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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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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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    I will edit my question to say what I had in mind.2012-11-27
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    @Amr: Having seen your edit, I'm still a bit puzzled; what's the theorem you're talking about? If a topological property is defined to be a property which is invariant under homeomorphism, then the theorem stating that a topological property is invariant under homeomorphism is a tautology! [But I do think my response answers your question - if it doesn't, please let me know what else you want to know.]2012-11-27
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    The theorem puts conditions on the the structure of the well formed formula $\phi$ (positions of quantifiers ...) and the conclusion is that homeomorphisms preserve the property $\phi$2012-11-27
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    @Amr: But *any* first-order formula $\phi$, i.e. a property of sets, could potentially be a topological property. For instance, *any* property of sets preserved under bijections of sets is a topological property under this definition; as is any theorem of the predicate calculus. I very much doubt such a theorem on the structure of $\phi$ exists; if it did then topology would become trivial!2012-11-27
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    Here is another question (harder than the first one but hopefully clearer): can you characterize all properties of topological spaces that are preserved by homeomorphisms2012-11-27
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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.