The following fact is trivial to see:

Let $X$ be a separable and locally compact metric space, then for each compact set $K\subset X$ there is a continuous function with compact support and such that $f|K=1$.

Indeed, $X=\bigcup \limits_{n=1}^{\infty} U_n$, where $\{U_n\}$ is a increasing sequence of open and precompact subset of $X$ (from the Lindelöf theorem). So there is an $m\in \mathbb{N}$, such that $K\subset U_m$. Now, applying Urysohn's theorem to the sets $K$ and $X \setminus U$, we find the suitable function (with support contained in $\operatorname{cl} U_m$, with is compact).

If something like that (or similar) would be true, when $X$ was a $\sigma$-compact Polish space?

  • 1
    $\begingroup$ I'm not sure I understand what you mean by "something like that (or similarly)". Given $K$ there is a continuous function with $f|K = 1$ and compact support (if and) only if $K$ is contained in a locally compact open subset $U$ of $X$ (take $U = f^{-1}((1/2,\infty))$). Of course, in general you can take the support of $f$ to be contained in as small (in a metric sense) a neighborhood of $K$ as you want, but that's obvious. $\endgroup$ – t.b. Jul 11 '11 at 21:01
  • $\begingroup$ So reffering to your comment the question may formulated like this: Is there locally compact open subset $U$ of $\sigma$-compact polish space $X$ , such that $K \subset U$? $\endgroup$ – Dawid C. Jul 11 '11 at 21:40
  • $\begingroup$ Again, I'm not sure what you want to know. If you want this to hold for all $K$ then your space is necessarily locally compact, as my argument shows. (Apply it to $K = \{x\}$ for each $x \in X$). $\endgroup$ – t.b. Jul 11 '11 at 21:43
  • $\begingroup$ Ok, now i understand, unfortunately i need this for all compact set $K$, thanks very much. $\endgroup$ – Dawid C. Jul 11 '11 at 21:57

For the sake of having an answer:

Let $X$ be a Hausdorff space. Let $C_{c}(X)$ be the space of continuous functions $f$ with compact support. Put $$Y = \bigcup_{f \in C_{c}(X)} \{|f| \gt 0\}.$$ Then $Y \subset X$ is an open locally compact subspace.

Indeed if $Y = \emptyset$ this is clear. Otherwise for each $y \in Y$ we have $|f(y)| \gt 0$ for some $f \in C_{c}(X)$. But then $U = \{|f| \geq |f(y)|/2\}$ is a compact neighborhood of $y$.

In other words, if $K \subset X$ is compact there exists a continuous function $f$ with compact support such that $f\,|_{K} = 1$ if and only if $K \subset Y$.

Since you said in a comment that you want to have such a function for all compact $K \subset X$ we must have $Y = X$ and thus $X$ must be locally compact.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.