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If $X$ is a $\sigma$-compact, locally compact Hausdorff topological space and $\varphi \colon X \to [0, \infty]$ is some function, then is it always true that there exists a sequence of non-negative, continuous, compactly supported functions $(\varphi_k)_{k \in \mathbb{N}} \subset \mathcal{C}_c(X,[0,\infty))$ such that $$\varphi \leq \sum\limits_{k \in \mathbb{N}}\varphi_k$$

I know that it would be enough, if we were able to find a sequence of ascending compact subsets $K_n$ of $X$ with $K_n \subset \text{int}(K_{n+1})$ and $\bigcup K_n =X$, but how would we go about constructing such a sequence under the given assumptions?

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  • $\begingroup$ You should add your definition of $\sigma$-compact. Does it include locally compact? $\endgroup$
    – Paul Frost
    Jul 18 '20 at 15:58
  • $\begingroup$ @Paul Frost In my lecture notes it is defined without assuming the space to be locally compact, i.e. a space is defined to be $\sigma$-compact if there exists a sequence of compact subsets $K_n$ so that $X =\bigcup K_n$. $\endgroup$
    – h3fr43nd
    Jul 18 '20 at 16:06
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    $\begingroup$ Does compact include Hausdorff? $\endgroup$
    – Paul Frost
    Jul 19 '20 at 7:51
  • $\begingroup$ @Paul Frost He didn't explicitly define compactness of sets, in that case I just assumed the definition to be the standard one. I think what we would need to make this work is, that $X$ is a locally compact Hausdorff space, that admits to a sequence of ascending compact sets $K_n$ which satisfy $\bigcup K_n=X$ and $K_n \subset \text{int}(K_{n+1})$. We would then be able to apply a version of Urysohn that actually gives us the existence of compactly supported functions $\varphi_n$ which would do the job. $\endgroup$
    – h3fr43nd
    Jul 19 '20 at 10:46
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    $\begingroup$ @Paul Frost I may overlook something completely trivial now, but how does local compactness, that is each $x \in X$ has a compact neighborhood, along with $\sigma$-compactness imply that we may find such a sequence of $K_n$ with $K_n \subset \text{int}(K_{n+1})$? $\endgroup$
    – h3fr43nd
    Jul 19 '20 at 11:26
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A space $X$ is $\sigma$-compact if it can be written as the countable union of compact subsets. Therefore one can find a sequence of compact subsets $K_n$ of $X$ with $K_n \subset K_{n+1}$ and $\bigcup K_n =X$. However, we do not necessarily find a sequence of compact subsets $K_n$ of $X$ with $K_n \subset \text{int}(K_{n+1})$ and $\bigcup K_n =X$. Call such a sequence strongly ascending. Clearly, the existence of a strongly ascending sequence of compact subsets implies $\sigma$-compactness.

Note that there is a subtle point concerning the notion of compactness. Some authors understand this to include Hausdorff, other authors do not. In my answer I shall not assume it.

Let us now define a space $X$ to be locally compact if each $x \in X$ has an open neighborhood whose closure is compact and to be strongly locally compact if for each $x \in X$ and each open neighborhood $U$ of $x$ there exists an open neighborhood $V$ of $x$ such that $\overline V$ is compact and $\overline V \subset U$. In my opinion the latter is the better concept, but there is no real standard. Anyway, if $X$ is Hausdorff, then both concepts agree. An example where they differ is the Alexandroff compactification $\alpha(\mathbb Q)$ of the rationals $\mathbb Q$. This space is compact, hence locally compact, but not strongly locally compact. See my answer to An example of a compact topological space which is not the continuous image of a compact Hausdorff space?.

Let us prove

Lemma. $X$ admits a strongly ascending sequence of compact subsets if and only if $X$ is $\sigma$-compact and locally compact.

Proof. (1) Let $X$ admit a strongly ascending sequence of compact subsets $K_n$. Then $X = \bigcup \text{int}K_{n+1}$, hence each $x \in X$ is contained in some $\text{int}K_{n+1}$ so that $K_{n+1}$ is a compact neigborborhood of $x$. Therefore $X$ is locally compact.

(2) Let $X$ is $\sigma$-compact and locally compact. Write $X= \bigcup C_n$ with $C_n$ compact. By induction we construct a sequence of compact $K_n \subset X$ such that $$(*) \quad K_i \cup C_i \subset \text{int}(K_{i+1})$$ for all $i$. Then $(K_n)$ is strongly ascending.

For $n=1$ let $K_1 = C_1$. Assume we have found $K_1,\ldots, K_n$ such that $(*)$ is satisfied for all $i < n$. We now construct $K_{n+1}$. By local compactness each $x \in K_n \cup C_n$ has an open neighborhood $U(x)$ such that $D(x) =\overline{U(x)}$ is compact. Since $K_n \cup C_n$ is compact, there are finitely many $x_i \in K_n \cup C_n$ such that $K_n \cup C_n \subset V = \bigcup_{i=1}^r U(x_i)$. Let $K_{n+1} = \bigcup_{i=1}^r D(x_i)$. This is a compact set such that $K_n \cup C_n \subset V \subset K_{n+1}$. But since $V$ is open, we get $K_n \cup C_n \subset \text{int}(K_{n+1})$.

For your purposes you need a strongly ascending sequence of compact subsets. To apply Urysohn's Lemma, you must assume that the $K_n$ are closed and normal. This is automatically satisfied if $X$ is Hausdorff. If it is not, then you get fairly technical assumptions concerning a very special strongly ascending sequence.

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  • $\begingroup$ Thanks a lot, this is a great answer. Very enlightening! :) $\endgroup$
    – h3fr43nd
    Jul 20 '20 at 10:53

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