A. MISHCHENKO

ON SPACES WITH A POINT-COUNTABLE BASE*

(Presented by Academician P. S. Aleksandrov on 26 I 1962)

In the present note I prove (§ 1) that every bicompactum with a point-countable base is metrizable and that, on the other hand, there exist: 1) a nonmetrizable paracompactum (i.e., a paracompact Hausdorff space) with a point-countable base, 2) a Hausdorff (nonregular) hereditarily finally compact space with a point-countable base.

§ 1. Theorem. Let \(R\) be a bicompact \(T_1\)-space and \(B\) a base such that for every point \(x \in R\) the set \(B_x\) of all elements of the base \(B\) containing the point \(x\) has cardinality \(\leq a\), where \(a\) is some cardinal number. Then \(\operatorname{card} B \leq a\).

Proof. We shall call a covering \(\omega\) of the space \(R\) minimal if no proper part \(\omega' \subset \omega\) covers \(R\). It is easy to see that every covering \(\omega\) of a bicompact space contains a minimal subcovering. Let \(V\) be the set of all possible minimal coverings of the bicompact space \(R\) composed of elements of the base \(B\). Since every minimal covering is finite, the cardinality of the set \(V\) does not exceed the cardinality of \(B\). But the cardinality of the set \(V\) cannot be less than the cardinality of \(B\): this follows from the fact that for any \(\sigma \in B\) there exists a covering \(\omega \in V\) such that \(\sigma \in \omega\). To prove the latter assertion, take some point \(x \in \omega\). For each point \(y \in R \setminus x\) choose an element \(U(y)\) of the base \(B\) containing it and lying in \(R \setminus x\). Denote by \(\Omega\) the covering consisting of all the just selected \(U(y)\) and of \(\sigma\). Let \(\omega\) be a minimal covering contained in \(\Omega\). Obviously, \(\sigma \in \omega\). Thus, the cardinality of the set \(V\) is equal to the cardinality of \(B\). Suppose that \(\operatorname{card} B > a\). Then, if \(V_n\) denotes the set of all minimal coverings consisting of \(n\) distinct elements of the base, there exists an \(n_0\) such that \(\operatorname{card} V_{n_0} > a\). Let \(k \leq n_0\), and let \(\sigma_1, \ldots, \sigma_k\) be arbitrary mutually distinct elements of the base \(B\). Denote by \(S_{\sigma_1\sigma_2\ldots\sigma_k}\) the set of all those coverings \(\omega \in V_{n_0}\) for which \(\sigma_i \in \omega\) \((i = 1, 2, \ldots, k)\).

Let \(x\) be an arbitrary point of the space \(R\), and let \(B_x\) be the set of all elements of the base \(B\) containing it. Then the following equalities hold:

\[ V_{n_0} = \bigcup_{\sigma \in B_x} S_\sigma . \tag{1} \]

If \(\sigma_1, \ldots, \sigma_k\) are any (pairwise distinct) elements of the base \(B\) not containing the point \(x\), then

\[ S_{\sigma_1\ldots\sigma_k} = \bigcup_{\sigma \in B_x} S_{\sigma_1\ldots\sigma_k\sigma}. \tag{2} \]

Let \(x_1 \in R\) be an arbitrary point. There exists a \(\sigma_1 \in B_{x_1}\) such that \(\operatorname{card} S_{\sigma_1} > a\). Suppose that for \(k < n_0\) such (pairwise distinct)

* A system \(S\) of sets \(A_\alpha\) is called point-countable if each \(x \in \bigcup_\alpha A_\alpha\) belongs to no more than a countable number of sets \(A_\alpha \in S\).

elements $\sigma_1,\ldots,\sigma_k$ of the base $B$, such that $\operatorname{card} S_{\sigma_1\ldots\sigma_k}>a$. Since $k<n_0$, there exists a point $x_{k+1}$ not contained in any of the sets $\sigma_1,\ldots,\sigma_k$. But then, by virtue of equality (2), there also exists a $\sigma_{k+1}\in Bx_{k+1}$ such that $\operatorname{card} S_{\sigma_1\ldots\sigma_k\sigma_{k+1}}>a$. Moreover, obviously, $\sigma_{k+1}$ does not coincide with any of $\sigma_1,\ldots,\sigma_k$. As a result, for all $k<n_0$ the constructed sets $S_{\sigma_1\ldots\sigma_k}$ have cardinality greater than $a$. In particular, the cardinality of the set $S_{\sigma_1\ldots\sigma_{n_0}}$ exceeds the cardinal number $a$. But $S_{\sigma_1\ldots\sigma_{n_0}}\subset V_{n_0}$, and therefore there exists only one cover $\omega\in S_{\sigma_1\ldots\sigma_{n_0}}$, namely $\omega=\{\sigma_1,\sigma_2,\ldots,\sigma_{n_0}\}$. The contradiction obtained proves the theorem.

