that the
rows and columns corresponding to those indices have nonzero elements only
where they intersect, then these rows and columns belong to a block of the
matrix. Such blocks can be treated separately when solving the eigenvalue
problem.

The first double sum of the symmetric part of the operator, which is given by
Eq. (17), is diagonal (proportional to the identity operator), while the
second one is a $d\times d$ block with equal matrix elements.

The pair of positive terms of Eq. (16),
$(1/d)|\mu,\nu+1\rangle\langle\nu,\mu+1|$ and
$(1/d)|\nu,\mu+1\rangle\langle\mu,\nu+1|$, and similarly the
$(1/d)|\mu+1,\nu\rangle\langle\nu+1,\mu|$ and
$(1/d)|\nu+1,\mu\rangle\langle\mu+1,\nu|$ correspond to off-diagonal elements
of $2\times 2$ blocks if $\nu\neq\mu$ (they exist if $d\geq 4$). Each of these
blocks of $\hat{\sigma}^{\Gamma}_{0}$, including the diagonal elements coming
from the symmetric matrix given by Eq. (17) look like:

$\left(\begin{array}[]{cc}\frac{1-p}{d(d+1)}&\frac{p}{d}\\\
\frac{p}{d}&\frac{1-p}{d(d+1)}\\\ \end{array}\right).$ (18)

The solutions of the eigenvalue equation of this block are $(1+pd)/[d(d+1)]$ ,
which is always positive, and $[1-(d+2)p]/[d(d+1)]$, which is non-negative if
$p\leq 1/(d+2)$. There are $d(d-2)/4$ such blocks. Each of the positive terms
of Eq. (16) with $\nu=\mu$ are diagonal elements. Together with the
contribution coming from Eq. (17) each of them gives the always positive
eigenvalue of $(1+pd)/[d(d+1)]$ again. There are $d$ such diagonal elements.

The pair of negative terms of Eq. (16):
$-(1/d)|\mu,\nu\rangle\langle\nu+1,\mu+1|$ and
$-(1/d)|\nu+1,\mu+1\rangle\langle\mu,\nu|$ also correspond to off-diagonal
elements of $2\times 2$ blocks. If $\nu=\mu$ ($d/2$ pairs) they fall within
the $d\times d$ block of the matrix corresponding to Eq. (17) whose eigenvalue
problem we will solve in the next step. The $\nu\neq\mu$ cases ($d(d-2)/4$
pairs) lead to the same $2\times 2$ matrix as given in Eq. (18) but with
negative signs for the off-diagonal elements, giving the same two eigenvalues
as we have already got.

Now we will calculate the eigenvalues of the $d\times d$ block. We will
demonstrate the calculations on the example of $d=8$. The determinant to be
calculated is the following:

$D_{8}=\left|\begin{array}[]{cccccccc}a&c&b&b&b&b&b&b\\\ c&a&b&b&b&b&b&b\\\
b&b&a&c&b&b&b&b\\\ b&b&c&a&b&b&b&b\\\ b&b&b&b&a&c&b&b\\\ b&b&b&b&c&a&b&b\\\
b&b&b&b&b&b&a&c\\\ b&b&b&b&b&b&c&a\\\ \end{array}\right|,$ (19)

where

$\displaystyle a$ $\displaystyle=2\frac{1-p}{d(d+1)}-\lambda,$ $\displaystyle
b$ $\displaystyle=\frac{1-p}{d(d+1)},$ $\displaystyle c$
$\displaystyle=b-\frac{p}{d}=\frac{1-p(d+2)}{d(d+1)}.$ (20)

The first term of $a$ and $b$ comes from Eq. (17), while the $-p/d$ term of
$c$ comes from Eq. (16), and $\lambda$ is the eigenvalue. Let us subtract the
$\nu$th row from the $(\nu-1)$th one then the $\nu$th column from the
$(\nu-1)$th one for all odd $\nu$, and introduce the notation $f\equiv a-c$ to
arrive at:

$D_{8}=\left|\begin{array}[]{cccccccc}a&-f&b&0&b&0&b&0\\\ -f&2f&0&0&0&0&0&0\\\
b&0&a&-f&b&0&b&0\\\ 0&0&-f&2f&0&0&0&0\\\ b&0&b&0&a&-f&b&0\\\
0&0&0&0&-f&2f&0&0\\\ b&0&b&0&b&0&a&-f\\\ 0&0&0&0&0&0&-f&2f\\\
\end{array}\right|.$ (21)

Let us subtract the 5th row from the 7th one, then the 3th row from the 5th
one and finally the first row from the third one. Then let us do the same with
the corresponding columns. Introducing the notation $e\equiv a-b$ we get:

$D_{8}=\left|\begin{array}[]{cccccccc}a&-f&-e&0&0&0&0&0\\\
-f&2f&f&0&0&0&0&0\\\ -e&f&2e&-f&-e&0&0&0\\\ 0&0&-f&2f&f&0&0&0\\\
0&0&-e&f&2e&-f&-e&0\\\ 0&0&0&0&-f&2f&f&0\\\ 0&0&0&0&-e&f&2e&-f\\\
0&0&0&0&0&0&-f&2f\\\ \end{array}\right|.$ (22)

