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I know that topological spaces over a given finite set are classified. But what about sets with cardinality $\omega$? The only restriction I would require is for the spaces to be $T_0$, i.e. have no topologically indistinguishable points.

I initially thought that it would be sufficient to look at connected spaces, but then I realized that just because a space is disconnected, it need not be completely decomposable into connected components: Consider, for instance, $\{1/n\mid n\in \mathbb N_{>0}\}\cup \{0\}\subseteq \mathbb R$ equipped with the subspace topology: Every set of the form $[0,1/n]$ is clopen, but as their intersection $\{0\}$ is not open, there is no minimal clopen set around $0$, i.e. no connected subspace $S\subseteq X$ that admits a decomposition $X \simeq S \sqcup X\setminus S$.

Lukas Juhrich
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  • In your example the components are all singletons and the space is just $\omega+1$, the second countabel ordinal. You seem to be confusing quasicomponents with components? The components are clear: no subset of more than one point is connected, so the maximal connected sets are singletons. – Henno Brandsma Jan 29 '19 at 22:36
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    A connected component of a space isn't a minimal clopen subset, it's a maximal connected subset. Trivially every singleton is connected; in your example, the connected components are exactly the singletons. Moreover, every space can be decomposed into connected components, since the union of two connected sets with a point in common is again connected. – Noah Schweber Jan 29 '19 at 22:42
  • @NoahSchweber thanks for clearing that up, I should have known that. My line of intuition was that clopen sets are the ones that enable you to describe decompose your topological space into a coproduct of two substructures, so I somehow assumed that the “minimal version” of that corresponded to a connected component. Sorry about that. – Lukas Juhrich Jan 29 '19 at 22:58
  • However, in my remark I am not concerned about whether something is a connected component or not, but about if I can decompose the original topology with respect to that subspace. Which seems to me to be more interesting in a classification problem :-) – Lukas Juhrich Jan 29 '19 at 23:00

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There is a classification for countable metric spaces (I sketch it in my answer to this related question, they include all the countable ordinals and the rationals and combos of those), but beyond that it's isolated results: there are the 4 countable Toronto spaces that are not indiscrete, included and excluded point topologies, many different types of spaces based on an filters on $\omega$ (we already have more than continuum many homeomorphism types there), either using the filter as the topology, or its elements for a single non-isolated point aded to $\omega$ (but there could be more of those of Hausdorff is not required) etc etc. I also don't know what the homeomorphism type of a dense countable set of $\{0,1\}^\mathbb{R}$ is, and how many types there are (they are all crowded normal and of weight continuum). Etc. etc. Arens space is countable and quite interesting. The one-point compactification of the rationals, and countable connected Hausdorff spaces also exist. In $\pi$-base you can look at more examples (often introduced to show specific combinations of properties are possible). There is much more variation than you'd think at first. A full classification seems far off as yet.

PatrickR
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Henno Brandsma
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    +1. Note to the OP that even the countable closed subsets of $\mathbb{R}$ are interesting; see e.g. the Cantor-Bendixson derivative, which gives a way to "count the layers" in such a set (or many others!). – Noah Schweber Jan 29 '19 at 22:50
  • Thanks for the starting point, I wasn't aware of the metric classification! I wasn't ven sure whether I wanted to put the $T_1$-Restriction (let alone Hausdorff or metrizable or “countable subspace of $\mathbb R$”) into the question, but perhaps that was too ambitious. – Lukas Juhrich Jan 29 '19 at 23:14