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Suppose $X$ is a CW complex and $X_n$ is the $n$th skeleton. Suppose that for $k=0, \dots n$ we have defined partitions of unity $(\psi_\alpha^k)$ for $X_k$ subordinate to $(U_\alpha^k)$ satisfying the following property for each $\alpha \in A$ and each $k$: If $\psi_\alpha^{k-1} \equiv 0$ on an open subset $V \subset X_{k-1}$, then there is an open subset $V' \subset X_k$ containing $V$ on which $\psi_\alpha^k \equiv 0.$

This property is used in showing local finiteness of the partition of unity. If $x \in X_n,$ because $\psi_\alpha^n \equiv 0$ except when $\alpha$ is one of finitely many indices, and then the above property shows that $\psi_\alpha^{n+1} \equiv 0$ on $V'$ except when $\alpha$ is one of the same indices.

However, I cannot see how this is true. If the index set $A$ is infinite, then we cannot take intersections of the open sets. How can we ensure this?

Eric Wofsey
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nomadicmathematician
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  • I have not looked closely but your question might be answered by one of these past questions about the same proof: https://math.stackexchange.com/questions/913200/proof-of-paracompactness-of-cw-complexes-j-lee-introduction-to-topological-ma https://math.stackexchange.com/questions/291571/paracompactness-of-cw-complexes-rather-long – Eric Wofsey Jan 19 '20 at 03:34
  • Looking more closely your question is exactly the same as the first of those and the answer there appears to be correct, so I've marked this as a duplicate. – Eric Wofsey Jan 19 '20 at 03:46

1 Answers1

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The supports of the $\psi^{k-1}_{\alpha}$ form a locally finite cover of $X_{k-1}$ so for each $x\in X_{k-1}$ there is a nbhd $W$ of $x$ such that $W$ intersects only finitely many $\text{supp}\ \psi^{k-1}_{\alpha}$, say $\text{supp}\ \psi^{k-1}_{\alpha_{1}},\cdots,\text{supp}\ \psi^{k-1}_{\alpha_{1}}.$ For any index $\alpha\ \textit{not}$ among the $\{\alpha_i:1\le i\le n\},\ \psi_{\alpha}(W)=0.$ Then, by hypothesis, there is an open set $O\in \tau_{X_k}$ on which $\psi_{\alpha}^k=0$. So the same indices for which $\psi_{\alpha}^{k-1}=0$ on $W$, satisfy $\psi_{\alpha}^k=0$ on $O$. The remaining indices $\{\alpha_i:\ 1\le i\le n\}$ are then the only ones that can intersect $O$. This proves that the supports of the $\psi_{\alpha}^k$ form a locally finite family.

Matematleta
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  • Why is there an open set $O$ disjoint from every supp$\psi_{\alpha_i}^{k-1}$ on which $\psi_\alpha^{k-1}=0$? Also, you find an open set $O'$ for a fixed index $\alpha$ not among the ${\alpha_i : 1 \le i \le n}$. So how do you ensure that this $O'$ is also the open set in which $\psi_\alpha^k = 0$ for other indices $\alpha$ not among $\alpha_i, 1\le i \le n$? – nomadicmathematician Jan 19 '20 at 02:27
  • I made a mistake. Please see my edit! – Matematleta Jan 19 '20 at 02:42
  • So the problem I have is the third sentence. What we have from the hypothesis is that for each index $\alpha$, we have an open set $O_\alpha \in \tau_{X_k}$ such that $\psi_\alpha^k = 0$ on $O_\alpha$ and $W \subset O_\alpha$. I cannot see how on this $O_\alpha$, other $\alpha'$ not among ${\alpha_i : 1 \le i \le n}$ will satisfy $\psi_{\alpha'}^k = 0$. – nomadicmathematician Jan 19 '20 at 02:52
  • No, I think we have for each point $x$ the situation of my answer, right? Seems ok to me, but maybe I am missing something. I will go back to Lee's proof and see exactly what he does. I read this book but a long time ago. – Matematleta Jan 19 '20 at 02:56
  • All we have for each point $x$ is a nbhd $W$ that intersects only finitely many supp $\psi_\alpha^{k-1}$ by local finiteness of the cover of $X_{k-1}$ as you wrote, and then we use the hypothesis, which works for each given index. I cannot see from other given conditions in the proof that we can use the same open set for other indices. – nomadicmathematician Jan 19 '20 at 02:59
  • I think the point is that the only indices that can intersect $O$ are those that intersect $W$, and there are only finitely many of these, so all the other indices do not. So, for each point $x$ we have found an open set $O$ (in $X_k$) containing $x$ for which only a finite number of supports intersect it. Which is what we want right? Let me check Lee's proof. I do not see where I am making a mistake! – Matematleta Jan 19 '20 at 03:09
  • $W$ is a subset of $O$, so isn't it possible that some support intersects $O$ but not $W$? – nomadicmathematician Jan 19 '20 at 03:12