Embeddings into countably compact Hausdorff spaces

TL;DR

This paper investigates which topological spaces can be embedded into countably compact Hausdorff spaces, highlighting limitations related to separation axioms and providing specific counterexamples.

Contribution

It characterizes embeddings into countably compact Hausdorff spaces and constructs an example of a space with specific separation properties that cannot embed into Urysohn spaces.

Findings

Regular separable scattered space cannot embed into Urysohn countably compact space

Such spaces can embed into Hausdorff countably compact spaces

Separation axioms influence embeddability into countably compact spaces

Abstract

In this paper we consider the problem of characterization of topological spaces that embed into countably compact Hausdorff spaces. We study the separation axioms of subspaces of countably compact Hausdorff spaces and construct an example of a regular separable scattered topological space which cannot be embedded into an Urysohn countably compact topological space but embeds into a Hausdorff countably compact space.

Equations

$$\omega$$

$$⟨U⟩=\{\mathcal{F}\in W(X):\exists F\in \mathcal{F},F\subset U\}$$

$$W_{\omega }X=\bigcup \{\overline{j_X(C)}:C\subset X,|C|\le \omega \},$$

$$V=\bigcup_{n\in\omega}\big{(}\{A_{n}\}\cup A_{n}\setminus\bigcup_{k\leq n}\{k\}\times\omega\big{)}$$

$$V=\bigcup _{n\in \omega }(\{A_n\}\cup A_n\setminus \bigcup _{k\le n}{B}_k)\text{mbox}andW=\bigcup _{n\in \omega }(\{B_n\}\cup B_n\setminus \bigcup _{k\le n}{A}_k)$$

$$\mbox{strongly discrete $\Rightarrow$ strictly discrete $\Rightarrow$ discrete}.$$

$$x_m\in I\cap {\overline{U}}_{x_0,m}\cap {\overline{U}}_{x_1,m}=\{x_0\}\cap {\overline{U}}_{x_1,m}=\varnothing ,$$

$$(2n,\lambda )\sim (2n+1,\lambda )$$

$$q(l,f)=q(l+1,f)\in {\overline{U}}_0\cap {\overline{U}}_1=\varnothing .$$

$$(2n,\lambda )\sim (2n+1,\lambda )$$

Full text

The work of the second author is supported by the Austrian Science Fund FWF (Grant I 3709 N35).

It is well-known that a topological space $X$ is homeomorphic to a subspace of a compact Hausdorff space if and only if the space $X$ is Tychonoff.

In this paper we discuss the following problem.

A topological space $X$ is

• •

compact if each open cover of $X$ has a finite subcover;

• •

$\omega$-bounded if each countable set in $X$ has compact closure in $X$;

• •

countably compact if each sequence in $X$ has an accumulation point in $X$;

• •

totally countably compact if each infinite set in $X$ contains an infinite subset with compact closure in $X$.

These properties relate as follows:

[TABLE]

More information on various generalizations of the compactness can be found in [5, 6, 7, 8, 9, 10].

In this paper we establish some properties of subspaces of countably compact Hausdorff spaces and hence find some necessary conditions of embeddability of topological spaces into Hausdorff countably compact spaces. Also, we construct an example of regular separable first-countable scattered topological space which cannot be embedded into a Urysohn countably compact topological space but embeds into a totally countably compact Hausdorff space.

First we recall some results [1] on embeddings into $\omega$-bounded spaces.

We recall [4, §3.6] that the Wallman compactification $W(X)$ of a topological space $X$ is the space of closed ultrafilters, i.e., families $\mathcal{U}$ of closed subsets of $X$ satisfying the following conditions:

• •

$\emptyset\notin\mathcal{U}$;

• •

$A\cap B\in\mathcal{U}$ for any $A,B\in\mathcal{U}$;

• •

a closed set $F\subset X$ belongs to $\mathcal{U}$ if $F\cap U\neq\emptyset$ for every $U\in\mathcal{U}$.

The Wallman compactification $W(X)$ of $X$ is endowed with the topology generated by the base consisting of the sets

[TABLE]

where $U$ runs over open subsets of $X$. By (the proof of) Theorem [4, 3.6.21], for any topological space $X$ its Wallman compactification $W(X)$ is compact.