Finally, let us add one half times the 8th column to the 7th one, subtract one
half times the 6th row from the 7th one, and then subtract one half times the
6th column from the 7th one to arrive at:

$D_{8}=\left|\begin{array}[]{cccccccc}a&-f&-e&0&0&0&0&0\\\
-f&2f&f&0&0&0&0&0\\\ -e&f&2e&-f&-e&0&0&0\\\ 0&0&-f&2f&f&0&0&0\\\
0&0&-e&f&2e&-f&-e+\frac{f}{2}&0\\\ 0&0&0&0&-f&2f&0&0\\\
0&0&0&0&-e+\frac{f}{2}&0&2e-f&-f\\\ 0&0&0&0&0&0&0&2f\\\ \end{array}\right|.$
(23)

It is easy to check that the manipulations we have done have not changed the
values of the subdeterminants $D_{6}$, $D_{4}$ and $D_{2}$. Then it is not
difficult to see that $D_{8}$ can be written as:

$D_{8}=2f\left[(2e-f)D_{6}-\left(e-\frac{f}{2}\right)^{2}2fD_{4}\right].$ (24)

We could have arrived at an analogous recurrence relation for any even $d\geq
6$. Using $e\equiv a-b$ and $f\equiv a-c$ the general recurrence relation is:

$D_{d+4}=(a-c)(a-2b+c)[2D_{d+2}-(a-c)(a-2b+c)D_{d}].$ (25)

The solution of the relation with the correct initial values is:

$D_{d}=(a-c)^{\frac{d}{2}}(a-2b+c)^{\frac{d}{2}-1}[a+(d-2)b+c].$ (26)

This can be proven by induction: $D_{2}$ and $D_{4}$ can explicitly be
calculated and compared, and it is easy to check that the expression satisfies
the recurrence relation. Using Eq. (20) the factors appearing in Eq. (26) are
the following:

$\displaystyle a-c$ $\displaystyle=\frac{1+pd}{d(d+1)}-\lambda,$ (27)
$\displaystyle a-2b+c$ $\displaystyle=\frac{1-p(d+2)}{d(d+1)}-\lambda,$ (28)
$\displaystyle a+(d-2)b+c$ $\displaystyle=\frac{1+d-p(d+4)}{d(d+1)}-\lambda.$
(29)

A root of $(1+pd)/[d(d+1)]$ with multiplicity $d/2$ of the eigenvalue equation
$D_{d}=0$ comes from $a-c=0$. This root is the always positive solution we
have already got. If $d\geq 4$ we get root $[1-(d+2)p]/[d(d+1)]$ with
multiplicity $(d/2-1)$ from requiring $a-2b+c=0$. We have already got this
root as well from the $2\times 2$ blocks. The remaining root,
$[1+d-p(d+4)]/[d(d+1)]$ comes from $a+(d-2)b+c=0$.

To summarize the results above, for $d\geq 4$ there are three different
eigenvalues of operator $\hat{\sigma}^{\Gamma}_{0}$. The always positive
$(1+pd)/[d(d+1)]$ has a multiplicity of $d(d+1)/2$, coming from the $d\times
d$ block, the $2\times 2$ blocks and the single diagonal elements. Eigenvalue
$[1-(d+2)p]/[d(d+1)]$ has multiplicity $(d+1)(d-2)/2$ coming from the $d\times
d$ block and the $2\times 2$ blocks. This is non-negative if $p\leq 1/(d+2)$.
The last root $[1+d-p(d+4)]/[d(d+1)]$ is a single one, and it is non-negative
if $p\leq(d+1)/(d+4)$, a less strict condition than the previous one.
Therefore, for $d\geq 4$ the largest value of $p$ such that
$\hat{\sigma}^{\Gamma}_{0}$ is positive semidefinite is $p=1/(d+2)$. For $d=2$
the eigenvalue giving this condition does not exist, therefore we get the
appropriate condition $p=0.5$ from the last eigenvalue.

The last step is to prove that $p^{PPT}(|\psi_{{\cal
A}}\rangle\langle\psi_{{\cal A}}|)\geq p^{PPT}(|\psi_{0{\cal
A}}\rangle\langle\psi_{0{\cal A}}|)$. Let us define operator $\hat{\tau}$ as:

$\hat{\tau}\equiv\sum_{i=1}^{d}\sum_{j=1}^{d}t_{i}t_{j}|i,j\rangle\langle
i,j|,$ (30)

where

$t_{\mu}=t_{\mu+1}\equiv\sqrt{c_{\mu}}d^{1/4}\quad\quad\mu\hphantom{0}{\rm
odd},$ (31)

and $c_{\mu}$ are the coefficients appearing in Eq. (10), the definition of
$|\psi_{{\cal A}}\rangle$. From Eqs. (10)-(12) it is easy to see that

$\displaystyle|\psi_{\cal A}\rangle$ $\displaystyle=\hat{\tau}|\psi_{0\cal
A}\rangle,$ (32) $\displaystyle\sum_{i=1}^{d}t_{i}^{4}$ $\displaystyle=d.$
(33)

