# Dominate any positive function from a topological space by a series of compactly supported functions

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?

• You should add your definition of $\sigma$-compact. Does it include locally compact? Jul 18 '20 at 15:58
• @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 $X =\bigcup K_n$. Jul 18 '20 at 16:06
• Does compact include Hausdorff? Jul 19 '20 at 7:51
• @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. Jul 19 '20 at 10:46
• @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 $K_n \subset \text{int}(K_{n+1})$? Jul 19 '20 at 11:26

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.

• Thanks a lot, this is a great answer. Very enlightening! :) Jul 20 '20 at 10:53