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A property $P$ of topological spaces is called a topological property if for any two topological spaces $X$ and $Y$ which are homeomorphic, $X$ has property $P$ if and only if $Y$ has property $P$. Topological properties are useful because if one space has a certain topological property and another space does not, then the two spaces are not homeomorphic.

But I’d like to go in the other direction, namely showing that if two spaces have certain topological properties in common, then they are homeomorphic. My question is, for a topological space $X$, does there exist a nontrivial collection of topological properties such that any space which has all these properties is homeomorphic to $X$?

By nontrivial I mean excluding properties like “homeomorphic to $X$”.

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    $\begingroup$ To clarify: Are you looking for examples of spaces $X$ with interesting characterizations by topological properties? Or are you asking about a general theorem along the lines of "Any space $X$ can be characterized by a set of nontrivial topological properties"? If it's the latter, you'll have to be much more precise about what you mean by "nontrivial". For example, is "not homeomorphic to $Y$" a nontrivial property of $X$? $\endgroup$ Oct 13, 2018 at 14:46
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    $\begingroup$ Hmm... I think there are two ways to go here. One reasonable definition of "topological property" is just "class of topological spaces closed under homeomorphism". Then you don't have to worry about how the property is expressed, you just worry about its extension (the spaces which satisfy the property). But then there is no way to distinguish your "trivial" properties from non-trivial ones. $\endgroup$ Oct 13, 2018 at 14:59
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    $\begingroup$ On the other hand, you could take properties to be expressed in english, or in some logic, e.g. the first-order language of set theory. But now you get into the complicated business of determining whether a topological property "makes reference to homeomorphism". This is more complicated than you might think. $\endgroup$ Oct 13, 2018 at 14:59
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    $\begingroup$ For example, if I say "there is a set of ordered pairs $F\subseteq Y\times X$ such that for all $y\in Y$ there is a unique $x\in X$ with $(y,x)\in F$ and for all $x\in X$ there is a unique $y\in Y$ with $(y,x)\in F$, and for every open set $U\subseteq Y$, etc." I think you would agree that this property of $Y$ makes reference to homeomorphism, despite the word homeomorphism not appearing. I've expanded the definition. But now I could obfuscate the definition in various ways, e.g. by replacing ordered pairs by ordered triples $(x,y,x)$, or adding an irrelevant clause to the definition. $\endgroup$ Oct 13, 2018 at 15:02
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    $\begingroup$ @KeshavSrinivasan What's a difference between a property and set of properties? It's just an illusion. Note that "$X$ satisfies all properties I've defined" is a property as well. But I guess that disecting "being homeomorphic" to smaller pieces is a valuable thing. $\endgroup$
    – freakish
    Oct 13, 2018 at 15:35

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For some classical spaces in topology there has been a long research tradition of finding nice unique characterisations of a space, i.e. a finite list of "real" properties (directly defined in terms of open sets and set theory, like compactness and connectedness (and their local versions) and various ones) such that a space that satisfies that finite list of properties is homeomorphic to the space in question. To be useful, it should be non-circular ones (so being a simple-closed curve is not a "good" property to characterise the circle, because that's already defined as being homeomorphic to a circle, basically).

For zero-dimensional spaces we have the classic ones for the Cantor set (the unique zero-dimensional compact metrisable space without isolated points), the rationals (the only countable metrisable space without isolated points), the irrationals (the only completely metrisable separable zero-dimensional, nowhere locally compact space) and a few more.

For connected metric spaces (often continua) we have classic characterisations of $[0,1]$ (metrisable Peano-continuum with exactly two non-cutpoints), $\mathbb{R}$ (separable metrisable, connected, locally connected and every point a strong cut-point (exactly two components of the space minus that point)), $S^1$( metrisable Peano continuum, with no cutpoints and such that every pair is a cutset) and a few others.

For infinite-dimensional spaces we have the topological classification of completely metrisable separable vector spaces (they're all homeomorphic, so purely topologically there is no difference between $C([0,1])$ and separable Hilbert space, e.g.)

But for general spaces there is no hope of such results, they are just too wild and have lots of connections with complicated set theory. By transfinite recursion one can sometimes construct non-homeomorphic spaces of very similar type (so that they'd be the same for a list of standard properties, at least that I could think of (I did this kind of thing in my PhD thesis) but that nonetheless are non-homeomorphic because they are constructed that way: by a diaginalisation argument one can often enumerate in advance all possible candidate functions, and then during the construction ensure that none of them will be a homeomorphism of the eventually resulting space; it's a common trick topologists use, Sierpinski already did it back in the 1920's). So sometimes spaces are not homeomorpjic because they aren't by construction. And if I made $2^{\aleph_1}$ distinct spaces in some construction, there is no way I could distinguish them even using a finite list from $\aleph_1$ properties, if could invent that many "properties" in the first place.. The only option left would be to give up and allow "homeomorphic to $X$" as a (useless) property for all $X$.

Just my thoughts..

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  • $\begingroup$ To the proposer: There are also some known "universal" spaces, that is, any space with a certain list of properties is homeomorphic to a subspace of one of them. For examples: (1) Any countable linear space is homeomorphic to a subspace of $\Bbb Q.$ (2) Every completely regular space is homeomorphic to a subpace of the Tychonoff product $[0,1]^S$ for some (index-set) S....... $\endgroup$ Oct 14, 2018 at 21:59
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Every perfect, 2nd countable, zero dimensional, compact Hausdorff space is homeomorphic to the Cantor set.
$\{0,1\}^\mathbb{N}$ for example.

Any order topology of a countable, order dense linear topology without end points is homeomorphic to the rationals. The dyadic rationals for example.

All discrete spaces of cardinality $\kappa$ are homeomorphic.
Same for indiscrete spaces.

All Hausdorff spaces of finite cardinality $n$ are homeomorphic.

All not empty, discrete and connected spaces are homeomorphic.

All spaces with exactly one open set are homeomorphic.

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    $\begingroup$ I think you can characterize the rationals as a topological space more simply, without mentioning order. Isn't every nonempty countable metrizable space without isolated points homeomorphic to the rationals? For instance the rational points in the plane? $\endgroup$
    – bof
    Oct 14, 2018 at 9:58
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    $\begingroup$ @bof yes indeed. Just countably infinite, $T_3$, first countable, and no isolated points will do. The second and third property together imply metrisability (for countable spaces). $\endgroup$ Oct 14, 2018 at 13:54

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