The matrix of operator $\hat{\tau}$ is diagonal with non-negative entries. If
operator $\hat{R}$ is positive semidefinite, so is
$\hat{\tau}\hat{R}\hat{\tau}$, because if the expectation value of the latter
with a state $|\xi\rangle$ were negative, so were the expectation value of the
former with $\hat{\tau}|\xi\rangle$. Furthermore, it commutes with the partial
transposition, as it multiplies both $|i,j\rangle\langle k,l|$ and
$|i,l\rangle\langle k,j|$ with the same factor of $t_{i}t_{j}t_{k}t_{l}$. It
also preserves the symmetricity or antisymmetricity of a state. Let us take
$\hat{\tau}\hat{\sigma}_{0}\hat{\tau}$ (see Eq. (15)), that is:

$p|\psi_{\cal A}\rangle\langle\psi_{\cal
A}|+(1-p)\frac{\hat{\tau}\hat{P}_{\cal S}\hat{\tau}}{d_{\cal S}}$ (34)

with $p=1/(d+2)$, which is the lower bound for $p^{PPT}(|\psi_{0\cal
A}\rangle\langle\psi_{0\cal A}|)$. Due to the properties of operator
$\hat{\tau}$ stated above, the partial transpose of the operator is positive
semidefinite with this value of $p$. The second term is a symmetric positive
semidefinite operator. However, the operator in Eq. (34) is not a density
operator, because $\hat{\tau}\hat{P}_{\cal S}\hat{\tau}/d_{\cal S}$ in the
second term is not one. Projector $\hat{P}_{\cal S}$ is an equal combination
of the swap and the identity operators (see Eq. (6)). For the transformed swap
operator $Tr(\hat{\tau}\sum_{ij}|i,j\rangle\langle
j,i|\hat{\tau})=\sum_{i}t_{i}^{4}=d$, which is the same as for the swap
operator itself. However, for the transformed identity operator the trace is
$\sum_{ij}t_{i}^{2}t_{j}^{2}=(\sum_{i}t_{i}^{2})^{2}\leq
d\sum_{i}t_{i}^{4}=d^{2}$, that is if $t_{i}$ are not all equal, it is smaller
than the trace of the identity operator. Therefore, to make the operator in
Eq. (34) a density operator that is a proper mixture of the antisymmetric
operator and a symmetric one to provide a lower bound for $p^{PPT}$, the
second term has to be renormalized by multiplying it by a factor larger than
one. This renormalization will not only preserve the PPT property of the
operator with the same $p$, but it will even allow $p$ to be increased
somewhat and still having a PPT operator. Therefore, $p^{PPT}(|\psi_{{\cal
A}}\rangle\langle\psi_{{\cal A}}|)\geq p^{PPT}(|\psi_{0{\cal
A}}\rangle\langle\psi_{0{\cal A}}|)$, indeed.


# On a generalization of compact and connected spaces

Nebojša Elez University of East Sarajevo<EMAIL_ADDRESS>Ognjen
Papaz University of East Sarajevo<EMAIL_ADDRESS>
###### Abstract

In this paper we will give two different natural generalizations of compact
spaces and connected spaces simultaneously. We will show that these
generalizations coincide for the subspaces of the real line and that they
differ for subspaces of $\mathbb{R}^{2}$.

##  1 Introduction

Classes of compact and connected spaces, although not very related, share some
similar properties. They are closed with respect to taking continuous image
and topological product.

Intersection of class of compact and class of connected spaces, the class of
continua, is well known. Class of continua is closed with respect to taking
continuous image and topological product.

Union of class of compact and class of connected spaces is not particularly
interesting.

The smallest class of spaces that contains both compact and connected spaces
and that is closed with respect to taking continuous image and topological
product is class of spaces that are continuous image of product of compact and
connected space. The following subspace of the real line is an example of such
space

$\\{0\\}\cup\bigcup_{n=1}^{+\infty}\left(\frac{1}{n+1},\frac{1}{n}\right).$

This space has infinitely many connected components but it doesn’t have
infinitely many open connected components. This property can inspire the
following definition.

######  Definiton 1.

We say that a topological space is a GCC space (Generalized compact and
connected space) if it cannot be represented as a union of infinitely many of
its disjoint non-empty open subsets.

In the first part of the paper we will give characterizations of the class of
GCC spaces.

It’s easy to see that continuous image of a GCC space is a GCC space. It turns
out that product of two GCC spaces doesn’t have to be a GCC space.

Similar class of spaces, having property that every clopen cover of a space
has finite subcover, was introduced in [2].

In the second part we will consider the following subclass of GCC spaces that
is closed with respect to taking continuous image and topological product.

######  Definiton 2.

We say that a topological space is a CCC space if it is a continuous image of
a product of compact and connected space.

As we have mentioned earlier, the class of CCC spaces is a minimal class of
spaces that contains all compact and connected spaces and that is closed with
respect to product and taking continuous image.

We will give intrinsic property that characterizes CCC spaces. Namely, we will
show that a space $X$ is a CCC space if and only if it has compact subspace
that intersects every connected component of $X$.

In the last part of the paper we will show that these two classes are equal on
the real line, and we will give an example of a subspace of $\mathbb{R}^{2}$
which is a GCC space but not a CCC space.

