(X,d) be a compact metric space. For every open cover, show there exists ε > 0 such that ∀ ∈ X, B(x,ε) is contained in some member of the cover.

Question

Let (X,d) be a compact metric space. For every open cover, show there exists ε > 0 such that for every x ∈ X, B(x,ε) is contained in some member of the cover.

My attempt:

(X,d) is compact. Therefore there exists a finite subcover of X.

Any element x in X must lie in some member of the cover, say x ∈ Ui. Otherwise they would not constitute a cover.

Since Ui is open, by definition every point is interior, so there exists ε > 0 such that B(x,ε) is contained in Ui.

I haven't used the fact the subcover is finite, or the fact X is a metric space rather than just topological space, so I feel my reasoning is flawed.

Any help is greatly appreciated!

Answer 1

The problem with that proof is you have an $\epsilon$ defined for each $x$. You need to prove there is an $\epsilon$ that works for all $x$ but is independent of the value of $x$

You are give an open cover $U$ consisting of open sets $U_\alpha$. We know this has a finite subcover but we should resist assuming we know it actually work on finding it in terms of the $x \in X$.

As you point out for each $x\in X$ there is a $U_\zeta \in U$ so $x\in U_\zeta$. And there is an $\epsilon_x$ (for that $x$, not nesc for all $x$) so that $B(x,\epsilon_x) \subset U_\zeta$. Fine.

(Hi... I'm a time traveler from a the future. I'm sticking in an extra step right now. We also have $B(x,\frac {\epsilon_x}2) \subset B(x,\epsilon)\subset U_\zeta$. I'll explain why I'm doing that later.)

If we collect these neighborhoods, $B(x,\epsilon_x)$ into a collection, $\mathcal B=\{B(x,\epsilon_x)|x\in X\}$. Well.... it's pretty easy to show that $\mathcal B$ is an open cover of $X$! So $\mathcal B$ has a finite subcover.

(Hi.... Time traveler again. We also have $\mathcal C=\{B(x,\frac {\epsilon_x}2)|x\in X\}$ is also a open cover with a finite subcover.)

Which if we put it in other words.... there is a finite subset $\{x_1,........, x_n\}\subset X$ so that $\{B(x_i \epsilon_{x_i})|x_i \in \{x_1,.....,x_n\}\}\subset \mathcal B$ and $X\subset \cup_{i=1}^n B(x_i \epsilon_{x_i})$

(Hi... remember me? The time traveler? Just noting that there is also subset of $\mathcal C$ that will cover $X$ and a subset $\{w_1,......., w_m\}\subset X$ that acts as an index.)

Now there are a finite number of $\epsilon_x$ so there must exist a $E=\min\{\epsilon_{x_i}\} > 0$ (Hi.... there also must be a $E'=\min\{\frac {\epsilon_{w_j}}2\}$) and thus as every $x$ must be in some $B(x_i\epsilon_{x_i})$. So $B(x, E) \subset B(x_i\epsilon_{x_i}) \subset U_\zeta$ for some $U_\zeta$ and we are done and...

Oh SHHHHHugar!!!!!! Although $d(x,x_i) < \epsilon_{x_i}$ and $E< \epsilon_{x_i}$ that doesn't mean that for any $y\in B(x,E)$ that $y \in B(x,\epsilon_{x_i})$ as $d(y,x_i) \le d(y,x)+d(x,x_i)< E + \epsilon_{x_i} \not < \epsilon_{x_i}$. If only there were some way I could travel back in time and fix my mistake.

warping weird music.

Hi. For every $x\in X$ then $x\in B(w_j, \frac {\epsilon_{w_j}}2)$ for some $w_j$. Thus for any $y\in B(x, E')$ then $d(y, w_j) \le d(y,x)+ d(x,w_j) < E' + \frac {\epsilon_{w_j}}2 < \frac {\epsilon_{w_j}}2+\frac {\epsilon_{w_j}}2 < \epsilon_{w_j}$ and so ....

$B(x, E')\subset B(w_j, \epsilon_{w_j})\subset U_\zeta$ for some $U_\zeta\in U$. And we are done.

Answer 2

The Wiki proof linked in the comments uses the fact that a continuous function on a compact set reaches its extrema. If you want a proof from scratch and closer to what you are trying to do, here are a few hints:

$1).\ $ Let $\mathcal A$ be an open cover of $X$. For each $x\in X$ there is an open neighborhood $B_{\epsilon_x}(x)$ such that each such $B$ is contained in an element of $\mathcal A$.