Let $j_{X}:X\to W(X)$ be the map assigning to each point $x\in X$ the principal ultrafilter consisting of all closed sets $F\subset X$ containing the point $x$. It is easy to see that the image $j_{X}(X)$ is dense in $W(X)$. By [4, 3.6.21], for a $T_{1}$-space $X$ the map $j_{X}:X\to W(X)$ is a topological embedding.

In the Wallman compactification $W(X)$, consider the subspace

[TABLE]

which is the union of closures of countable subsets of $j_{X}(X)$ in $W(X)$. The space $W_{\omega}X$ will be called the Wallman $\omega$-compactification of $X$.

Following [1], we define a topological space $X$ to be $\overline{\omega}$-normal if for any closed separable subspace $C\subset X$ and any disjoint closed sets $A,B\subset C$ there are disjoint open sets $U,V\subset X$ such that $A\subset U$ and $B\subset V$.

The properties of the Wallman $\omega$-compactification are described in the following theorem whose proof can be found in [1].

A topological space $X$ is called

• •

first-countable at a point $x\in X$ if it has a countable neighborhood base at $x$;

• •

Fréchet-Urysohn at a point $x\in X$ if for each subset $A$ of $X$ with $x\in\bar{A}$ there exists a sequence $\{a_{n}\}_{n\in\omega}\subset A$ that converges to $x$;

• •

regular at a point $x\in X$ if any neighborhood of $x$ contains a closed neighborhood of $x$.

• •

completely regular at a point $x\in X$ if for any neighborhood $U\subset X$ of $x$ there exists a continuous function $f:X\to[0,1]$ such that $f(x)=1$ and $f(X\setminus U)\subset\{0\}$.

If for each point $x$ of a topological space $X$ there exists a countable family $\mathcal{O}$ of open neighborhoods of $x$ such that $\bigcap\mathcal{O}=\{x\}$, then we shall say that the space $X$ has countable pseudocharacter.

The following example shows that Theorem 2 cannot be generalized over Fréchet-Urysohn spaces with countable pseudocharacter.

1. (1)

$X$ is locally countable and hence has countable pseudocharacter; 2. (2)

$X$ is separable and Fréchet-Urysohn; 3. (3)

$X$ is not regular; 4. (4)

$X$ is a subspace of a totally countably compact Hausdorff space.

Let $\mathcal{F}$ be the family of closed infinite subsets of $Y$ that do not contain the point $\infty$. The definition of the topology on $Y$ implies that for every $F\in\mathcal{F}$ and $n\in\omega$ the intersection $(\{n\}\times\omega)\cap F$ is finite. By the Kuratowski-Zorn Lemma, the family $\mathcal{F}$ contains a maximal almost disjoint subfamily $\mathcal{A}\subset\mathcal{F}$. The maximality of $\mathcal{A}$ guarantees that each set $F\in\mathcal{F}$ has infinite intersection with some set $A\in\mathcal{A}$.

Consider the space $X=Y\cup\mathcal{A}$ endowed with the topology consisting of the sets $U\subset X$ such that $U\cap Y$ is open in $Y$ and for any $A\in\mathcal{A}\cap U$ the set $A\setminus U\subset\omega\times\omega$ is finite.

We claim that the space $X$ has the properties (1)–(4). The definition of the topology of $X$ implies that $X$ is separable, Hausdorff and locally countable, which implies that $X$ has countable pseudocharacter. Moreover, $X$ is first-countable at all points except for $\infty$. At the point $\infty$ the space $X$ is Fréchet-Urysohn (because its open subspace $Y$ is Fréchet-Urysohn at $\infty$).

The maximality of the maximal almost disjoint family $\mathcal{A}$ guarantees that each neighborhood $U\subset Y\subset X$ of $\infty$ has an infinite intersection with some set $A\in\mathcal{A}$, which implies that $A\in\overline{U}$ and hence $\overline{U}\not\subset Y$. This means that $X$ is not regular (at $\infty$).