##  2 Main results

######  Theorem 1.

For a topological space $X$, the following conditions are equivalent:

(1) $X$ is a GCC space;

(2) Every disjoint open cover of $X$ has a finite subcover;

(3) Every countable clopen cover of $X$ has a finite subcover;

(4) Every decreasing sequence of non-empty clopen subsets of $X$ has a non-
empty intersection;

(5) There is no continuous function from $X$ onto $\mathbb{N}$.

(6) $X$ is not a direct sum of infinitely many of its nonempty subspaces.

###### Proof.

$(1)\Rightarrow(2)$. Let $(U_{i})_{i\in I}$ be a disjoint open cover of $X$.
Since $X$ is GCC space, it follows that all but finitely many of the sets
$U_{i}$ are empty. Hence, $(U_{i})_{i\in I}$ has finite subcover.

$(2)\Rightarrow(3)$. Let $(U_{n})_{n\in\mathbb{N}}$ be a clopen cover of $X$.
For every $n\in\mathbb{N}$ let $V_{n}=U_{n}\setminus\bigcup_{k<n}U_{k}$.
Family $(V_{n})_{n\in\mathbb{N}}$ is a disjoint open cover of $X$, hence it
has a finite subcover. Since $(V_{n})_{n\in\mathbb{N}}$ is a refinement of the
cover $(U_{n})_{n\in\mathbb{N}}$, it follows that $(U_{n})_{n\in\mathbb{N}}$
also has a finite subcover.

$(3)\Rightarrow(4)$. Let $\\{F_{n}\\}_{n=1}^{+\infty}$ be a decreasing
sequence of non-empty clopen subsets of $X$. If
$\bigcap_{n=1}^{+\infty}F_{n}=\emptyset$, then the family $(X\setminus
F_{n})_{n\in\mathbb{N}}$ is a countable clopen cover of $X$ that has no finite
subcover, which is impossible by the assumption of the theorem. Therefore,
$\bigcap_{n=1}^{+\infty}F_{n}\neq\emptyset$.

$(4)\Rightarrow(5)$. Assume to the contrary that there exists a continuous
function $f$ from $X$ onto $\mathbb{N}$. For every $n\in\mathbb{N}$ let
$F_{n}=\mathbb{N}\setminus\\{1,2,\ldots,n\\}$. Since each of the sets $F_{n}$
is clopen in $\mathbb{N}$, it follows that
$\\{f^{-1}(F_{n})\\}_{n=1}^{+\infty}$ is a decreasing sequence of non-empty
clopen subsets of $X$. But,
$\bigcap_{n=1}^{+\infty}f^{-1}(F_{n})=f^{-1}(\bigcap_{n=1}^{+\infty}F_{n})=f^{-1}(\emptyset)=\emptyset$,
which is a contradiction.

$(5)\Rightarrow(6)$. Let us assume that $X$ is the direct sum of the sequence
$\\{X_{n}\\}_{n=1}^{+\infty}$ of disjoint non-empty subspaces of $X$. The
function $f:X\to\mathbb{N}$, defined with $f(X_{n})=\\{n\\}$, is a continuous
function from $X$ onto $\mathbb{N}$. Thus, we have a contradiction.

$(6)\Rightarrow(1)$. This implication is obviously true. ∎

One readily proves the following proposition.

######  Proposition 1.

Each of the following statements holds.

1\. All countably compact, compact and connected spaces are GCC spaces.

2\. An image of a GCC space under a continuous mapping is a GCC space.

3\. A clopen subspace of a GCC space is a GCC space.

4\. A space $X$ is a GCC space if it has a dense GCC subspace.

5\. A space $X$ is a GCC space if it can be represented as a union of finitely
many of its GCC subspaces.

6\. If $(X,\tau)$ is a GCC space and if $\sigma\subseteq\tau$, then
$(X,\sigma)$ is also a $GCC$ space.

The CLP-compact spaces, that were introduced in [2], are defined by having the
property that every clopen cover has finite subcover. Every CLP-compact space
is a GCC space. In the realm of second countable spaces every GCC space is
CLP-compact. The following example shows that class of GCC spaces is strictly
larger then the class of CLP-compact spaces.

######  Example 1.

First uncountable ordinal $\omega_{1}$ is a GCC space that is not CLP-compact.
Indeed, $\omega_{1}$ is countably compact so it is a GCC space. And
$\omega_{1}$ is not CLP-compact since clopen cover of $\omega_{1}$ consisting
of the sets of the form $[0,a],a<\omega_{1}$, has no finite subcover.

The following example shows that product of two GCC spaces doesn’t need to be
a GCC space.

######  Example 2.

Famous example of Novak [1, 3.10.19 Example] exhibits two countably compact
subspaces $X$ and $Y$ of $\beta\mathbb{N}$ whose product $X\times Y$ contains
$\\{(1,1),(2,2),\ldots\\}$ as an open, closed and discrete subset. Thus, $X$
and $Y$ are GCC spaces and their product is not a GCC space.

We will now consider the class of $CCC$ spaces.

######  Proposition 2.

Every CCC space is a GCC space.

