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The Shrinking Lemma: Let $X$ be a normal space; let $\{U_1, U_2, \dots \}$ be a finite-indexed open covering of $X$. Then there exists an indexed open covering $\{V_1, V_2, \dots \}$ of $X$ such that $\overline{V_n} \subset U_n$ for each $n$.

In particular, I was reading this post Counter Example for The Shrinking Lemma, but I have a question. Why is the finite-indexed condition required? Isn't the induction itself enough to transform an open covering $\{U_1, U_2, \dots\}$ to a "shrunk" open covering $\{V_1, V_2, \dots\}$?

MJD
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Jonathan McDonald
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    Induction gives you for each $k$ an open covering ${V_1,\dots,V_k,U_{k+1},U_{k+2},\dots}$. This alone does not guarantee that ${V_1, V_2, \dots}$ covers $X$. – George Giapitzakis Jul 16 '24 at 21:13
  • I don't really understand why that is.... when we do induction, for example, proving that $\sum_{i = 1}^n i = n(n + 1)/2$, we do the same thing, prove it for $n = k$ then show that from there we can also prove it for $n = k + 1$. Essentially, I don't see whats wrong with the proof if we don't have the finite-indexed condition. – Jonathan McDonald Jul 17 '24 at 04:53
  • $X$ is normal iff any point-finite open cover of $X$ has a shrinking – Jakobian Aug 25 '24 at 01:20
  • Yes, so inductive proofs show that a proposition $P(n)$ holds for every finite $n$, nothing in the infinite case. – Ted Shifrin Aug 25 '24 at 01:21

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This lemma requires “point-finite indexed open covering”. Basically, you can have countably infinitely many $U_i$ in your starting cover (therefore « indexed »), however any point $x\in X$ can belong to only finitely many $U_i$ (therefore « point-finite »).

The question you refer to states a proof, so I will simply walk through it.

The first part of the proof allows you to construct one by one open subsets $V_i$’s such that for each $i, \overline{V_i} \subset U_i$. Their existence is guaranteed by the normality of the space $X$. And this is what the induction does, it replaces $U_i$’s by $V_i$’s while preserving the fact that $\{V_1,…,V_k,V_{k+1},U_{k+2}…\}$ is as much of an open cover as $\{V_1,…,V_k,U_{k+1}…\}$ is.

Now, to answer your question try to do this construction assuming that there is a point $x$ in $X$ that belongs to every $U_i$, basically assuming that the condition « point-finite indexed open covering » isn’t fulfilled. Since in the construction you are asked to consider « open neighbours » and « closures », it is tough to be rigorous about that when doing a drawing, nonetheless it will maybe give a feeling for why this condition is necessary.

The tricky thing with induction here is that our proposition thus far is $P(n)= « \{V_1,…,V_n,U_{n+1}…\} \ \text{is an open cover} »$. And we showed it to be true for all $n$, but we want $P(\infty)$ to be true, and that jump isn’t actually always legal. Here is a link to a nice discussion on that Why doesn't induction extend to infinity? (re: Fourier series) .

The second part of the proof is exactly about that. It is about showing that eventually every point $x$ of the space will be in some $V_i$ and that the $U_i$’s containing it are redundant. We can say « eventually » exactly because of the fact that pass some index $k$ there is no $U_i$ with $ i \geq k$ containing $x$ (this is the consequence of « point-finite »). Therefore by construction, the point $x$ has to be already in (thus covered by) some $V_i$. And only now we can finally say that if $\{V_1,…,V_k,V_{k+1},U_{k+2}…\}$ is an open cover for all $k$, therefore $\{V_1,…,V_k,V_{k+1},V_{k+2}…\}$ is one as well.

To me different infinities arising are quite problematic. The infinity of the points of the space, of the possible covers, whether the cover is infinite, and if so, whether some points can appear in infinitely many covering subsets. It’s only natural to get a bit overwhelmed…

Thank you for your question.

Cobra
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