§ 2. In this section a Hausdorff strongly paracompact nonmetrizable space with a point-countable base is constructed.

Let $\alpha$ be an ordinal number. As usual, denote by $W(\alpha)$ the set of all ordinal numbers $\beta<\alpha$. Let $N$ be the set of all natural numbers. Denote by $X_\alpha$ the set of all mappings $x=x(\gamma)$, $\gamma<\alpha$, of the set $W(\alpha)$ into $N$ (i.e., the set of all well-ordered sequences of type $\alpha$ whose elements are natural numbers: $\{x_1,x_2,\ldots,x_\gamma,\ldots\}$; $\gamma<\alpha$, $x_\gamma\in N$). Put
\[ X=\bigcup_{1<\alpha<\omega_0+\omega_0} X_\alpha, \]
where $\omega=\omega_0$ is the first infinite ordinal number.

The length of an element $x\in X_\alpha$ will be called the ordinal number $\alpha$. Thus every $x\in X$ has length $<\omega_0+\omega_0$. Let $x\in X$, $y\in X$. We shall say that the element $x$ is an extension of the element $y$ if $\operatorname{len} x=\alpha>\beta=\operatorname{len} y$ and for every $\gamma<\beta$ we have $x(\gamma)=y(\gamma)$. Suppose the length of $x$ is equal to $\alpha$. Denote by $U_n(x)$ the set consisting of the point $x$ and of all those $y\in X$ which are extensions of the element $x$ and for which $y(\alpha)\ge n$. Then $B=\{U_n(x)\}_{n=1}^{\infty}$, $x\in X$, is a base of some topology on $X$. This follows from the fact that if $y\ne x$ and $y\in U_k(x)$, then for every $n$ we have $U_n(y)\subset U_k(x)$.

Let us establish a number of properties of the base $B$.

1) If neither of the two elements $x$ and $y$ is an extension of the other, then $U_n(x)\cap U_m(y)=\Lambda$ for all $n$ and $m$.

2) If $x$ is an extension of $y$, but $x\notin U_m(y)$, then $U_n(x)\cap U_m(y)=\Lambda$ for all $n$.

Let us show that the base $B$ is point-countable. If $x\in U_k(y)$, then $x$ is an extension of $y$. The set of all such $y$ that $x$ extends $y$ is at most countable. Hence the set of all such $U_k(y)$ for which $x\in U_k(y)$ is at most countable.

Let us show that into any cover $\omega$ one can inscribe a disjoint cover. For each point $x\in X_2$ there is a $U_n(x)$ lying in some element of the cover $\omega$. The set of such neighborhoods, as $x$ ranges over the set $X_2$, will be denoted by $V_2$. Then $V_2$ covers $X_2$ and is a disjoint system (for no two points $x,y\in X_2$ extend one another). Suppose that for every $\alpha<\beta<\omega_0+\omega_0$ there has been constructed such a disjoint system of neighborhoods $V_\alpha$ of points of the set $\bigcup_{\gamma\le\alpha}X_\gamma$, that:

1) $V_\alpha$ covers the set $\displaystyle \bigcup_{\gamma\le\alpha}X_\gamma$; 2) if $\alpha<\alpha'<\beta$, then $V_\alpha\subset V_{\alpha'}$.

Consider the set
\[ G=\bigcup\{\,U_k(x):U_k(x)\in \bigcup_{\alpha<\beta}V_\alpha\,\}. \]

For each point $x\in X_\beta\setminus G$ there is a $U_k(x)$ lying in some element of the cover $\omega$. Let $y\in X$ be an arbitrary point such that $U_k(y)\in\bigcup_{\alpha<\beta}V_\alpha$. Then $\operatorname{len} x>\operatorname{len} y$. Hence it follows that $U_k(x)\cap G=\Lambda$.