###### Proof.

Let $X$ be a CCC space and let $X=f(K\times C)$, where $f$ is continuous, $K$
is compact and $C$ is connected. To prove that $X$ is a GCC space it is enough
to prove that $K\times C$ is a GCC space. Let $(U_{i})_{i\in I}$ be a disjoint
open cover of $K\times C$. We pick some element $c\in C$. Subspace
$K\times\\{c\\}$ of $K\times C$ is compact so there is a finite subfamily
$\\{U_{i_{1}},U_{i_{2}},\ldots,U_{i_{n}}\\}$ of $(U_{i})_{i\in I}$ that covers
$K\times\\{c\\}$. We will prove that
$\\{U_{i_{1}},U_{i_{2}},\ldots,U_{i_{n}}\\}$ covers $K\times C$. Let $k\in K$.
Subspace $\\{k\\}\times C$ of $K\times C$ is connected and there is some
$U_{i_{k}}\in\\{U_{i_{1}},U_{i_{2}},\ldots,U_{i_{n}}\\}$ such that
$(\\{k\\}\times C)\cap U_{i_{k}}\neq\emptyset$. Since $U_{i_{k}}$ is clopen
set we have that $\\{k\\}\times C\subseteq U_{i_{k}}$. ∎

######  Proposition 3.

The class of CCC spaces is closed with respect to product and taking
continuous image.

###### Proof.

One easily sees that a continuous image of a CCC space is a CCC space. We
prove that the class of CCC spaces is closed with respect to product. Let
$(X_{i})_{i\in I}$ be a family of CCC spaces and let each
$X_{i}=f_{i}(K_{i}\times C_{i})$ where $f_{i}$ is continuous, $K_{i}$ is
compact, and $C_{i}$ is connected. We define a mapping

$f:\prod_{i\in I}(K_{i}\times C_{i})\to\prod_{i\in I}X_{i}$

in the following way

$f(((k_{i},c_{i})_{i\in I}))=(f_{i}(k_{i},c_{i}))_{i\in I}.$

The mapping $f$ is a continuous surjection and since

$\prod_{i\in I}(K_{i}\times C_{i})\cong\left(\prod_{i\in
I}K_{i}\right)\times\left(\prod_{i\in I}C_{i}\right),$

we have that $\prod_{i\in I}X_{i}$ is a CCC space. ∎

Now we give an intrinsic characterization of $CCC$ spaces.

######  Theorem 2.

A space $X$ is a CCC space if and only if $X$ has a compact subspace that
intersects every connected component of $X$.

###### Proof.

Let $X$ be a CCC space $X=f(K\times C)$ where $f$ is continuous, $K$ is
compact and $C$ is connected. We pick some element $c\in C$. Then
$K\times\\{c\\}$ is a compact subspace of $K\times C$ that intersects every
connected component of $K\times C$ and $f(K\times\\{c\\})$ is a compact
subspace of $X$ that intersects every connected component of $X$.

Now we prove the converse. Let $K$ be a compact subspace of $X$ that
intersects every connected component of $X$.

For every $k\in K$ let $C_{k}$ be connected component of $X$ that contains
$k$. We will regard each $C_{k}$ as a based space $(C_{k},k)$. Let

$C=\bigvee_{k\in K}C_{k}$

be the wedge sum of the spaces $(C_{k},k)$. The space $C$ is connected because
every $C_{k}$ is connected.

Let $A(K)=K\cup(K\times\\{1\\})$ be Alexandroff duplicate of $K$ (see [3]).
The family

$\\{U\cup((U\setminus\\{x\\})\times\\{1\\}):U\ \text{is open in}\ K,x\in
U\\}\cup\\{(x,1):x\in K\\}$

is subbasis of topology of $A(K)$. Since $K$ is compact space, $A(K)$ is also
compact. We will construct a continuous surjection

$f:A(K)\times C\to X.$

For every $k,k^{\prime}\in K$ we define $g_{k,k^{\prime}}:C_{k^{\prime}}\to X$
in the following way

$g_{k,k^{\prime}}(x)=\begin{cases}x,k=k^{\prime}\\\ k,k^{\prime}\neq
k.\end{cases}$

Each $g_{k,k^{\prime}}$ is continuous map and sends base point of
$C_{k^{\prime}}$ to $k$. Thus, there exists unique continuous map

$g_{k}:C\to X,$

such that $g_{k}\circ i_{k^{\prime}}=g_{k,k^{\prime}}$ for every $k^{\prime}$,
where $i_{k^{\prime}}:C_{k^{\prime}}\to C$ is a natural embedding.

We now define $f$. Let $k\in K$. We define

$f_{|\\{(k,1)\\}\times C}=g_{k},$

and

$f_{|\\{k\\}\times C}=\text{constant map with the value}\ k.$

Its easy to see that $f$ is a surjection. Now we prove the continuity of $f$.
Since each of the sets $\\{(k,1)\\}\times C$ is open in $A(K)\times C$ and
each $g_{k}$ is continuous we only need to check the continuity of $f$ at an
arbitrary point $(k,c)\in K\times C$.

Let $k_{0}\in K$ and $c\in i_{k_{1}}(C_{k_{1}})$. We have $f(k_{0},c)=k_{0}$.
Let $V$ be an open neighborhood of $k_{0}$ in $X$. For every $k\in K$ let

$U_{k}=\begin{cases}i_{k}(V\cap C_{k}),k\in V\setminus\\{k_{1}\\},\\\
i_{k}(C_{k}),k\in(K\setminus V)\cup\\{k_{1}\\}.\end{cases}$

Let $U=\bigcup_{k\in K}U_{k}$. We have that $U$ is an open neighborhood od $c$
in $C$.