$2).\ $ The $B's$ give you $\textit{another}$ open cover of $X$.

$3).\ $ Take a finite subcover of the cover from $2)$ and note that you also get a finite number of $\epsilon_x's$.

$4).\ $ Using the conclusion in $3)$, define $\delta>0$ appropriately to conclude.

Answer 3

Let $\mathcal{A}$ be an open cover of $X$. For each $x \in X$ we can find $A_x \in \mathcal{A}$ such that $x \in A_x$, and as $A_x$ is open there is some $r(x)>0$ such that $$B(x,2r(x)) \subseteq A_x\tag{1}$$

(Note that we use $2r(x)$, to take some space; openness guarantees us some $s>0$ and we just use half of it..)

Then $\{B(x, r(x)): x \in X\}$ is an open cover for $X$, so by compactness we have a finite subcover $\{B(x_1, r(x_1), \ldots, B(x_n, r(x_n)\}$.

Now define $\delta=\min_{i=1}^n r(x_i) > 0$ (a minimum of finitely many positive reals) and I claim this $\delta$ is as required: let $x \in X$, and we have to show $B(x, \delta)$ is some member of $\mathcal{A}$.

Firstly, note that $x \in B(x_i, r(x_i))$ for some $i \in \{1,\ldots, n\}$ as the subcover is a cover. So $d(x,x_i) < r(x_i)$ and if now $y \in B(x, r(x_i))$ is arbitrary, $d(y,x) < r(x_i)$ and the triangle inequality then tells us that

$$d(y,x_i) \le d(y,x)+d(x,x-i) < r(x_i) + r(x_i)=2r(x_i)\text{, so } y \in B(x_i, 2r(x_i)$$

And as $y$ was arbitrary, $B(x, r(x_i)) \subseteq B(x_i, 2r(x_i))$. Now it's obvious that $\delta \le r(x_i)$ and so

$$B(x, \delta) \subseteq B(x, r(x_i)) \subseteq B(x_i, 2r(x_i)) \subseteq A_{x_i} \in \mathcal{A}$$ and this finishes the proof.

Answer 4

You have proved that for all $x$, there exists a $\varepsilon$ so that $B(x,\varepsilon)$ is contained in some member of the cover - which does not rely on the fact that the chosen subcover is finite or on compactness. This is a much weaker statement than the one you wished to prove. Indeed, you can convince yourself that a proof has to be more intricate by considering that, even if your cover were finite, that does not mean it has the desired property without compactness; for instance, in the space $[0,1/2)\cup (1/2,1]$, the cover consisting of the two components does not satisfy the desired property. Since your proof apparently would apply to any finite cover, you can see that something must be wrong with it; more generally, we can see that taking a finite subcover of the cover we were given is probably not going to help us.

A very fast way to do this, however, is to start with your open cover $\mathscr U$. For each $x\in X$, you can define $$f(x)=\max \{\varepsilon > 0 : B(x,\varepsilon) \subseteq U\text{ for some }U\in \mathscr U\}$$ You are trying to show that $f(x)\geq \varepsilon$ for some $\varepsilon$ for all $x$. It's worth taking a moment to consider why this can be assumed to be a maximum rather than supremum and to convince yourself that this function is continuous. You can also convince yourself that this function is positive everywhere, since every $x$ has some $B(x,\varepsilon)$ contained in some $U$.

A standard way to finish this proof is to apply the extreme value theorem to $f$ and find a minimum - but there's an easier way that uses the definition of compactness directly. In particular, let $A_{n}$ be the set of $x\in X$ that satisfy $f(x)>1/n$. Clearly, every $x$ is in some $A_n$ - so the set $\{A_1,A_2,\ldots\}$ is a cover of $X$. Note also that $A_1\subseteq A_2\subseteq A_3 \subseteq \ldots$, so the union of any finite subset of this cover is an element of the cover. By compactness, this has a finite subcover - and by the prior note, this means that $A_n=X$ for some $n$. Then you have the desired statement for $\varepsilon = 1/n$.

Generally, this is a good approach to have to such questions: if you're trying to prove a quantity can be taken to be uniform across the space (i.e. to turn "$\forall\exists$" into $"\exists\forall"$), you can often proceed by looking at sets where some fixed constant suffices, and seeing if you can use compactness (or whatever other property you have) to show that one of those sets is the whole space. Note also that here we are taking a finite subcover of a cover we constructed - it would not help us to take a finite subcover of $\mathscr U$, even though it seems like a good idea.