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I have a question about a proof in John Lee's Introduction to Topological Manifolds (5.22).

Given CW-complex $X$ with skeletons $X_n$ and open cover $\left(U_\alpha\right)_{\alpha\in A}$, we inductively define partitions of unity $\left\{\psi_\alpha^n\right\}$ on $X_n$ subordinate to cover $\left(U_\alpha^n=U_\alpha \cap X_n\right)_{\alpha\in A}$, such that

  1. $\psi_\alpha^k\mid_{X_{k-1}}=\psi_\alpha^{k-1}$
  2. 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 $\psi_\alpha^k\equiv 0$

At the end of the induction step, prof. Lee proceeds to argue that supports of $\left\{\psi_\alpha^{n+1}\right\}$ form locally finite family.

If $x$ is in the interior of an $(n+1)$-cell, then that cell is a neighborhood of $x$ on which only finitely many of the functions $\psi_\alpha^{n+1}$ are nonzero by construction. On the other hand, if $x\in X_n$, because $\left\{\psi_\alpha^n\right\}$ is a partition of unity there is some neighborhood $V$ of $x$ in $X_n$ on which $\psi_\alpha^n\equiv 0$ except when $\alpha$ is one of finitely many indices, and then (2) shows that $\psi_\alpha^{n+1}\equiv 0$ on $V'$ except when $\alpha$ is one of the same indices.

I don't understand why does this imply local finiteness. Because of (2), there is a $V'$ for each such $\alpha$, but unless I'm missing something obvious, there can be infinitely many such alphas. How can we conclude there is a single neighborhood of $x$ in $X_n$ on which all these functions vanish?

Eric Wofsey
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Marcin Łoś
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1 Answers1

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Upon second reading, I realized that in fact, a property slightly stronger than (2) is both established and used in the proof. Namely, with the change:

If $J\subset A$ is a subset of indices, $V\subset X_{k-1}$ is open and $\psi_\alpha^{k-1}\equiv 0$ on $V$ for all $\alpha\in J$, then there exists an open set $V'\subset X_k$ containing $V$ such that $\psi_\alpha^k\equiv 0$ on $V$ for all $\alpha\in J$

(and the errata) I believe I can follow the proof and justify each step.

Marcin Łoś
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