For every $k\in G=V\setminus\\{k_{1}\\}$ we have

$g_{k}(U_{k^{\prime}})=g_{k}(i_{k^{\prime}}(V\cap
C_{k^{\prime}}))=g_{k,k^{\prime}}(V\cap C_{k^{\prime}})\subseteq V,$

for $k^{\prime}\in G$, and

$g_{k}(U_{k^{\prime}})=g_{k}(i_{k^{\prime}}(C_{k^{\prime}}))=g_{k,k^{\prime}}(C_{k^{\prime}})=\\{k\\},$

for $k^{\prime}\in(K\setminus V)\cup\\{k_{1}\\}$.

Thus, for every $k\in G$ we have $g_{k}(U)\subseteq V$.

Let

$V^{\prime}=V\cup(G\times{1}).$

We have that $V^{\prime}$ is an open neighborhood of $k_{0}$ in $A(K)$.

Now, we have that

$\displaystyle f(V^{\prime}\times U)$ $\displaystyle=f((V\times
U)\cup((G\times{1})\times U))=$ $\displaystyle f(V\times U)\cup
f((G\times\\{1\\})\times U)=$ $\displaystyle V\cup\bigcup_{(k,1)\in
G\times\\{1\\}}f((k,1)\times U)=$ $\displaystyle V\cup\bigcup_{k\in
G}g_{k}(U)\subseteq V.$

∎

The following example is an example of a subspace of $\mathbb{R}^{2}$ which is
a GCC space but not a CCC space.

######  Example 3.

For every $n\in\mathbb{N}$ let $A_{n}=\bigcup_{m\geq
n+1}([0,1]\times\\{1/n+1/m\\})$ and $x_{n}=(1,1/n)$. Let

$X=\\{x_{n}:n\in\mathbb{N}\\}\cup\left(\bigcup_{n\in\mathbb{N}}A_{n}\right)\cup\\{\mathbf{0}\\}.$

We will first show that $X$ is not a CCC space. If $K$ is compact subspace of
$X$ that intersects every connected component of $X$ then
$\\{x_{n}:n\in\mathbb{N}\\}\subseteq K$. But $\\{x_{n}:n\in\mathbb{N}\\}$ is
infinite closed and discrete subspace of $X$, and we have a contradiction.

We will now prove that $X$ is a GCC space. Let $(U_{i})_{i\in i}$ be a clopen
cover of $X$. Let $\mathbf{0}\in U_{i_{0}}$. There exists $n_{0}\in\mathbb{N}$
such that $A_{n}\subseteq U_{i_{0}}$ for all $n>n_{0}$. We have that
$x_{n}\in\overline{A_{n}}$ for every $n\in\mathbb{N}$, thus $x_{n}\in
U_{i_{0}}$ for all $n>n_{0}$. Let

$x_{1}\in U_{i_{1}},x_{2}\in U_{i_{2}},\ldots,x_{n_{0}}\in U_{i_{n_{0}}}.$

Each of the sets $U_{i_{k}}$ contains all but finitely many connected
components of $A_{k}$. Thus $\bigcup_{k=0}^{n_{0}}U_{i_{k}}$ contains all but
finitely many connected components of $X$. From here we can conclude that
$(U_{i})_{i\in I}$ has finite subcover of $X$.

We note that every connected component of the space $X$ from the previous
example is a quasicomponent. So, $X$ is also an example of a GCC subspace of
$\mathbb{R}^{2}$ that doesn’t have compact subspace that intersects every
quasicomponent of $X$.

Now we restrict our attention to the subspaces of $\mathbb{R}$.

######  Theorem 3.

A space $X\subseteq\mathbb{R}$ is a GCC space if and only if every monotone
sequence $\\{x_{n}\\}_{n=1}^{+\infty}$ of real numbers, such that

$x_{2n}\in X\ \text{and}\ x_{2n-1}\in\mathbb{R}\setminus X\ \text{for every}\
n\in\mathbb{N},$

is convergent and $\lim_{n\to+\infty}x_{n}\in X$.

###### Proof.

Let $X$ be a GCC space and let $\\{x_{n}\\}_{n=1}^{+\infty}$ be a monotone
sequence of real numbers such that $x_{2n}\in X$ and
$x_{2n-1}\in\mathbb{R}\setminus X$ for every $n\in\mathbb{N}$. Let us first
prove that the $\lim_{n\to+\infty}x_{n}$ exists. If $\lim_{n\to+\infty}x_{n}$
does not exist then the sequence $\\{x_{n}\\}_{n=1}^{+\infty}$ is unbounded,
and the family

$\\{(-\infty,x_{1})\cap X\\}\cup\\{(x_{2n-1},x_{2n+1})\cap
X:n\in\mathbb{N}\\},$

i.e. the family

$\\{(x_{2n-1},x_{2n+1})\cap X:n\in\mathbb{N}\\}\cup\\{(x_{1},+\infty)\cap
X\\},$

depending on whether the sequence $\\{x_{n}\\}_{n=1}^{+\infty}$ is increasing
or decreasing, is a disjoint open cover of $X$ that has no finite subcover,
which cannot exist because $X$ is a $GCC$ space. Thus, the
$\lim_{n\to+\infty}x_{n}$ exists. Let $x=\lim_{n\to+\infty}x_{n}$. If $x\notin
X$, then the family

