Prove the following statement:

If $X$ is compact and $C = \{U_\alpha : \alpha \in A\}$ is an open finite cover of $X$ then $C$ has a maximum Lebesgue number.

Is the following proof correct?

Proof: Let $E$ be the set of all Lebesgue numbers of $C$, that is for each $\epsilon \in E$, for all $x$, $B(x, \epsilon)$ is contained in some $U_\alpha$. The cover $C$ has a maximum Lebegue number if $\sup E$ exists and is contained in $E$. Since $X$ is compact and since $C$ is finite there exists some $M > 0$ such that all $U_\alpha$ are bounded by $M$ which implies $E$ is bounded above by $M$. Thus $\sup E$ exists. To show $\sup E \in E$ consider sequence $r_n \in E$ that converges to $\sup E$. For each $r_n$ there is some $U_\alpha$ such that $B(x, r_n) \subseteq U_\alpha$ Thus for all $x \in X$ the union $\bigcup_{n=1}^{\infty}B(x, r_n) = B(x, \sup E) \subseteq U_\alpha$ which makes $\sup E$ a Lebesgue number by definition. $\square$

This relies on the fact that the union of all balls less than radius $r$ around $x$ is equal to $B(x, r)$ which can be shown easily.


You were doing fine until you got to the point of showing that $\sup E\in E$. Picking the sequence $\langle r_n:n\in\Bbb Z^+\rangle$ in $E$ converging to $\sup E$ is find but then you write this:

For each $r_n$ there is some $U_\alpha$ such that $B(x, r_n) \subseteq U_\alpha$

The set $U_\alpha$ must surely depend on $r_n$ and on $x$, but these dependences aren’t evident in what you’ve written. Worse, this $x$ is nowhere defined. Is it a specific $x$? If so, it can’t justify the next assertion:

Thus for all $x \in X$ the union $\bigcup_{n=1}^{\infty}B(x, r_n) = B(x, \sup E) \subseteq U_\alpha$ ...

If not, you need to say more. I suspect that you really meant something like this:

For each $n\in\Bbb Z^+$ and each $x\in X$ there is some $U_{\alpha(x,n)}$ such that $B(x,r_n)\subseteq U_{\alpha(x,n)}$.

If so, then you’re almost in business, but there’s still one small tricky spot. What you can conclude directly at this point is that for each $x\in X$,

$$B(x,\sup E)=\bigcup_{n\ge 1}B(x,r_n)\subseteq\bigcup_{n\ge 1}U_{\alpha(x,n)}\;.$$

This isn’t quite what you want, but recall that the cover $C$ is finite. Thus, for each $x\in X$ there are an $\alpha_x\in A$ and an infinite $N_x\subseteq\Bbb Z^+$ such that $\alpha(x,n)=\alpha_x$ for each $n\in N_x$. Thus, the sequence $\langle r_n:n\in N_x\rangle$ converges to $\sup E$, and for each $x\in X$ you have

$$B(x,\sup E)=\bigcup_{n\in N_x}B(x,r_n)\subseteq U_{\alpha_x}\;,$$

and you can conclude that $\sup E\in E$.

Added: I should note that you need to know that each member of $C$ is a proper subset of $X$: $B(x,r)\subset X$ for every $x\in X$ and $r>0$, so if $U=X$ for some $U\in C$, then $B(x,r)\subseteq U$ for all $x\in X$ and $r>0$, and $\sup E$ does not exist (or is $\infty$, if you’re working in the extended reals).

  • $\begingroup$ Is it correct to say that for all $r_n$ and for all $x \in X$ there is some $U_{\alpha(x, n)} \in C$ that contains $B(x, r_n)$. The reason being that $r_n$ is a Lesbegue number. $\endgroup$ – Cairn O. Aug 30 '15 at 20:56
  • $\begingroup$ @CairnO.: That’s essentially what I wrote in the third yellow section. I prefer each to all here, however, since each pair of $x$ and $r_n$ in principle requires its own $\alpha(x,n)$; when you write all, it sounds as if there might be one $\alpha(x,n)$ that works simultaneously for all choices of $x$ and $r_n$. $\endgroup$ – Brian M. Scott Aug 30 '15 at 20:58
  • $\begingroup$ Ah, that makes sense. I'm still unsure why we can't conclude $\bigcup_{n\ge1} B(x, r_n) \subseteq U$ for some $U \in C$. Also what do you mean by the notation $\alpha_x$ is this the index for the $U$ containing all $B(x, r_n)$? $\endgroup$ – Cairn O. Aug 30 '15 at 21:10
  • $\begingroup$ @CairnO.: You can’t immediately conclude that $\bigcup_{n\ge 1}B(x,r_n)\subseteq U$ for some $U\in C$ because you don’t know that the sets $U_{\alpha(x,n)}$ are all the same member of $C$. You know that each $B(x,r_n)$ is contained in some member of $C$, but not that there is one member of $C$ containing all of them. However, there are infinitely many $n$ and only finitely many members of $C$, so infinitely many of the sets $U_{\alpha(x,n)}$ have to be the same member of $C$; I call that member $U_{\alpha_x}$. (There may not be a largest member of $C$ containing all $B(x,r_n)$.) $\endgroup$ – Brian M. Scott Aug 30 '15 at 21:14
  • $\begingroup$ Thank you so much, that makes perfect sense. The assignment that I received actually specified that the members of $C$ are strict subsets of $X$, is there a more straightforward proof that relies on that? I couldn't find a way to work it in. $\endgroup$ – Cairn O. Aug 30 '15 at 21:23

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