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Thinking about the question whether there is something like a “maximal compactification” (the one-point compactification could be seen as “minimal” compactification), I've come across the possibility to recursively compactify a set (so that you get multiple copies of the original space, but it is still a compactification of one of them).

The basic construction goes like this: Be $X$ a non-compact topological space (this is the space that shall be compactified). Be $C$ an arbitrary compact topological space. Then I define a topology on the disjoint union of $X$ and $C$ as follows:

A set on the disjoint union is open if it is either an open subset of $X$ or the union of the X-complement of a closed compact set of $X$ and a non-empty open subset of $C$.

It is not hard to show that this gives indeed a compactification of $X$, and in addition the subspace topology on $C$ is the original topology on $C$.

This of course answered my original question, as $C$ could be made arbitrary large, making the compactifiation of $X$ arbitrary large, too.

But then, I noticed that since the only condition on $C$ was that it is compact, I could choose as $C$ a compactification of $X$. That is, I then get a compactification of $X$ that contains a second copy of $X$.

Indeed, I could in turn use that compactification of $X$ as $C$, so I now get a compactification of $X$ that in total contains three copies of $X$.

Now I can continue this recursively, and since each of the compactifications is (homeomorphic to) a proper subspace of the next one, I can take the direct limit to get a space with (countably) infinitely many copies of $X$.

Now that direct limit is clearly not a compactification of $X$ (as the closure of each copy of $X$ only covers the earlier copies of $X$), however is it at least still a compact set (and thus a proper “seed” for another round of recursive compactifications)?

celtschk
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    I notice that if $C$ contains more than one point then your compactification is not Hausdorff, even if both $X$ and $C$ are Hausdorff: for all $x,y \in C$, every open set that contains $x$ and every open set that contains $y$ have nonempty intersection. Constructions like yours may indeed be the reason that compactification operations are usually required to preserve the Hausdorff property. – Lee Mosher Oct 30 '19 at 20:21
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    To add to what Lee Mosher said, if you require your compactifications to be Hausdorff, then there is a very natural "maximal compactification", the Stone-Cech compactification. – Eric Wofsey Oct 30 '19 at 21:02
  • @LeeMosher: Well, it was not my intention to make it Hausdorff (or any other condition other than being a compactification). Note that I also didn't put any conditions (other than noncompactness) on $X$. – celtschk Oct 30 '19 at 22:49
  • @EricWofsey: However Stone-Cech has quite stringent conditions on $X$ for the construction to even be a compactification. – celtschk Oct 30 '19 at 22:52
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    @celtschk: Well, that depends on your perspective. The conditions are just exactly the minimum conditions for there to exist any embedding of $X$ into a compact Hausdorff space at all. – Eric Wofsey Oct 30 '19 at 23:00

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Sequential direct limits are almost never compact. In particular, a direct limit of a sequence of inclusions of $T_1$ spaces is never compact (unless the sequence eventually stabilizes); see Compact subset in colimit of spaces. So, if the initial compactification you use is $T_1$, then your direct limit will not be compact.

Eric Wofsey
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