In the Wallman compactification $W(X)$ of the space $X$ consider the subspace $Z:=X\cup W_{\omega}\mathcal{A}=Y\cup W_{\omega}\mathcal{A}$. We claim that the space $Z$ is Hausdorff and totally countably compact. To prove that $Z$ is Hausdorff, take two distinct ultrafilters $a,b\in Z$. If the ultrafilters $a,b$ are principal, then by the Hausdorff property of $X$, they have disjoint neighborhoods in $W(X)$ and hence in $Z$. Now assume that one of the ultrafilters $a$ or $b$ is principal and the other is not. We lose no generality assuming that $a$ is principal and $b$ is not. If $a\neq\infty$, then we can use the regularity of the space $X$ at $a$ and prove that $a$ and $b$ have disjoint neighborhoods in $W(X)\supset Z$. So, assume that $a=\infty$. It follows from $b\in Z=X\cup W_{\omega}\mathcal{A}$ that the ultrafilter $b$ contains some countable set $\{A_{n}\}_{n\in\omega}\subset\mathcal{A}$. Consider the set

[TABLE]

and observe that $V$ has finite intersection with every set $\{k\}\times\omega$, which implies that $Y\setminus V$ is a neighborhood of $\infty$. Then $\langle Y\setminus V\rangle$ and $\langle V\rangle$ are disjoint open neighborhoods of $a=\infty$ and $b$ in $W(X)$.

Finally, assume that both ultrafilters $a,b$ are not principal. Since $a,b\in W_{\omega}\mathcal{A}$ are distinct, there are disjoint countable sets $\{A_{n}\}_{n\in\omega},\{B_{n}\}_{n\in\omega}\subset\mathcal{A}$ such that $\{A_{n}\}_{n\in\omega}\in a$ and $\{B_{n}\}_{n\in\omega}\in b$. Observe that the sets

[TABLE]

are disjoint and open in $X$. Then $\langle V\rangle$ and $\langle W\rangle$ are disjoint open neighborhoods of the ultrafilters $a,b$ in $W(X)$, respectively.

To see that $Z$ is totally countably compact, take any infinite set $I\subset Z$. We should find an infinite set $J\subset I$ with compact closure $\bar{J}$ in $Z$. We lose no generality assume that $I$ is countable and $\infty\notin I$. If $J=I\cap W_{\omega}\mathcal{A}$ is infinite, then $\bar{J}$ is compact by the $\omega$-boundedness of $W_{\omega}\mathcal{A}$, see Theorem 1. If $I\cap W_{\omega}\mathcal{A}$ is finite, then $I\cap Z\setminus W_{\omega}\mathcal{A}=I\cap Y=I\cap(\omega\times\omega)$ is infinite. If for some $n\in\omega$ the set $J_{n}=I\cap(\{n\}\times\omega)$ is infinite, then $\bar{J}_{n}=J_{n}\cup\{\infty\}$ is compact by the definition of the topology of the space $Y$. If for every $n\in\omega$ the set $I\cap(\{n\}\times\omega)$ is finite, then $I\cap(\omega\times\omega)\in\mathcal{F}$ and by the maximality of the family $\mathcal{A}$, for some set $A\in\mathcal{A}$ the intersection $J=A\cap I$ is infinite, and then $\bar{J}=J\cup\{A\}$ is compact. ∎

A topological space $X$ is called weakly $\infty$-regular if for any infinite closed subset $F\subset X$ and point $x\in X\setminus F$ there exist disjoint open sets $V,U\subset X$ such that $x\in V$ and $U\cap F$ is infinite.

A subset $D$ of a topological space $X$ is called

• •

discrete if each point $x\in D$ has a neighborhood $O_{x}\subset X$ such that $D\cap O_{x}=\{x\}$;

• •

strictly discrete if each point $x\in D$ has a neighborhood $O_{x}\subset X$ such that the family $(O_{x})_{x\in D}$ is disjoint in the sense that $O_{x}\cap O_{y}=\emptyset$ for any distinct points $x,y\in D$;

• •

strongly discrete if each point $x\in D$ has a neighborhood $O_{x}\subset X$ such that the family $(O_{x})_{x\in D}$ is disjoint and locally finite in $X$.

It is clear that for every subset $D\subset X$ we have the implications

[TABLE]

$x_{n}\in V_{n}\subset U_{n-1}$;

• 2)

$y\in U_{n}\subset U_{n-1}$;

• 3)

$V_{n}\cap U_{n}=\emptyset$.

The inductive conditions imply that the sets $V_{n}$, $n\in\omega$, are pairwise disjoint, witnessing that the set $D=\{x_{n}\}_{n\in\omega}\subset I$ is strictly discrete in $X$. ∎

For closed discrete subspaces in Lindelöf subspaces, the strict discreteness of the set $D$ in Theorem 3 can be improved to the strong discreteness. Let us recall that a topological space $X$ is Lindelöf if each open cover of $X$ contains a countable subcover.

