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Given an open set $U \subset \mathbb R ^n $, there exists an exhaustion by compact sets, i.e. a sequence of compact sets $K_i$, s.t.

$\cup _{i=0}^{\infty} K_i = U$ and $\forall i \in \mathbb N : K_i \subset K_{i+1} ^{\circ}$

We can imagine that different exhaustions by compact sets 'propagate' at different speeds for the various parts $U$.

Let us call an exhaustion $(K_i)_i$ 'stronger' than another exhaustions $(L_i)_i$, whenever we have

$\forall i \in \mathbb N \exists j \in \mathbb N : L_i \subset K_j$.

We call two exhaustions equivalent, if each one is stronger than the other - i.e. for each compact set in the first, we have a compact superset in the second, and vice versa.

Question: Are all exhaustions of $U$ equivalent?

Purpose: You encounter various settings, where you define a certain structure by such an exhaustion. If the above question has a positive answer, this would spare from wondering whether your construction is independent of such an exhaustion.

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What is K_{i+1}^{\circ}? Does this denote the interior? –  Qiaochu Yuan Nov 16 '10 at 22:00
    
Yes - I did not find any better style. –  shuhalo Nov 16 '10 at 22:09
    
@user3557: I think Qiaochu just wanted to confirm that notation with you as it looks very small. –  anonymous Nov 16 '10 at 22:13

2 Answers 2

up vote 6 down vote accepted

Yes. Let $\{L_i\}_{i=1}^\infty$, $\{K_i\}_{i=1}^\infty$ be two exhaustions. It follows that $\bigcup_{i=1}^\infty K_i^\circ = U$, so for each $n$, $\{K_i^\circ\}_{i=1}^\infty$ is an open cover of $L_n$. By compactness, it has a finite subcover $\{K_{i_1}^\circ, \dots, K_{i_l}^\circ\}$. But if $m = \operatorname{max}(i_1, \dots, i_l)$, we have $K_{i_j}^\circ \subset K_{i_j + 1} \subset K_{m+1}$ for all $j=1,\dots, l$. Thus $L_n \subset K_{m+1}$, and so $K_i$ is stronger than $L_i$. By symmetry they are equivalent.

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what properties of R^n are you using when you assert the first equality? Local compactness or something? It seems like it might fail in general. –  Qiaochu Yuan Nov 17 '10 at 2:00
    
Do you mean the claim, that the interiors of the $K_i$ are an open cover of $U$?. Let x \in U and i s.t. x \in K_i. Then x lies in the interior of K_{i+1}, too. –  shuhalo Nov 17 '10 at 10:27
    
This should be good in any topological space. Of course, some spaces may not admit an exhaustion at all. –  Nate Eldredge Nov 17 '10 at 13:48
    
@Qiaochu: He is just using that $K_i^\circ \subseteq K_i \subseteq K_{i+1}^\circ$, for each $i$ and that $U$ is covered by $\{K_i\}_{i \in \mathbb{N}}$. Local compactness is one of the properties used to guarantee the existence of exhaustions. In fact, if $X$ is a locally compact, Hausdorff and second countable topological space, then each open set in $X$ has an exhaustion. I don't know if it's a characterization of spaces with exhaustions. Probably not. –  Nuno Nov 17 '10 at 20:02
    
Ah, sorry; I forgot that part of the definition of an exhaustion. –  Qiaochu Yuan Nov 17 '10 at 20:59

Ok, one version on my own, more technical.

Let $V$ be any compact set included in $U$. It remains to show $\exists i \in \mathbb N : V \subset K_i$.

Note that $\forall i : V \cap K_i$ bounded again.

Suppose, there is no index $i$ s.t. $V \subset K_i$.

Then there is a nested sequence $V_i$ of non-empty bounded sets, $V = V_0, \forall i : V_i \cap K_i = \{\}$, .

Let $x_i \in V_i$ be a sequence of points, which exists by assumption. Then this sequence is bounded and w.l.o.g it is convergent.

Hence let $x_i \rightarrow x$. By assumption, $\exists k \in \mathbb N : \exists \epsilon > 0 : B_{\epsilon}(x) \subset {K_k}^{\circ}$.

As $x_i$ convergence, we skip a leading number of elements and w.l.o.g. $\forall i : x_i \in B_{\epsilon/2}(x) \subset B_{\epsilon}(x) \subset K_k$.

After reindexing the $V_i$ according to our choices of subsequences, we obtain a contradiction.

Hence, such there is an $i$ s.t. $V \subset K_i$. As $V$ has been any compact set, we are done.

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