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"A topological property or topological invariant is a property of a topological space which is invariant under homeomorphisms. That is, a property of spaces is a topological property if whenever a space X possesses that property every space homeomorphic to X possesses that property. Informally, a topological property is a property of the space that can be expressed using open sets." (I copied it from Wikipedia)

Now my question is: What is the definition of a topological property ? Of course you can define it as wiki defines it. But I am more concerned about the the part of wiki's "definition" which says that "Informally, a topological property is a property of the space that can be expressed using open sets."

Is there a definition of a topological property that says which well formed formulas are well formed formulas of topological properties and which are not ?

Because of what I read in wikipedia, I was expecting to see a definition of a topological property that talks about the internal structure of the well formed formula of the property. Then, I also expected that there was a theorem that says that if $(X_1,T_1),(X_2,T_2)$ are any two homeomorphic topological spaces and the well formed formula $\phi(X,T)$ is a topological property, then:

$\phi(X_1,T_1)$ iff $\phi(X_2,T_2)$

Is there such a definition and such a theorem ?

Such a definition and such a theorem will enable one to spot many topological properties easily.

Here is a similar question: Can you characterize all properties of topological spaces which are preserved by homeomorphisms

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    I think I just asked a very similar question: http://math.stackexchange.com/q/16504602016-02-14

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We can formalise topology using only the language of set theory. [For instance, a topological space is a pair $\langle X, \tau \rangle$ where $X$ and $\tau$ are sets satisfying various properties, and we can define a homeomorphism $\langle X, \tau \rangle \to \langle Y, \sigma \rangle$ as a function $X \to Y$ (which is itself a set) satisfying some conditions, etc. All this can be formalised.]

So we can define a unary predicate $\text{TS}$ defined by $\forall x[\text{TS}(x) \leftrightarrow x\ \text{is a topological space}]$ where '$x\ \text{is a topological space}$' is shorthand for... $\begin{align}\exists X \exists \tau( \hspace{53pt}\\ x= \langle X, \tau \rangle \wedge &\tau \subseteq \mathcal{P}(X) \wedge \varnothing \in \tau \wedge X \in \tau\\ \wedge & \forall U\ \forall V\ [U \in \tau \wedge V \in \tau \to U \cap V \in \tau]\\ \wedge & \forall A\ [A \subseteq \tau \to \bigcup A \in \tau]\\ ) \hspace{78pt} \end{align}$

Now suppose $\phi$ is a formula with one free variable, $x$ say. Then $\phi$ is a topological property (i.e. is preserved under homeomorphism) if $\forall x [\text{TS}(x) \wedge \phi(x) \rightarrow \forall y[\text{TS}(y) \wedge x \cong y \rightarrow \phi(y)]]$ That is, if $\phi$ holds for any space $x$ then for any space $y$ homeomorphic to $x$, $\phi$ holds for $y$.

Here I've used $x \cong y$ as shorthand for the formula expressing that $x=\langle X, \tau \rangle$ and $y=\langle Y, \sigma \rangle$ are homeomorphic.

Is this what you were after?

Frankly, I don't see how it's any more enlightening to put yourself through all this than it is to just say "a topological property is one that is preserved by homeomorphism", as so succinctly put by Thomas Andrews in the comments.

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    Ofcourse, I wouldn't be satisfied if you tell me that a property is preserved by all homeomorphisms iff it is preserved by all homeomorphisms2012-11-27
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There is a fundamental misunderstanding about what topology is actually capabable of doing. An invariant is a way to calculate whether two space could be homeomorphic. Take the Euler characteristic. Just because given two spaces that have the same Euler characteristic doesn't mean that they are homeomorphic. However, if two space don't have the same Euler characteristic then the are definitely not homeomorphic. There is very much a zen thing that happens in topology. To answer your question more directly you could make an exhaustive list of invariants but you would not know if it was complete.