$\\{(-\infty,x_{1})\cap X\\}\cup\\{(x_{2n-1},x_{2n+1})\cap
X:n\in\mathbb{N}\\}\cup\\{(x,+\infty)\cap X\\},$

i.e. the family

$\\{(-\infty,x)\cap X\\}\cup\\{(x_{2n-1},x_{2n+1})\cap
X:n\in\mathbb{N}\\}\cup\\{(x_{1},+\infty)\cap X\\},$

depending again on whether the sequence $\\{x_{n}\\}_{n=1}^{+\infty}$ is
increasing or decreasing, is a disjoint open cover of $X$ that has no finite
subcover.

Now let us prove the converse. If $X$ is not a GCC space then there is a
sequence $\\{U_{n}\\}_{n=1}^{+\infty}$ of disjoint non-empty open subsets of
$X$ such that $X=\bigcup_{n=1}^{+\infty}U_{n}$. Let
$\\{a_{n}\\}_{n=1}^{+\infty}$ be a sequence in $X$ such that $a_{n}\in U_{n}$
for every $n\in\mathbb{N}$ and let $\\{a_{n_{k}}\\}_{k=1}^{+\infty}$ be a
monotone subsequence of the sequence $\\{a_{n}\\}_{n=1}^{+\infty}$. We will
assume that $\\{a_{n_{k}}\\}_{k=1}^{+\infty}$ is increasing. For every
$k\in\mathbb{N}$ we have $(a_{n_{k}},a_{n_{k+1}})\setminus X\neq\emptyset$, so
there is an increasing sequence $\\{b_{n}\\}_{n=1}^{+\infty}$ such that
$b_{2k}=a_{n_{k}}$ and $b_{2k+1}\in(a_{n_{k}},a_{n_{k+1}})\setminus
X\subseteq\mathbb{R}\setminus X$ for every $k\in\mathbb{N}$. By the assumption
of the theorem we have $\lim_{n\to+\infty}b_{n}\in X$, but that is impossible
because every set $U_{n}$ contains at most one of the members of the sequence
$\\{b_{n}\\}_{n=1}^{+\infty}$. We analogously reach a contradiction if
$\\{a_{n_{k}}\\}_{k=1}^{+\infty}$ is decreasing. ∎

######  Corollary 1.

If $X$ is a GCC subspace of $\mathbb{R}$ and if $(C_{i})_{i\in I}$ is a family
of components of the space $X$, then $\overline{X}=\bigcup_{i\in
I}\overline{C_{i}}$.

###### Proof.

Let $x\in\overline{X}\setminus X$ and let $\\{x_{n}\\}_{n=1}^{+\infty}$ be a
strictly increasing sequence in $X$ such that $\lim_{n\to+\infty}x_{n}=x$.
Suppose that $x\notin\bigcup_{i\in I}\overline{C_{i}}$. There exists a
subsequence $\\{x_{n_{k}}\\}_{k=1}^{+\infty}$ of $\\{x_{n}\\}_{n=1}^{+\infty}$
and a sequence $\\{y_{k}\\}_{k=1}^{+\infty}$ such that

$y_{k}\in(x_{n_{k}},x_{n_{k+1}})\setminus X$

for every $k\in\mathbb{N}$. Now, the sequence $\\{z_{k}\\}_{k=1}^{+\infty}$,
defined with

$z_{2k+1}=x_{n_{k}}\ \text{and}\ z_{2k}=y_{k},$

contradicts theorem 3. ∎

Similar argument can be used to prove the following corollaries.

######  Corollary 2.

If $X$ is a GCC subspace of $\mathbb{R}$ then $\mathbb{R}\setminus X$ is
locally connected. If $X$ is bounded subspace of $\mathbb{R}$ such that
$X\setminus\mathbb{R}$ is locally connected then $X$ is a GCC space.

######  Corollary 3.

Let $X$ be a GCC space and let $f$ be a continuous mapping form $X$ to
$\mathbb{R}$. If $a=\sup f(X)(\inf f(X))\in\mathbb{R}$, then $a\in f(X)$ or
$(a-\epsilon,a)((a,a+\epsilon))\subseteq f(X)$ for some $\epsilon>0$. If $f$
is unbounded, then $f(X)$ contains an unbounded interval.

We will now prove that for every GCC subspace $X$ of $\mathbb{R}$ is
continuous image of the product $C\times\mathbb{R}$, where $C$ is Cantor set.
We begin with the following lemma.

######  Lemma 1.

Let $X$ be a subspace of $\mathbb{R}$ and let $(C_{i})_{i\in I}$ be the family
of all connected components of $X$. Let $\varphi_{X}:(C_{i})_{i\in I}\to
2^{\mathbb{R}}$ be a mapping that satisfies the following two conditions:
1) $C_{i}\setminus C_{i}^{\circ}\subseteq\varphi_{X}(C_{i})\subseteq C_{i}$;
2) $C_{i}^{\circ}\neq\emptyset\Rightarrow\varphi_{X}(C_{i})\cap C_{i}^{\circ}$
is a singleton.
Then $X$ is a GCC space if and only if $\bigcup_{i\in I}\varphi_{X}(C_{i})$ is
compact subspace of $\mathbb{R}$.