Choose any point $y_{0}\in I\setminus\{y\}$ and using the Hausdorff property of $Y$, find a disjoint open neighborhoods $V_{0}$ and $U_{0}\subset W_{0}$ of the points $y_{0}$ and $y$, respectively. Choose any point $y_{1}\in U_{0}\cap W_{1}\cap I\setminus\{y\}$ and using the Hausdorff property of $Y$ choose open disjoint neighborhoods $V_{1}\subset U_{0}$ and $U_{1}\subset U_{0}\cap W_{1}$ of the points $y_{1}$ and $y$, respectively. Proceeding by induction, we can construct a sequence $(y_{n})_{n\in\omega}$ of points of $X$ and sequences $(V_{n})_{n\in\omega}$ and $(U_{n})_{n\in\omega}$ of open sets in $Y$ such that for every $n\in\mathbb{N}$ the following conditions are satisfied:

$y_{n}\in V_{n}\subset U_{n-1}\cap W_{n}$;

• 2)

$y\in U_{n}\subset U_{n-1}\cap W_{n}$;

• 3)

$V_{n}\cap U_{n}=\emptyset$.

The inductive conditions imply that the family $(V_{n})_{n\in\omega}$ are pairwise disjoint, witnessing that the set $D=\{y_{n}\}_{n\in\omega}\subset I$ is strictly discrete in $X$. To show that $D$ is strongly discrete, it remains to show that the family $(V_{n})_{n\in\omega}$ is locally finite in $X$. Given any point $x\in X$, find $n\in\omega$ such that $x\in V_{x_{n}}$ and observe that for every $i>n$ we have $V_{i}\cap V_{x_{n}}\subset W_{i}\cap V_{x_{n}}\subset W_{n}\cap V_{x_{n}}=\emptyset$. ∎

A topological space $X$ is called $\ddot{\omega}$-regular if it for any closed discrete subset $F\subset X$ and point $x\in X\setminus F$ there exist disjoint open sets $U_{F}$ and $U_{x}$ in $X$ such that $F\subset U_{F}$ and $x\in U_{x}$.

Proceeding by induction, we can construct sequences of open sets $(V_{n})_{n\in\omega}$ and $(W_{n})_{n\in\omega}$ in $X$ such that for every $n\in\omega$ the following conditions are satisfied:

• •

$z_{n}\in V_{n}\subset W_{n-1}$;

• •

$\{z_{k}\}_{k>n}\subset W_{n}\subset W_{n-1}$;

• •

$V_{n}\cap W_{n}=\emptyset$.

These conditions imply that the family $(V_{n})_{n\in\omega}$ is disjoint, witnessing that the set $D$ is strictly discrete in $X$.

Now assume that the space $X$ is Lindelöf and let $V=\bigcup_{n\in\omega}V_{n}$. By the $\ddot{\omega}$-regularity of $X$, each point $x\in X\setminus V$ has a neighborhood $O_{x}\subset X$ whose closure $\bar{O}_{x}$ does not intersect the closed discrete subset $D$ of $X$. Since $X$ is Lindelöf, there exists a countable set $\{x_{n}\}_{n\in\omega}\subset X\setminus V$ such that $X=V\cup\bigcup_{n\in\omega}O_{x_{n}}$. For every $n\in\omega$ consider the open neighborhood $U_{n}:=V_{n}\setminus\bigcup_{k\leq n}\bar{O}_{x_{k}}$ of $z_{n}$ and observe that the family $(U_{n})_{n\in\omega}$ is disjoint and locally finite in $X$, witnessing that the set $D$ is strongly discrete in $X$. ∎

The following proposition shows that the property described in Theorem 3 holds for $\ddot{\omega}$-regular spaces.

A topological space $X$ is called superconnected [2] if for any non-empty open sets $U_{1},\dots,U_{n}$ the intersection $\overline{U}_{1}\cap\dots\cap\overline{U}_{n}$ is not empty. It is clear that a superconnected space containing more than one point is not regular. An example of a superconnected second-countable Hausdorff space can be found in [2].

