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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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    "Closure" only makes sense in a parent space. Homeomorphism is about the spaces themselves, separate from any superset they might have been constructed from.2012-11-27
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    Here's how you can "fix" the definition from Wikipedia: a topological property of a space $X$ is that which can be expressed using open sets in $X$ itself. In your case, closure of the unit ball uses open sets in the whole $\mathbb{R}^2$, not open subsets of the unit ball itself.2012-11-27
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    For example, if $X$ is a space, then what does it mean to say the closure is compact? We don't know anything about $X$, so what does the closure mean? We can only talk about closure of $X$ (usefully) inside a space $Y$ with $X\subset Y$. And, indeed, we can talk about topology of pairs of spaces, $(X,Y)$ with $X\subset Y$. In that sense, $(D,\mathbb R^2)$ is not homeomorphic to $(\mathbb R^2,\mathbb R^2)$.2012-11-27
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    I was thinking about my justification again. The closure of the space of the open unit ball with the inherited topology is the unit open ball itself not the closed ball (A mistake that I did). The open ball is closed in the inherited topology thus its closure is itself.2012-11-27
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    Do you agree with me ?2012-11-27
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    Thus it might be true that the compactness of the closure of a space is a topological property. (The closure of the space is the space itself really).2012-11-27
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    @Amr Yep. Exactly.2012-11-27
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    @Amr well, yes. But the closure of a space in its own topology is the same space again. So the property you've formulated is just compactness of the space. It is of course a topological property.2012-11-27
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    Yes, @Amr, the closure of any space $X$, when considered a subset itself is always just $X$. (Which is why I said it isn't really "useful" to talk about closure of spaces separate from any super-set.)2012-11-27
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    OK I will remove the part about the question but I am still intrested in seeing a "FORMAL" definition of a topological prperty.(One that involves the quantifiers...)2012-11-27
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    Closure is more like an "external topological property", ie. something that depends on how the topological space is placed within the other space. For example, you could have homeomorphic subspaces with non-homeomorphic closures. Also, it does not make sense to talk about closure until you specify the space you are taking the closure in.2012-11-27
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    A topological property is one that is preserved by homeomorphism :)2012-11-27
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    @ Thomas Andrews, true but I consider this definition useless. I'd like to see a definition that talks about the well formed formula of the property2012-11-27
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    Now, I know nothing of logic so I don't know what well formed formulas are, but are you essentially asking for some definition that tells you the theorems automatically? I mean, if you want to know if a given property is a topological property, then you prove that that given property is invariant under homeomorphisms. What else do you want?2012-11-27
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    @Graphth: My understanding is that the OP is looking for a _syntactic_ criterion such that all formulas that fit within the restricted syntax will define topological properties, and such that a usefully large variety of topological properties can be expressed within the restricted syntax. Some appropriate type discipline could undoubtedly work, but I'm not sure whether it is syntactic enough for the OP.2012-11-27
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    If $\Phi(X,T)$ involves no constants, then $\Phi$ is a topological property of the space $X$ with topology $T$. However, one will also want to express properties involving topologial spaces as constants/parameters (e.g. "contains a subspace homeomorphic to $S^1$") and there one would have to make sure that such parameters are not used "directly" (giving e.g. "contains $S^1$ as subspace"). How about: If we have $\Phi(X,T,X',T')$, then $\Psi(X,T,\xi',\tau')\equiv \forall X',T'\colon (\xi',\tau')\cong(X',T')\rightarrow \Phi(X,T,X',T')$ is a topological property?2012-11-27
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    The properties that are considered "topological" are fairly complex. You are going to have a tough time coming up with a formal language that will establish properties that are "topological" that will include homotopy and homology and other things that use maps to other categories. That said, I think you see the error of your interpretation of the Wikipedia page.2012-11-27
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    @Thomas Andrews This may be true. I will edit my question to say what I had in mind2012-11-27
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    Thank you all. Many of you have no idea about what the theorem that I thought it existed, which probably means that it does not exist. Thank you all for your efforts.2012-11-27
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    You can certainly come up with a set of statements such that $X\cong Y$ implies $\phi(X)\iff \phi(Y)$ for all of the statements, but I doubt it is possible to go the other way without a very "big" language capable of handling all set theory. That is, $\forall \phi: \phi(X)\iff \phi(Y)$ would not imply $X\cong Y$.2012-11-27
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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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