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.

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 $Hom (X,-)\cong Hom (Y,-)$.