Choose any two distinct points $x_{0},x_{1}\in X$ and for every $n\geq 2$ choose a point $x_{n}\in\bigcap_{k<n}\overline{U}_{x_{k},n}$. We claim that the set $I=\{x_{n}\}_{n\in\omega}$ is infinite. In the opposite case, we use the Hausdorff property and find a neighborhood $V$ of $x_{0}$ such that $\overline{V}\cap I=\{x_{0}\}$. Find $m\in\omega$ such that $U_{x_{0},m}\subset V$ and $x_{0}\notin\overline{U}_{x_{1},m}$. Observe that

[TABLE]

which is a desired contradiction showing that the set $I$ is infinite.

Next, we show that any infinite subset $D\subset I$ is not strictly discrete in $X$. To derive a contradiction, assume that $D$ is strictly discrete. Then each point $x\in D$ has a neighborhood $O_{x}\subset X$ such that the family $(O_{x})_{x\in D}$ is disjoint. Choose any point $x_{k}\in D$ and find $m\in\omega$ such that $U_{x_{k},m}\subset O_{x_{k}}$. Replacing $m$ by a larger number, we can assume that $m>k$ and $x_{m}\in D$. Since $x_{m}\in\overline{U}_{x_{k},m}\subset\overline{O}_{x_{k}}$, the intersection $O_{x_{m}}\cap O_{x_{k}}$ is not empty, which contradicts the choice of the neighborhoods $O_{x}$, $x\in D$. ∎

Next, we establish one property of subspaces of functionally Hausdorff countably compact spaces. We recall that a topological space $X$ is functionally Hausdorff if for any distinct points $x,y\in X$ there exists a continuous function $f:X\to[0,1]$ such that $f(x)=0$ and $f(x)=1$.

A subset $U$ of a topological space $X$ is called functionally open if $U=f^{-1}(V)$ for some continuous function $f:X\to\mathbb{R}$ and some open set $V\subset\mathbb{R}$.

A subset $K\subset X$ of topological space is called functionally compact if each open cover of $K$ by functionally open subsets of $X$ has a finite subcover.

Finally, we construct an example of a regular separable first-countable scattered space that embeds into a Hausdorff countably compact space but does not embed into Urysohn countably compact spaces. We recall that a topological space $X$ is Urysohn if any distinct points of $X$ have disjoint closed neighborhoods in $X$.

1. (1)

$X$ is regular, separable, and first-countable; 2. (2)

$X$ can be embedded into a Hausdorff totally countably compact space; 3. (3)

$X$ cannot be embedded into an Urysohn countably compact space.

We say that a family of function $D\subset\omega^{\omega}$ is almost disjoint if for any distinct $x,y\in D$ the intersection $x\cap y$ is finite. Here we identify a function $x\in\omega^{\omega}$ with its graph $\{(n,x(n)):n\in\omega\}$ and hence identify the set of functions $\omega^{\omega}$ with a subset of the family $[\omega\times\omega]^{\omega}$ of all infinite subsets of $\omega\times\omega$ .

By Claim 1, there exists an almost disjoint dominating subset $D\subset\omega^{\omega}\subset[\omega\times\omega]^{\omega}$. For every $n\in\omega$ consider the vertical line $\lambda_{n}=\{n\}\times\omega$ and observe that the family $L=\{\lambda_{n}\}_{n\in\omega}$ is disjoint and the family $D\cup L\subset[\omega\times\omega]^{\omega}$ is almost disjoint.

Consider the space $Y=(D\cup L)\cup(\omega\times\omega)$ endowed with the topology consisting of the sets $U\subset Y$ such that for every $y\in(D\cup L)\cap U$ the set $y\setminus U\subset\omega\times\omega$ is finite. Observe that all points in the set $\omega\times\omega$ are isolated in $Y$. Using the almost disjointness of the family $D\cup L$, it can be shown that the space $Y$ is regular, separable, locally countable, scattered and locally compact.

Choose any point $\infty\notin\omega\times Y$ and consider the space $Z=\{\infty\}\cup(\omega\times Y)$ endowed with the topology consisting of the sets $W\subset Z$ such that

• •

for every $n\in\omega$ the set $\{y\in Y:(n,y)\in W\}$ is open in $Y$, and

• •

if $\infty\in W$, then there exists $n\in\omega$ such that $\bigcup_{m\geq n}\{m\}\times Y\subset W$.

It is easy to see $Z=\{\infty\}\cup(\omega\times Y)$ is first-countable, separable, scattered and regular.

