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I found at the beginning of tom Dieck's Book the following (non proved) result

Suppose $X$ is the colimit of the sequence $$ X_1 \subset X_2 \subset X_3 \subset \cdots $$ Suppose points in $X_i$ are closed. Then each compact subset $K$ of $X$ is contained in some $X_k$

Now I really don't know how to prove this fact. The idea would be to find a suitable open cover to it and after taking a finite sub cover trying to claim that $K$ lies in one of the $X_k$. I'm able to do this reasoning in some more specific cases, where I've more control on how open subsets looks like, but in this full generality I don't see which open cover I can take.

My Attempt: The only idea or approach I'm able to cook up so far is to try use some kind of sequence of points $x_n \in K\cap X_n \setminus X_{n-1}$ which can be assumed to exist by absurd. Being $K$ compact, there must be an accumulation point $k\in K$. Clearly $k \in X_k$ (little abuse of notation here) and for every neighbourhood of $k$, there is a tail of this sequence entirely contained in it. Now everything seems to boil down to find the right nbhd to find the counterexample. It seems doable, but I don't have any idea on how to choose it, because the only open I have for sure are complements of points, but they seems a little bit coarse for what I want to do.

As a side note, May claims at page $67$ of his "Concise Course (revised)" that this result holds for any based spaces. The proof seems to use the above result without T1 assumption. How one can prove this result in such generalities? (no details where provided, only the rough idea.

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Luigi M
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  • Throwing a simple idea out there, haven't worked out the details: isn't the union of the interiors of the $X_i$ an open cover of $X$? Then by compactness $K$ would have to belong to some $X_k$. – Alex Provost Dec 21 '15 at 19:32
  • @A.P. mmh I think you get some problem in proving that $K$ is contained in this cover let's say. Morally, I Think it can happen that all the $X_i$'s share a common boundary let's say, and if $K$ contains points of this common boundary, then your cover wouldn't contain them – Luigi M Dec 21 '15 at 19:35
  • I'm not sure I understand your concern. By hypothesis we have $\operatorname{int}(X_1) \subset \operatorname{int}(X_2) \subset \operatorname{int}(X_3) \subset \cdots$. Now assuming the union of these is an open cover of $X$, $K$ has to admit a finite subcover, $K \subset \operatorname{int}(X_{i_1}) \cup \cdots \cup \operatorname{int}(X_{i_n})$, and then in particular $K \subset \operatorname{int}(X_{\max(i_j)}) \subset X_{\max(i_j)}$ . – Alex Provost Dec 21 '15 at 19:39
  • @A.P. yes, my concern is exactly assuming the union of these is an open cover, I think it doesn't hold in general. I mean think of the union of $[0,1/n$] running over $n$. if you take interior points, you'd never cover $0$, and if $K$ contained $0$, then what you define is not an open cover of $K$ – Luigi M Dec 21 '15 at 19:41
  • Hmm, yes, that is true. – Alex Provost Dec 21 '15 at 19:43
  • But still, m claim hold in the particular case I gave you for other reason (the inclusions are closed embeddings and spaces are $T2$... But tom Dieck's result doesn't make use of these, and May doesn't make use of anything – Luigi M Dec 21 '15 at 19:45
  • Let us continue this discussion in chat. – Alex Provost Dec 21 '15 at 19:45
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    May is working in compactly generated weak Hausdorff spaces, which are $T_1$, so he doesn't need to make those assumptions explicit. –  Dec 21 '15 at 23:49

1 Answers1

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As you suggest, choose a sequence of points $x_n\in K\cap X_n\setminus X_{n-1}$ (possibly replacing $(X_n)$ with a subsequence). Let $A=\{x_n\}$. Then if $B\subseteq A$, then $B\cap X_n$ is finite for each $n$, so since points are closed in $X_n$, $B\cap X_n$ is closed in $X_n$. Since $X$ is the colimit, this means $B$ is closed in $X$. In particular, $A$ is a closed subset of $X$, and every subset of $A$ is closed so it has the discrete topology. But a closed subset of a compact space is compact, and a compact discrete space must be finite. This is a contradiction.

Without any hypotheses about points being closed, the result is definitely not true. For instance, let $X=\mathbb{N}$, topologized by saying the sets $\{n:n\geq m\}$ are open for each $m\in\mathbb{N}$. Then $X$ is the colimit of the subspaces $X_n=\{0,\dots,n-1\}$, but $X$ itself is compact. However, I believe that in May's book all "spaces" are assumed to be compactly generated weak Hausdorff, which implies points are closed.

(Note that if $X$ is a colimit of a sequence of maps that are not necessarily injective, the hypothesis you need is not that points are closed in each $X_n$ but that points are closed in $X$; see this answer of mine on MO. I recall that when I wrote that answer I came up with a counterexample where points are closed in each $X_n$, but I don't remember the details at the moment.)

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