Since for any points $x,x'\in X_\beta\setminus G$ we have $\operatorname{len} x=\operatorname{len} x'$, it follows that $U_k(x)\cap U_s(x')=\Lambda$.

Denote then by \(V_\beta\) the sum of the set \(\bigcup_{\alpha<\beta} V_\alpha\) and of all the neighborhoods \(U_k(x)\) just considered, where \(x\) runs through the whole set \(X_\beta \setminus G\). Then the system \(V_\beta\) is disjoint and covers \(\bigcup_{\alpha\leqslant\beta} X_\alpha\), and moreover \(V_\beta \supset V_\alpha\) for every \(\alpha<\beta\). We have constructed \(V_\beta\) for every \(\beta<\omega_0+\omega_0\). The system \(\bigcup_{\alpha<\omega_0+\omega_0} V_\alpha\) is disjoint; it covers \(X\) and is inscribed in the cover \(\omega\).

This proves that \(X\) is a strongly paracompact (Hausdorff and, consequently, normal) space.

Let us show that \(X\) is not metrizable. Suppose the contrary. Then in the space \(X\) there exists a uniform base \(B_0\) (in the sense of P. S. Aleksandrov \((^1)\)). Let \(x_2\) be an arbitrary point of \(X_2\). There exists an \(\sigma_2\in B_0\) such that \(x_2\in\sigma_2\). Then there also exists \(U_{n_2}(x_2)\subset\sigma_2\). Suppose that for all \(k\leqslant m\) such \(x_k,\sigma_k,U_{n_k}(x_k)\) have been constructed that \(x_k\in X_k,\ \sigma_k\in B_0\),

\[ U_{n_{k-1}}(x_{k-1}) \supset \sigma_k \supset U_{n_k}(x_k). \]

In particular, we have

\[ U_{n_{m-1}}(x_{m-1}) \supset \sigma_m \supset U_{n_m}(x_m). \]

Choose some point \(x_{m+1}\in X_{m+1},\ x_{m+1}\in U_{n_m}(x_m)\). There exist such \(\sigma_{m+1}\in B_0\) and \(U_{n_{m+1}}(x_{m+1})\) that

\[ U_{n_m}(x_m) \supset \sigma_{m+1} \supset U_{n_{m+1}}(x_{m+1}). \]

Thus \(x_k,\sigma_k,U_{n_k}(x_k)\) are constructed by induction for every natural number \(k\). Obviously, \(x_k\) is a continuation of \(x_{k-1}\); but then there also exists a point \(x\in X_{\omega_0}\) that is a continuation of \(x_k\) for every \(k\). This point \(x\), for every \(k\), is contained in \(U_{n_k}(x_k)\). Hence it follows that every neighborhood \(U_n(x)\) of the point \(x\) lies in \(U_{n_k}(x_k)\). Since \(\sigma_k\supset U_{n_k}(x_k)\), we have \(\sigma_k\supset U_n(x)\) for all numbers \(k,n\). Thus, contrary to the assumption of uniformity of the base \(B_0\), we have an infinite set of elements of this base \(\{\sigma_k\}\), containing the point \(x\) and not forming a base at this point. The contradiction obtained proves the nonmetrizability of the space \(X\).

§ 3. In this section we construct a hereditarily finally compact Hausdorff (but nonregular) space \(X\) with a point-countable base. The space has no countable base. Represent the interval \([0,1]\) as the sum of a disjoint system, of cardinality \(\aleph_1\), of nonempty everywhere dense sets \(X_\alpha\):

\[ [0,1]=\bigcup_{\alpha<\omega_1} X_\alpha,\quad X_\alpha\cap X_\beta=\Lambda,\quad \text{where } \alpha\ne\beta. \]

Let \(p_1,p_2\) be arbitrary rational numbers, \(p_1<p_2\). By \((p_1,p_2)\) we denote the interval of the number line with endpoints \(p_1\) and \(p_2\). Let \(x\) be an arbitrary point of the interval \([0,1]\), \(x\in X_\alpha\). Define the neighborhood \(U_{p_1p_2}(x)\) as the set

\[ U_{p_1p_2}(x)=(p_1,p_2)\cap\bigcup_{\beta\geqslant\alpha} X_\beta. \]

