4

A space is Lindelöf if every open cover admits a countable subcover.

A space is Menger if given a countable sequence of open covers $\mathcal U_n$ for $n<\omega$ there exist finite subcollections $\mathcal F_n\subseteq\mathcal U_n$ such that $\bigcup\{\mathcal F_n:n<\omega\}$ is a cover.

Is every Lindelöf space Menger?

(The answer to this question is asserted in the comments of this MO question - I'm posting here as I don't believe it's been answered on the StackExchange network directly.)

RobPratt
  • 50,938
Steven Clontz
  • 8,773

1 Answers1

3

No. Consider the Baire space $\omega^\omega$, the product of countably-many copies of $\omega$ with the discrete topology. This space is second-countable and therefore Lindelöf.

Let $\mathcal U_n=\{[t]:t\in\omega^{n}\}$, where $[t]=\{f\in\omega^\omega:f\text{ extends }t\}$. Then given any choices $\mathcal F_n\subseteq\mathcal U_n$ for $n<\omega$, we will find $f\in\omega^\omega$ that's not covered. If $f(0),\dots,f(n-1)$ have been chosen such that $[\langle f(0),\dots,f(i)\rangle]\not\in\mathcal F_i$ for each $i < n$, we may also choose $f(n)<\omega$ such that $[\langle f(0),\dots,f(n)\rangle]\not\in\mathcal F_{n}$.

Now consider this $f$. If $f$ extended $t$ for some $[t]\in\mathcal F_n$, this would violate the choice made for $f(n)$. Thus $f\not\in [t]$ for every $[t]\in\mathcal F_n$, so $f$ is not covered. Thus the space is not Menger.

Steven Clontz
  • 8,773
  • 1
    Having trouble with the induction step. $\mathcal U_n$ is defined as ${[t]:t\in\omega^{\leq n}}$, but I think also allowing $t$ partial function of any length less than $n$ will break the argument. Or maybe I am just confused by the notation $\omega^{\leq n}$. What is it exactly? – PatrickR Mar 16 '25 at 22:20
  • 1
    Thanks, it should be $\omega^n$ – Steven Clontz Mar 18 '25 at 14:13