I'm working on this proof in Gamelin "Introduction to Topology" and I think I'm almost at the result, I'm just a little stuck with how to proceed.

It is this. Let $X$ be a be compact Hausdorff space and let {$U_\alpha$}$_{\alpha \in A}$ be an open cover of $X$. Show that there exist a finite number of continuous valued functions $h_1, . . ., h_2$ on $X$ with the following properties:

(a) $0\leq h_j \leq1$, $1\leq j \leq m$,

(b) $\Sigma h_{j} = 1$

(c) For each $1\leq j \leq m$, there is an index $\alpha_{j}$ s.t. the closure of the set {$x : h_{j}(x) > 0$} is contained in $U_{\alpha_{j}}$

So I know by a theorem in the book that compact Hausdorff spaces are normal. I took a point $x\in X$ and noted that the {$x$} is closed since the space is Hausdorff. By definition of open cover, $\exists$ some $U_{\alpha}$ in the open cover s.t. $x\in U_{\alpha}$. The complement, $X-U_{\alpha}$ is closed.

Further by normality, $\exists$ open sets $V,W$ s.t $V \cap W=\emptyset$ and s.t. {$x$}$\subseteq V$ and {$X-U_{\alpha}$}$\subseteq W$. $W$ is open hence $X-W$ is closed. Further $U_{\alpha}\subseteq X-W$ and $V\subseteq U_{\alpha} \subseteq X-W$. $\overline{V}$ is the smalest closed set containing V hence $\overline{V}\subseteq \overline{U_{\alpha}} \subseteq {X-W}$.

Now I want to apply Urysohn's Lemma which would say here that $\exists$ a continuous function $g$ s.t. $g$({x})=1 and $g$({$X-W$})=0. So I think I've shown properties (a) and (c), but I'm not sure where to go to show that there is only a finite number of these functions. Couldn't I just do this same process at all points $x \in X$ and find perhaps infinitely many of these functions?

Thanks for any help you can offer.

Edit: So I considered and thought about what you suggested and I think I can continue from where I left off with some of your input and some of the "Remark" from the textbook.

I think my construction would suggest that supp($g_{x}$)=$V_{x}$={$y\in X : g_{y}(x) > 0$}. For each $x\in X$ I can find another such function $g_{i}$ and another supporting set $V_{x_{i}}$. Since $X$ is compact, I can choose $x_{1}, x_{2}, ... , x_{n}$ s.t. the resulting collection {$V_{x_{i}}$} is a finite sub-cover of $X$. Hence, I have finite number of functions with these properties?

Thanks again.


It is usually better to start simplifying the hypothesis as much as you can. For example, start by taking a finite subcover of the $U_\alpha$. Also, think about the case where $U_\alpha=X$. Even though the result is trivial in this case, the procedure given by a general proof should work in this particular case. When we take functions $h_i$ but simply have $h_i(x_i)=1$ (that is, $h_i$ is $1$ at only one point), it is hard to deal with their sums, and modify them, to assure that $\sum h_i=1$.

Here's how I'd do it (which I believe is the usual way)

Denote by $\operatorname{supp}(h)$ (support of $h$) the closure of the set $\left\{x:h(x)>0\right\}$.

Since $X$ is compact, the cover $\left\{U_\alpha\right\}_{\alpha\in A}$ contains a finite subcover $U_1,\ldots,U_n$. Let's show that we can find functions $h_1,\ldots,h_n$ satisfying the properties 1. and 2. as you stated and $\operatorname{supp}(h_i)\subseteq U_i$.

Then we can proceed by induction. I'll leave the case $n=2$ for you (use Urysohn's Lemma, but with sets and not points: If $A,B$ are closed sets in $X$, then there is some continuous $f$ with $f|_A=1$, $f|_B=0$ and $0\leq f\leq 1$).

Now suppose the result is true for some $n$, and that we have a cover $U_1,\ldots,U_{n+1}$. Set $V_1=\bigcup_{i=1}^n V_i$ and $V_2=U_{n+1}$. By the case for two functions, there exists continuous functions $f,g$ with $f+g=1$, $0\leq f,g\leq 1$, $\operatorname{supp}(f)\subseteq V_1$ and $\operatorname{supp}(g)\subseteq V_2=U_{n+1}$.

Now, we apply the result for a cover of $n$ elements to $\operatorname{supp}(f)$, which is closed in $X$ and hence a compact Hausdorff with the induced topology, with the cover $U_1,\ldots,U_n$ (or more precisely the intersection of them with $\operatorname{supp}(f)$). Then we can find $f_1,\ldots,f_n$ continuous in $\operatorname{supp}(f)$, with $\sum f_i=1$, $\operatorname{supp}(f_i)\subseteq U_i$ and $0\leq f_i\leq 1$.

Set $g_i=f_if\in C(\operatorname{supp}(f))$. Then I'll leave it to you to prove the following claims:

Claim: $f|_{\partial\operatorname{supp}(f)}=0$, and hence $g_i|_{\partial\operatorname{supp}(f)}=0$, where $\partial\operatorname{supp}(f)=\operatorname{supp}(f)\cap\operatorname{closure}(X\setminus\operatorname{supp}(f))$ is the boundary of $\operatorname{supp}(f)$.

Claim: Extend $g_i$ to $X$ by setting $g_i=0$ in $X\setminus\operatorname{supp}(f)$. Then $g_i$ is continuous on $X$ (use the previous claim) and $\sum_{i=1}^n g_i=f$.

Therefore, $g_1,g_2,\ldots,g_n,g_{n+1}=g$ satisfy the desired properties.

  • $\begingroup$ Thanks for the feedback. I edited my original post. $\endgroup$ – Leif Ericson May 27 '16 at 23:20

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