\(1^\circ\). The neighborhood space \(R\) obtained in this way is Hausdorff. Indeed, for any two distinct points \(x\in R,\ y\in R\), one can find disjoint rational intervals \((p'_1,p'_2), (p''_1,p''_2)\), containing, respectively, the points \(x\) and \(y\). Then, all the more,

\[ U_{p'_1p'_2}(x)\cap U_{p''_1p''_2}(y)=\Lambda. \]

\(2^\circ\). The space \(R\) is hereditarily finally compact. Let
\(\omega=\{\sigma_\gamma\}\) be a system of open sets. Then for every \(x\in\tilde{\omega}\) (where \(\tilde{\omega}\) denotes \(\bigcup_\gamma \sigma_\gamma\)) there is some \(U_{p_1p_2}(x)\subset \sigma_\gamma\in\omega\).

The set of all pairs \((p_1,p_2)\) is countable; therefore the set of those pairs for which there exists an \(x\in R\) such that \(U_{p_1p_2}(x)\subset \sigma_\gamma\in\omega\) is at most countable. Denote this set by \(V=\{\gamma_i\}\), \(i=1,2,\ldots\), where \(\gamma_i=(p_1^{(i)},p_2^{(i)})\). For each \(i\) there exists an \(x_i\) such that \(U_{\gamma_i}(x_i)\subset \sigma_\gamma\in\omega\). Let \(x_i\in X_{\alpha_i}\). Take an ordinal \(\alpha_0<\omega_1\) exceeding all \(\alpha_i\).

Then

\[ \bigcup_{i=1}^{\infty} U_{\gamma_i}(x_i)\supset \bigcup_{\beta\ge \alpha_0} X_\beta\cap \tilde{\omega}. \]

Since \(X_\beta\cap\tilde{\omega}\) is finally compact in the usual topology of the interval \([0,1]\), for every \(\beta<\alpha_0\) there exists a countable system of intervals \(\{\sigma_i^\beta\}\) covering \(X_\beta\cap\tilde{\omega}\) and inscribed in the cover \(\omega_\beta\) generated by the system \(\omega\) on \(X_\beta\cap\tilde{\omega}\). Since the set of all \(\beta<\alpha_0\) is at most countable, the set of all \(\{\sigma_i^\beta\}\) is also at most countable.

Denote by \(\sigma_i^{*\beta}\) the interval of the number line cut out from \(X_\beta\) by the set \(\sigma_i^\beta\). Put \(\sigma_i^{**\beta}=\sigma_i^{*\beta}\cap\bigcup_{\gamma\ge\beta}X_\gamma\). Then the countable system of open sets in \(R\), \(\{\sigma_i^{**\beta}\}\cup\{U_{\gamma_i}(x_i)\}\), covers all of \(\tilde{\omega}\) and is inscribed in the system \(\omega\).

\(3^\circ\). The base \(\{U_{p_1p_2}(x)\}\) is point-countable. If \(y\in U_{p_1p_2}(x)\), and \(y\in X_\alpha,\ X\in X_\beta\), then \(\beta\le\alpha\). But the set of all \(\beta\le\alpha\) is at most countable; consequently, the set of all \(U_{p_1p_2}(x)\ni y\) is at most countable.

\(4^\circ\). \(R\) has no countable base. Let \(B\) be any base of the space \(R\). We shall show by contradiction that \(B\) is not a countable base, \(B=\{\sigma_i\}\). Let \(x_i\in\sigma_i,\ x_i\in X_{\alpha_i}\), and let \(\alpha_0\) exceed all \(\alpha_i\). Then, if \(x_0\in X_{\alpha_0}\), there exists no \(\sigma_i\) contained in \(U_{p_1p_2}(x_0)\), and the contradiction we need is obtained.

\(5^\circ\). The space \(R\) is nonregular. Let \(X\subset R\) be the set of all rational numbers. There exists an \(\alpha\) such that \(X\subset \bigcup_{\beta<\alpha}X_\beta=A\). The set \(A\) is closed and cannot be separated by neighborhoods from any point \(x\in R\setminus A\).

Received
26 I 1962

REFERENCES

1. P. S. Aleksandrov, Bull. Acad. Polon. Sci., Ser. Math., 8, No. 3, 135 (1960).