'Equivalent' Exhaustion by compact sets 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.
 A: 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.
A: 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.
