# Using functions to separate a compact set from a closed set in a completely regular space

Suppose $X$ is completely regular, $K\subset X$ compact and $C\subset X$ closed such that $K\cap C$ empty.

• Prove that there exists $f_x\in C(X,[0,1])$ such that $f_x(x)=1$ and $f_x=1$ on $C$ for $x\in K$ (this follows from my point of view from the $T_{3,5}$ property)
• Construct with help of the $f_x$ a cover of $K$ (can we use the inverse of $f_x$ vor every $x$ in $1$?)
• Use compactness to find a $g\in C(X,[0,\infty))$ such that $g=0$ on $C$ and $g>\frac{1}{2}$ on $K$ (can we use Urysohn?)
• Modify $g$ to get an $f\in C(X,[0,1])$ such that $f=1$ on $K$ and $f=0$ on $C$

I think it is not so difficult but I don't see it. Can someone help me? Thanks

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In the first bullet point you want $f_x=0$ on $C$. –  Brian M. Scott Jan 30 '13 at 21:13

Yes, you're right, the first bullet point follows from $T_{3.5}$ applied to the point $x \notin C$ and the closed set $C$. There is a typo: you want $f_x(x) = 1$ and $f_x = 0$ on $C$ (not $f_x = 1$ on $C$).

The sets $U_x = \{y \in X : f_x(y) \gt 1/2\}$, $x \in K$ form an open cover of $K$.

Take a finite subcover $U_{x_1},\dots,U_{x_n}$ and set $g = f_{x_1} + \dots + f_{x_n}$. Then $g \gt 1/2$ on $K$ and $g = 0$ on $C$.

Now set $f(x) = 2\min\{g(x),1/2\}$.

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The first remark is true: the existence of the $f_x$ follows from applying complete regularity to $x$ and $C$, which can be done as all $x \in K$ are not in $C$. (nitpick: we also use the axiom of choice, to simultaneously choose such $f_x$, but topology without choice gets hairy pretty fast.)

Use $U_x = (f_x)^{-1}[ [0, \frac{1}{2} ) ]$ as open set that contains $x$.

There are finitely many $U_{x_1},...,U_{x_n}$ that also cover $K$.

Use $g' = \min_{i=1 \ldots n} f_{x_i}$, which is continuous as a minimum of finitely many continuous functions (and the minimum is well defined as we map into $[0,1]$).

All $f_x$ are constantly 1 on $C$, and hence so is $g'$ and if $x \in K$, $x$ is in some $U_{x_i}$, so $g'(x) \le f_{x_i}(x) < \frac{1}{2}$, so $g = 1 - g'$ is as required for the third point.

Now try the last modification yourself...

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