###### Proof.

Let $Y=\bigcup_{i\in I}\varphi_{X}(C_{i})$. If $Y$ is compact then $X$ is a
GCC space because $Y$ intersects every component of $X$, hence $X$ is a GCC
space.

Let $X$ be a GCC space. To prove that $Y$ is compact it is enough to prove
that every strictly monotone sequence in $Y$ converges. Let
$\\{y_{n}\\}_{n=1}^{+\infty}$ be a strictly increasing sequence in $Y$. Let
$x_{2n}=y_{4n}$ for every $n\in\mathbb{N}$. We have that $y_{4n}\neq
y_{4n+4}$. Since every $\varphi_{X}(C_{i})$ contains at most 3 points we have
that

$C_{x_{2n}}=C_{y_{4n}}\neq C_{y_{4n+4}}=C_{x_{2n+2}},$

where $C_{x}$ is a connected component of $x$ in $X$. Since elements $x_{2n}$
and $x_{2n+2}$ belong to different connected components of $X$ we have that
$(x_{2n},x_{2n+2})\setminus X\neq\emptyset$. For every $n\in\mathbb{N}$ we
pick $x_{2n+1}\in(x_{2n},x_{2n+2})\setminus X$. Now, by the theorem 3, we have
that $\\{x_{n}\\}_{n=1}^{+\infty}$ converges to a point $x\in X$ and

$\lim y_{n}=\lim y_{4n}=\lim x_{2n}=x.$

∎

######  Theorem 4.

Every GCC subspace of $\mathbb{R}$ is a continuous image of the space
$C\times\mathbb{R}$, where $C$ is the Cantor set.

###### Proof.

Let $X$ be a GCC subspace of $\mathbb{R}$ and let $(C_{i})_{i\in I}$ be the
family of connected components of $X$. Let $\varphi_{X}$ be a mapping as in
the previous lemma and let $A=\bigcup_{i\in I}\varphi_{X}(C_{i})$. The space
$A$ is a compact subspace of $\mathbb{R}$ by the previous lemma. Since every
compact subspace of $\mathbb{R}$ is a continuous image of the Cantor space it
suffices to prove that $X$ is a continuous image of the space
$A\times\mathbb{R}$. Let $f$ be a mapping from $A\times\mathbb{R}$ to $X$
defined by the following conditions:

(1) if $x\in\varphi_{X}(C_{i})\cap C_{i}^{\circ}$ for some $i\in I$, then
$f_{|\\{x\\}\times\mathbb{R}}(x,y)=g(y)$, for some continuous mapping $g$ from
$\mathbb{R}$ onto $C_{i}$;

(2) if $x\in C_{i}\setminus C_{i}^{\circ}$ for some $i\in I$, then
$f_{|\\{x\\}\times\mathbb{R}}(x,y)=x$ for every $y\in\mathbb{R}$.

It is easy to see that $f$ is a surjection.

Let us prove that $f$ is continuous. Let $(x,y)\in A\times\mathbb{R}$.

If $x\in\varphi_{X}(C_{i})\cap C_{i}^{\circ}$ for some $i\in I$, then $x$ is
an isolated point in $A$ and $f$ is continuous at $(x,y)$ because
$\\{x\\}\times\mathbb{R}$ is an open neighborhood of $(x,y)$ in
$A\times\mathbb{R}$ and $f_{|\\{x\\}\times\mathbb{R}}$ is continuous.

Now let $x\in C_{i_{0}}\setminus C_{i_{0}}^{\circ}$ for some $i_{0}\in I$.
Then we have that $f(x,y)=x$. Let $\epsilon>0$.

Let

$J=\\{j\in I\setminus\\{i_{0}\\}:x-\epsilon\in C_{j}\vee x+\epsilon\in
C_{j}\\}.$

Let

$U=(x-\epsilon,x+\epsilon)\setminus\left((\varphi_{X}(C_{i_{0}})\cap
C^{\circ}_{i_{0}})\cup\bigcup_{j\in J}C_{j}\right).$

We note that the set $U$ is open since $|J|\leq 2$, and we also note that
$x\in U$. Further, for every component $C_{i}\ (i\neq i_{0})$ of $X$ we have

$C_{i}\cap U\neq\emptyset\Rightarrow C_{i}\subseteq U,$

and we have that

$f((U\cap\varphi_{X}(C_{i_{0}}))\times\mathbb{R})\subseteq U.$

Let $V=U\cap A$. We have that $V\times\mathbb{R}$ is an open neighborhood of
$(x,y)$ in $A\times\mathbb{R}$ and that

$f(V\times\mathbb{R})\subseteq U\subseteq(x-\epsilon,x+\epsilon).$

Thus, $f$ is continuous at $(x,y)$. ∎

## References

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  * [2] J. Sterpāns, A. Šostak, Restricted compactness properties and their preservation under products, Topology Appl., 101 (3) (2000), pp. 213-229
  * [3] R. Engelking, On the double circumference of Alexandroff, Bull. Acad. Pol. Sci. Ser. Astron. Math. Phys. 16, no 8 (1968), 629-634