Hurewicz's reformulation of Menger property

I’ve extracted the meat of the relevant part of Hurewicz’s original paper and recast it in modern terminology.

Let $X$ be any metric space with Hurewicz’s property, and let $\mathscr{B}$ be a base for $X$. For each $n\in\Bbb N$ let $\mathscr{U}_n=\{B\in\mathscr{B}:\operatorname{diam}B<2^{-n}\}$. Each $\mathscr{U}_n$ is an open cover of $X$, so there are finite $\mathscr{F}_n\subseteq\mathscr{U}_n$ for $n\in\Bbb N$ such that $\bigcup_{n\in\Bbb N}\mathscr{F}_n$ covers $X$. Enumerate $\bigcup_{n\in\Bbb N}\mathscr{F}_n=\{B_k:k\in\Bbb N\}$ arbitrarily; then $\langle\operatorname{diam}B_k:k\in\Bbb N\rangle\to 0$, since each $\mathscr{F}_n$ is finite, and hence $X$ has Menger’s property.

The other direction is harder and requires imposing some conditions on $X$, at least as Hurewicz did it. First we need a lemma.

Lemma. Let $\mathscr{U}$ be an open cover of a totally bounded metric space $X$, and let $r>0$. Then there is an open cover $\mathscr{V}$ of $X$ such that

  • each $V\in\mathscr{V}$ is contained in the union of a finite subset of $\mathscr{U}$, and
  • there is an $s>0$ such that $s<\operatorname{diam}V\le r$ for each $V\in\mathscr{V}$.

Proof. $X$ is separable, hence Lindelöf, so we may as well assume that $\mathscr{U}=\{U_n:n\in\Bbb N\}$. For $n\in\Bbb N$ let $W_n=\bigcup_{k\le n}U_k$. For $x\in X$ let $n(x)=\min\{k\in\Bbb N:x\in W_k\}$, and let $$V_x=W_{n(x)}\cap B\left(x,\frac{r}2\right)\;;\tag{1}$$ I claim that $\mathscr{V}=\{V_x:x\in X\}$ has the desired properties. The only one that isn’t obvious is the existence of an $s>0$ such that $\operatorname{diam}V_x>s$ for each $x\in X$.

Suppose not; then there is a sequence $\langle x_k:k\in\Bbb N\rangle$ in $X$ such that $\langle\operatorname{diam}V_{x_k}:k\in\Bbb N\rangle$ is a strictly decreasing sequence with limit $0$. By passing to a tail of a subsequence if necessary we may assume that $\operatorname{diam}V_{x_k}<\frac{r}4$ for all $n\in\Bbb N$ and that $\langle n(x_k):k\in\Bbb N\rangle$ is non-decreasing. Then $$V_{x_k}=W_{n(x_k)}\cap B\left(x_k,\frac{r}4\right)\tag{2}$$ for each $k\in\Bbb N$. Now let $k,\ell\in\Bbb N$ with $k<\ell$. Then $n(x_k)\le n(x_\ell)$, so $W_{n(x_k)}\subseteq W_{n(x_\ell)}$. If $d(x_k,x_\ell)\le\frac{r}4$, then $B\left(x_k,\frac{r}4\right)\subseteq B\left(x_\ell,\frac{r}2\right)$, and it follows from $(1)$ and $(2)$ that $V_{x_k}\subseteq V_{x_\ell}$. This is impossible, since $\operatorname{diam}V_\ell<\operatorname{diam}V_k$, so $d(x_k,x_\ell)>\frac{r}4$. Clearly $\langle x_k:k\in\Bbb N\rangle$ has no Cauchy subsequence, contradicting the total boundedness of $X$; this contradiction establishes the desired result. $\dashv$

Now let $X$ be a totally bounded metric space with Menger’s property, and let $\langle\mathscr{U}_n:n\in\Bbb N\rangle$ be a sequence of open covers of $X$. For each $n\in\Bbb N$ let $\mathscr{V}_n$ be an open cover of $X$ such that

  • each $V\in\mathscr{V}_n$ is contained in the union of a finite subset of $\mathscr{U}_n$, and
  • there is an $s_n>0$ such that $s_n<\operatorname{diam}V\le 2^{-n}$ for each $V\in\mathscr{V}_n$.

Then $\mathscr{V}=\bigcup_{n\in\Bbb N}\mathscr{V}_n$ is a base for $X$, so there is a sequence $\langle B_n:n\in\Bbb N\rangle$ in $\mathscr{V}$ such that $\langle\operatorname{diam}B_n:n\in\Bbb N\rangle\to 0$ and $\bigcup_{n\in\Bbb N}B_n=X$. Fix $n\in\Bbb N$; $\operatorname{diam}B_k<s_n$ for all sufficiently large $k\in\Bbb N$, so only finitely many of the sets $B_k$ can belong to $\mathscr{V}_n$. Each member of $\mathscr{V}_n$ is covered by a finite subset of $\mathscr{U}_n$, so there is a finite $\mathscr{F}_n\subseteq\mathscr{U}_n$ such that

$$\bigcup\{B_k:B_k\in\mathscr{V}_n\}\subseteq\bigcup\mathscr{F}_n\;,$$

and hence $\bigcup_{n\in\Bbb N}\mathscr{F}_n$ covers $X$. Thus, $X$ has Hurewicz’s property.

In short, the properties are equivalent for totally bounded metric spaces.

It’s easy to identify one nice class of metric spaces that have the properties.

Lemma. Suppose that $X=\bigcup_{n\in\Bbb N}X_n$, where each $X_n$ has Hurewicz’s property; then $X$ also has Hurewicz’s property.

Proof. Let $\langle\mathscr{U}_n:n\in\Bbb N\rangle$ be a sequence of open covers of $X$. Fix $n\in\Bbb N$; then there are finite $\mathscr{F}_k^{(n)}\subseteq\mathscr{U}_k$ for $k\ge n$ such that $\bigcup_{k\ge n}\mathscr{F}_k^{(n)}$ covers $X_n$. For $k\in\Bbb N$ let $\mathscr{F}_k=\bigcup_{n\le k}\mathscr{F}_k^{(n)}$; $\mathscr{F}_k$ is a finite subset of $\mathscr{U}_k$, and $\bigcup_{k\in\Bbb N}\mathscr{F}_k$ covers $X$. $\dashv$

Plainly every compact metric space has Hurewicz’s property, so the lemma implies that every $\sigma$-compact metric space has it.

He goes on to prove results that imply, among other things, that the space of irrationals in $(0,1)$ does not have Hurewicz’s property.


Please see Theorem 2 in the following paper:

On some properties of Hurewicz, Menger, and Rothberger,
authors Arnold Miller, David Fremlin,
Fundamenta Mathematicae (1988)
Volume: 129, Issue: 1, page 17-33
ISSN: 0016-2736

available online at
https://eudml.org/doc/211668