I thought of the following proof of the fact that a topological space $X$ in which any open cover has a partition of unity subordinated to it is paracompact. I would like to have feedback on wether my proof is correct or not.

Let $\{ U_\alpha:\alpha\in\mathcal{A}\}$ be an open cover of $X$ and $\{\phi_\alpha\}$ a partition of unity subordinated to it. By assumption any $p\in X$ satisfies $p\in \mathrm{supp}(\phi_{\alpha_p})\subset U_{\alpha_p}$ for some $\alpha_p\in\mathcal{A}$. Moreover, $p$ has a neighbourhood which intersects only finitely many of the supports of the $\phi_\alpha$s, say $\phi_{\alpha_1},\ldots ,\phi_{\alpha_n}$. Consider the continuous function $\phi=\phi_{\alpha_1}+\cdots +\phi_{\alpha_n}: X\rightarrow[0,1]$. Take any neighbourhood $U\neq [0,1]$ of $1$ in $[0,1]$, then $\phi^{-1}(U)$ is open, contains $p$ and is contained in $\bigcup_{i=1}^n \mathrm{supp}(\phi_i)$. Let $W_p=\Big (\phi^{-1}(U)\setminus \bigcup_{i =1,\alpha_{\,i}\ \neq \alpha_p}^n\ \mathrm{supp}(\phi_{\alpha_i}\,) \ \Big )\cap U_{\alpha_p}$ (I removed the support of all the $\phi_{\alpha_i}$s excluding $\phi_{\alpha_p}$ to ensure that the cover $\{W_p:p\in X\}$ is locally finite, and took the intersection with $U_{\alpha_p}$ to ensure that I get a refinement). Then $\{W_p:p\in X\}$ is an open cover of $X$ which is locally finite and refines $\{U_\alpha\}$.

This appears as an exercise in the book by Lee "Introduction to topological manifolds" - and, I presume, in many other books.

EDIT as pointed out by Stefan in the comments, $W_p$ as constructed above may be empty. Instead, since $\phi_{\alpha_p}(p)>0\Rightarrow p\in\mathrm{Int}(\mathrm{supp}(\phi_{\alpha_p}))\subset U_{\alpha_p}$, $p\in\phi^{-1}(U)$, define \begin{equation}W_p=\phi^{-1}(U)\cap \mathrm{Int}(\mathrm{supp}(\phi_{\alpha_p})). \end{equation} Is the corrected version of the proof correct?

  • 1
    $\begingroup$ Couldn't $W_p$ be empty? $\endgroup$ – Stefan Hamcke Aug 7 '15 at 16:10
  • $\begingroup$ @StefanHamcke I think you are right, it would work if for $\alpha_i\neq \alpha_p$ $p\notin \mathrm{supp}(\phi_{\alpha_i})$, but of course that needs not to be true. I'll see if it can be fixed. $\endgroup$ – GFR Aug 7 '15 at 16:27
  • $\begingroup$ Why don't you just take the sets $\{x\in X\mid \phi_\alpha(x)>0\}$ ? They form a locally finite open refinement cover. $\endgroup$ – Stefan Hamcke Aug 7 '15 at 17:00
  • $\begingroup$ @StefanHamcke Because I did not think of it :) Even if I have basically used the fact in the corrected proof, only to save the old one I wanted to intersect with $\phi^{-1}(U)$. Oh well, thanks for the help! $\endgroup$ – GFR Aug 7 '15 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.