I read the following claim in Z.Frolik's article "A generalization of realcompact spaces" on page 135.

Two subset $M$ and $N$ of a space $X$ are called completely seperated if there exists a real valued continuous function $f$ on $X$ with $f(M)\subset \{0\}$ and $f(N)\subset\{1\}$.

Claim: Let $X$ be a normal space. For every countable open covering $\mathfrak{U}$ of $X$, there exists a countable open covering $\mathfrak{B}$ of $X$ such that for every $B$ in $\mathfrak{B}$ there exists an $A$ in $\mathfrak{U}$ such that $B$ and $X-A$ are completely seperated.

I didn't show the proof of the claim.

  • $\begingroup$ What is the question? $\endgroup$ Jan 2, 2021 at 20:39
  • 1
    $\begingroup$ The existence of $\mathfrak{B}$ $\endgroup$ Jan 2, 2021 at 20:50

1 Answer 1


The claim is false it would imply that normal spaces are countably paracompact and hence that normality of $X$ would imply normality of $X\times[0,1]$. The latter is not the case, see Mary Ellen Rudin, A normal space $X$ for which $X\times I$ is not normal, Fundamenta Mathematicae, 73 (1971/72), 179-186.

To show that the property in the claim implies countable paracompactness we use Theorem 5.2.1 in Engelking's General Topology. Let $\{U_n:n\in\omega\}$ be an increasing open cover; we need to find open $O_n$ such that $\operatorname{cl}O_n\subseteq U_n$ for all $n$ and $\bigcup_nO_n=X$. Take an open cover $\{V_m:m\in\omega\}$ as in the claim; hence for every $m$ an $n$ such that $\operatorname{cl}V_m\subseteq U_n$. Now define $O_n=\bigcup\{V_m:m\le n$ and $\operatorname{cl}V_m\subseteq U_n\}$; then $\operatorname{cl}O_n\subseteq U_n$ for all $n$ and the $O_n$ form a cover (if $x\in V_m$ and $\operatorname{cl}V_m\subseteq U_n$ then $x\in O_n$).

  • $\begingroup$ Thank you for your answer. So, Theorem 11 in Frolik's article on 136 is not true . $\endgroup$ Jan 5, 2021 at 17:15
  • $\begingroup$ That I do not know, all this says is that the proof contains a gap: Rudin's space is not a counterexample to Theorem 11 as it is not almost realcompact. See P. Simon, A note on Rudin's example of Dowker space [eudml.org/doc/16467] $\endgroup$
    – KP Hart
    Jan 5, 2021 at 17:40
  • $\begingroup$ Would you expand on your answer to make it clearer what you state. (As it is, it is hard to parse the logic of your Answer). $\endgroup$
    – Wlod AA
    Jan 7, 2021 at 1:00

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