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Let $X$ be a locally compact Hausdorff space, and let $(U_\alpha)_{\alpha \in A}$ be an open cover of $X$. Show that there exist compactly supported continuous functions $f_\alpha: X \to [0,1]$ supported on $U_\alpha$ for each $\alpha \in A$ with $\sum_{\alpha \in A} f_\alpha(x)=1$ for all $x \in X$ (with only finitely many of the terms on the left-hand side non-zero for each $x$).”

I have no idea how to get started, to make sure that $\sum_{\alpha\in A}f_\alpha(X)$ is well-defined, it seems that we need to find a refinement $(V_\beta)_{\beta\in B}$ of $(U_\alpha)_{\alpha_\in A}$ such that every $x$ in $X$ belongs only finitely many $V_\beta$.

Xiang Yu
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  • I don't think this is true without second countability, see http://math.stackexchange.com/questions/1557117/is-every-locally-compact-hausdorff-space-paracompact and https://en.m.wikipedia.org/wiki/Paracompact_space, in particular "a Hausdorff space is paracompact if and only if it admits partitions of unity subordinate to any open cover." – PhoemueX Dec 15 '15 at 16:58
  • This is an exercise from tao's blog, you can see it here: https://terrytao.wordpress.com/books/an-epsilon-of-room-pages-from-year-three-of-a-mathematical-blog/. In this exercise, $f_\alpha$ is just compactly supported, it doesn't mean that $\text{supp}(f_\alpha)\subset U_\alpha$, so $(f_\alpha)_{\alpha\in A}$ is not partition of unity in the usual sense. – Xiang Yu Dec 15 '15 at 17:28

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