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Consider a compact Hausdorff space $A$. Suppose we have a family of compact Hausdorff spaces $\{B_a\}_{a\in A}$ indexed by the space $A$. Then is it true that the space $$X=\sqcup_{a\in A} B_a$$ is also compact Hausdorff? The Hausdorff part is clear. But I'm not sure how one might show compactness. It's clear when the spaces $B_a$ are identical, and when $A$ is finite. But how does one show in general? I'm not certain if the usual disjoint union topology is correct here, because I believe in the usual disjoint union, the indexing set is considered discrete, but here the indices also have a topology, so we may also need to consider the continuity of the map $g:X\to A$. So maybe the question is better phrased as --

Is there a way to topologize $X$ such that it is compact Hausdorff, which generalizes the disjoint union topology from wikipedia, when $A$ is discrete?

Perhaps one might need to use the finite intersection property, to show compactness, but it's not clear to me if that works because I'm not sure what the topology should be. Thanks for any help.

Edit : As posted in the comments, considering the usual disjoint union won't quite work. So my proposal is topologize $X$ by the basis of opens $$W=\{(x,a) : x\in U_a\subset B_a, a\in V\}$$ where each $U_a$ is an open subset of $B_a$, and $V$ is open in $A$. But I'm not sure if this correct.

shadow10
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    What role does the topology of $A$ play? Isn't it just an indexing set? Does the topology of $A$ enter into the definition of disjoint union? – Randall Sep 27 '24 at 20:41
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    For the "traditional" disjoint union, it is easy to see that compactness will fail when $A$ is infinite. for each $B_\alpha$ is open in the disjoint union topology. – Randall Sep 27 '24 at 20:46
  • It's not clear to me what role it should play, I think the basis for the topology should be opens of the form ${(x,a) : x\in U_a\subset B_a, a\in V}$, where $U_a$ is an open in $B_a$, and $V$ is an open in $A$. But this seems quite messy, and I was wondering if there is a good way to understand it. – shadow10 Sep 27 '24 at 20:59
  • Note, if $A$ is just treated as an indexing set, i.e. discrete, then when $B_a=B$ for every $a$, $X=A\times B$ which need not be compact for compact $B$, but it will be compact once we take the topology of $A$ into account. – shadow10 Sep 27 '24 at 21:00
  • Even if we generalize this construction when $B_a = B$ for all $a$, how do you do it when this isn't true? – Jakobian Sep 27 '24 at 21:19
  • Whoever is voting to close it, and/or downvoting it, I would appreciate if you comment and ask for more clarifications or edits if you feel like it. I am happy to add clarification if needed. And @Jakobian, this is precisely why I'm asking the question. How DO you do it in general? – shadow10 Sep 27 '24 at 21:30
  • You can always consider $X$ as $\bigcup_{a\in A} {a}\times B_a\subseteq A\times \bigsqcup_{a\in A} B_a$ – Jakobian Sep 27 '24 at 21:37
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    You said "But I'm not sure if this correct.". What is correct or incorrect about a construction? What does this mean? – Jakobian Sep 27 '24 at 21:40
  • @Jakobian I think OP is asking for a definition of a suitable disjoint-union-ish space so that when $A$ and all $B_\alpha$ are compact Hausdorff, so is the constructed space. By the comments, it cannot be the usual disjoint union topology. – Randall Sep 27 '24 at 21:42
  • I mean if this is the right way to topologize to get a compact Hausdorff space. Like I said before, there's different ways of topologizing it, and the usual disjoint union may not give us a compact Hausdorff space. I am not certain my way of topologizing is giving a compact Hausdorff space, which is what I mean by 'correct'. – shadow10 Sep 27 '24 at 21:42
  • I kind of feel like OP is creeping up on the notion of pushouts. – Randall Sep 27 '24 at 21:44
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    It seems like no one knows what OP is asking about – Jakobian Sep 27 '24 at 21:50
  • @Jakobian I'm sorry, what part of this discussion is unclear to you? I thought what Randall said and what I said was quite clear? Also Randall you're right, one way of looking at this would be that I'm asking for the construction of coproduct inside $\mathbf{CHaus}$. The usual disjoint union does the same for $\mathbf{Top}$. – shadow10 Sep 27 '24 at 21:53
  • In that case, relevant: https://math.stackexchange.com/questions/4570624/is-the-category-of-compact-hausdorff-spaces-cocomplete – Randall Sep 27 '24 at 21:56
  • @shadow10 its not that what you're saying is unclear to me but what you're actually asking is unclear, both to me and to user Randall. Although you provided a lot more context with your recent comment – Jakobian Sep 27 '24 at 22:05
  • I have given a set $X$, I'm asking if there is a topology(suitably canonical) on $X$, which makes it compact Hausdorff. Alternatively, it would be worthwhile to know, if the basis of open sets I have proposed, does that give $X$ a compact Hausdorff topology. Hope this clears things up. – shadow10 Sep 27 '24 at 22:12
  • The link Randall provided should be clear. To obtain coproduct in category $\text{CHaus}$ you take disjoint union and then Stone-Cech compactification. This will be larger as a set than just taking disjoint union. For instance for countable amount of singletons the resulting space will be of size $2^\mathfrak{c}$ whereas the disjoint union is countable – Jakobian Sep 27 '24 at 22:19
  • Perhaps you are confused what you are trying to ask as much as we are, since coproducts in $\text{CHaus}$ seem to have little to do with your question about the given construction on $X$ – Jakobian Sep 27 '24 at 22:30
  • I'm more interested in this space $X$, but I see the construction of coproduct is different as given in the link. So perhaps what I said about the coproduct is not correct, and should be ignored. – shadow10 Sep 27 '24 at 22:48
  • Your proposed topology certainly does not work. Even if $B_a$ is a singleton for all $a$, you just end up getting the discrete topology on $A$. – David Gao Sep 28 '24 at 06:03

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