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Usually one employs Urysohn's Lemma to proof the above identity.

See for example here in the first answer: Show that the closure of $C_c(X)$ is $C_0(X)$.

I find that proof rather complicated compared to the following one.

Additionally in the proof in the link we need $X$ to be Hausdorff.

Now I've seen the following proof which does not use any property of $X$ and is quite simple aswell. It doesn't even have to be a Hausdorff-space. This is weird to me and I think there must be a mistake in it but I can't find it.

$f\in C_0(x)$. Define $f_\epsilon(x) = \begin{cases}0\quad &\text{if }|f(x)|< \epsilon\\ f(x)-\epsilon\frac{f(x)}{|f(x)|}\quad &\text{if }|f(x)|\geq \epsilon.\end{cases}$

$f_\epsilon \to f$ uniformly follows immidiately by distinguishing cases if $|f(x)|$ is bigger or smaller than $\epsilon$.

Additionally $f_\epsilon$ has compact support because $\{|f|\geq \epsilon\}$ is compact.

By the https://en.wikipedia.org/wiki/Pasting_lemma $f_\epsilon$ is continuous.

If this really would be correct, I wonder why this simple and more general proof isn't the one one usually sees.


Edit: As requested, more details:

1) $f_\epsilon\to f$ uniformly: Let $x\in X$. If $|f(x)|<\epsilon$, then $|f(x)-f_\epsilon(x)| = |f(x)|<\epsilon$. If $|f(x)|\geq \epsilon$ then $|f(x)-f_\epsilon(x)| = \epsilon\left|\frac{f(x)}{|f(x)|}\right| = \epsilon$, hence $\|f-f_\epsilon\|_\infty\leq \epsilon$.

2) $\{|f|\geq \epsilon\}$ is compact: Since $f\in C_0(X)$, there is compact $A\subset X$ with $|f(x)|<\epsilon$ if $x\notin A$, therefore $\{|f|\geq \epsilon\}\subseteq A$ and $\{|f|\geq \epsilon\}$ is a closed subset of a compact set and therefore compact.

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  • $\begingroup$ “follows immidiately” can you do it immediately? $\endgroup$
    – user537667
    May 31 '18 at 19:03
  • $\begingroup$ @PraphullaKoushik Yes, i edited my question. $\endgroup$
    – Lukas Betz
    May 31 '18 at 19:13
  • $\begingroup$ Yeah, the proof seems to be correct. $\endgroup$
    – freakish
    Jun 1 '18 at 11:18

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