Topological space M with partition of unity--->M paracompact. John Lee Problems

Question

Suppose $M$ is a topological space with the property that for every open cover $X$ of $M$, there exists a partition of unity subordinate to $X$. Show that $M$ is paracompact.

Answer 1

Given any covering $ \{U_\alpha \} $ you have the partition of unity $ \phi_\alpha $ such that $ V_\alpha = supp(\phi_\alpha ) \subset U_\alpha $ is a locally finite covering of $M$ (which is why we have the sum of $\phi_\alpha $ as a finite sum). Clearly $ \{V_\alpha \} $ is a refinement of $\{U_\alpha \}$. Thus any open cover has a locally finite refinement that covers $M$, hence $M$ is paracompact.

Answer 2

$X=\lbrace X_a \rbrace$ and let $\lbrace P_a\rbrace$ be the associated partition of unity, the index set for both is $A$. Define $A'=\lbrace a|a\in A,P_a\neq 0\rbrace$ and finally define $V_a=P_a^{-1}(0,1]$ for $a\in A'$, open in $M$. It is easy to see that $\lbrace V_a|a\in A'\rbrace$ is a locally finite refinement.