# If two topological spaces have the same topological properties, are they homeomorphic?

Topological properties are investigated because we can show that two spaces are not homeomorphic by finding one property that holds in one space but not the other. But what if no topological property can distinguish two topological spaces? So I ask:

If two topological spaces have the same topological properties, must they be homeomorphic?

Edit: I don't really have any particular class of topological properties in mind, because what I am thinking is really every single topological property, whenever it is well-defined for a topological space. I just didn't know that the class of topological properties is so large, that even "homeomorphic to a space $X$" is itself a topological property, making my question trivial.

• What "topological properties" are you referring to? – Jack Oct 30 '16 at 1:15
• A little vague, but maybe you could make it more answerable by specifying a class of properties you're interested in, like topologically-invariant algebraic objects associated to the space. – Tim kinsella Oct 30 '16 at 1:16
• Is «being homeomorphic to the space Z?» a topological property? – Mariano Suárez-Álvarez Oct 30 '16 at 2:28
• So the class of topological property is too large that this question is not quite well-defined? – edm Oct 30 '16 at 2:51
• @edm: Under the usual definition of "topological property", the question is well-defined but has a trivial answer. To make the question more interesting, you can do one of two things (and you usually need both so that it will really be interesting): restrict your attention to certain topological properties and/or restrict your attention to some class of topological spaces (discrete, smooth manifolds, compact Hausdorff, etc). – levap Oct 30 '16 at 3:19

## 5 Answers

Yes.

One topological property of $X$ is belonging to the homeomorphism (equivalence) class of $X$.

Well, it usually goes the other way. A property $P$ of a topological space $X$ is deserved to be called topological if $P(X)$ holds if and only if $P(Y)$ holds whenever $X$ and $Y$ are homeomorphic. An example of a topological property is "$X$ is connected" while an example of a non-topological property is "$X$ is a subset of $\mathbb{R}^n$. The latter can be upgraded to a topological property by requiring instead "$X$ can be embedded in $\mathbb{R}^n$".

With this convention, given a topological space $X$ there is a smart-ass topological property one can define using $X$: "$P(Z)$ holds iff $Z$ is homeomorphic to $X$". Since homeomorphism is an equivalence relation, this is indeed a topological property and clearly $X$ satisfies $P$. If $Y$ is another space that satisfies the same topological properties as $X$, then $P(Y)$ must hold and so $X$ and $Y$ are homeomorphic.

For two arbitrary topological spaces $X$ and $Y$, there isn't a nice list of things you can check off on each of them to be able to say they are homeomorphic. The only thing you can really do to show $X$ and $Y$ are homeomorphic is demonstrate the existence of a homeomorphism.

On the other hand, there are certain classes of topological spaces for which there exist complete classifications. For example, a one dimensional, simply connected complex manifold $X$ is homeomorphic to either the complex plane or its one point compactification, and this is determined by whether or not $X$ is compact.

• I don't understand the downvote to this answer at all. – Jack Oct 30 '16 at 2:36
• It appears that someone has downvoted nearly every answer, so the downvote is likely not based on anything substantive. I agree that this is a good answer that does not deserve a downvote. – Alex Wertheim Oct 30 '16 at 2:43

If your definition of "same topological properties" includes knowledge of homomorphisms between spaces then this has a positive answer, although it requires no topology. Suppose that two spaces $X$ and $Y$ have the property that for every topological space $A$ one has $Hom (A, X) \cong Hom (A,Y)$, in a manner compatible with composition along any homomorphism $B \rightarrow A$, ie so each square of the following form commutes:

\begin{array}{ccc}Hom (A,X)& \xrightarrow{} & Hom(A,Y) \\ \downarrow & & \downarrow \\ Hom (B,X) & \xrightarrow{} & Hom (B,Y)\end{array}

Then there is an isomorphism between $X$ and $Y$. The proof of this is an application of the Yoneda lemma to the functors $Hom (-,X)$ and $Hom (-,Y)$, and in particular works for $Hom (X,-)\cong Hom (Y,-)$.

The claim is the following:

If two spaces $X$ and $Y$ are not homeomorphic, then one can find a property that holds in one of the spaces and not the other. The contropositive of this is:

If one cannot find a property that holds in once space but not the other then $X$ and $Y$ are homeomorphic. So, yes, thie claim is true.

• What are you stating as being a topological property? I think that your claim is somehow really vague. – Daniel Calderón Oct 30 '16 at 6:20
• @DanielCalderón The definition of topological property is not in dispute. thus his answer is unambiguous (but without proof) – Jacob Wakem Oct 30 '16 at 17:32