Let $\sim$ be the smallest equivalence relation on $Z$ such that

[TABLE]

for any $n\in\omega$, $\lambda\in L$ and $d\in D$.

Let $X$ be the quotient space $Z/_{\sim}$ of $Z$ by the equivalence relation $\sim$. It is easy to see that the equivalence relation $\sim$ has at most two-element equivalence classes and the quotient map $q:Z\to X$ is closed and hence perfect. Applying [4, 3.7.20], we conclude that the space $X$ is regular. It is easy to see that $X$ is separable, scattered and first-countable. It remains to show that $X$ has the properties (2), (3) of Example 2. This is proved in the following two claims.

Let us show that the number $l$ is well-defined. Indeed, by the Hausdorffness of the space $C$, there exists a neighborhood $W\subset C$ of $q(\infty)$ such that $c_{0}\not\subset\overline{W}$. By the definition of the topology of the space $Z$, there exists $m\in\omega$ such that $\bigcup_{k\geq m}\{k\}\times Y\subset q^{-1}(W)$. Then $c_{0}$ is not an accumulation point of the set $\bigcup_{k\geq m}q(\{k\}\times L)$ and hence the number $l$ is well-defined and $l<m$.

The definition of the equivalence relation $\sim$ implies that the number $l$ is odd. By the countable compactness of $C$, the infinite set $q(\{l+1\}\times L)$ has an accumulation point $c_{1}\in C$. The maximality of $l$ ensures that $c_{1}\neq c_{0}$. By the Urysohn property of $C$, the points $c_{0},c_{1}$ have open neighborhoods $U_{0},U_{1}\subset C$ with disjoint closures in $C$.

For every $i\in\{0,1\}$ consider the set $J_{i}=\{n\in\omega:q(l+i,\lambda_{n})\in U_{i}\}$, which is infinite, because $c_{i}$ is an accumulation point of the set $q(\{l+i\}\times L)=\{q(l+i,\lambda_{n}):n\in\omega\}$. For every $n\in J_{i}$ the open set $q^{-1}(U_{i})\subset Z$ contains the pair $(l+i,\lambda_{n})$. By the definition of the topology at $(l+i,\lambda_{n})$, the set $(\{l+i\}\times\lambda_{n})\setminus q^{-1}(U_{i})\subset\{l+i\}\times\{n\}\times\omega$ is finite and hence is contained in the set $\{l+i\}\times\{n\}\times[0,f_{i}(n)]$ for some number $f_{i}(n)\in\omega$. Using the dominating property of the family $D$, choose a function $f\in D$ such that $f(n)\geq f_{i}(n)$ for any $i\in\{0,1\}$ and $n\in J_{i}$. It follows that for every $i\in\{1,2\}$ the set $\{l+i\}\times f\subset\{l+i\}\times(\omega\times\omega)$ has infinite intersections with the preimage $q^{-1}(U_{i})$ and hence $\{(l+i,f)\}\in\overline{q^{-1}(U_{i})}\subset q^{-1}(\overline{U}_{i})$. Taking into account that the number $l$ is odd, we conclude that

[TABLE]

which is a desired contradiction completing the proof of the claim. ∎

Let $\tilde{Z}=\{\infty\}\cup(\omega\times\tilde{Y})$ where $\infty\notin\omega\times\tilde{Y}$. The space $\tilde{Z}$ is endowed with the topology consisting of the sets $W\subset\tilde{Z}$ such that

• •

for every $n\in\omega$ the set $\{y\in\tilde{Y}:(n,y)\in W\}$ is open in $\tilde{Y}$, and

• •

if $\infty\in W$, then there exists $n\in\omega$ such that $\bigcup_{m\geq n}\{m\}\times\tilde{Y}\subset W$.

Taking into account that the space $\tilde{Y}$ is Hausdorff and totally countably compact, we can prove that so is the the space $\tilde{Z}$.

Let $\sim$ be the smallest equivalence relation on $\tilde{Z}$ such that

[TABLE]

for any $n\in\omega$, $\lambda\in W_{\omega}L$ and $d\in W_{\omega}D$.

Let $\tilde{X}$ be the quotient space $\tilde{Z}/_{\sim}$ of $\tilde{Z}$ by the equivalence relation $\sim$. It is easy to see that the space $\tilde{X}$ is Hausdorff, totally countably compact and contains the space $X$ as a dense subspace. ∎

∎

However, we do not know the answer on the following intriguing problem: