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Let $X$ be a set and $\mathcal P \left({X}\right)$ be the set of subsets ordered with inclusion. Let $\mathcal{T}(X)$ be the set of all topologies on $X$ ordered with set inclusion. Let $\mathcal{T}(X)/ \sim$ denote the quotient set that identifies homeomorphic topologies (in other words the set of non-homeomorphic topologies on $X$).

(?) What is the cardinality of $\mathcal{T}(X)/ \sim$? ADDED: Answered by Brian. let $X$ be a set with infinite cardinality $k$. Then the cardinality of $\mathcal{T}(X)/ \sim$ is $2^{2^k}$

I'm trying to define and make sense of functions like this:

$$f:\mathcal P \left({X}\right) \to \mathcal{T}(X)/ \sim$$ If $U$ is a subset of $X$ then:

$$f(U)=\tau\quad\text{iff}\quad\tau\text{ is the coarsest Hausdorff topology for which }U\in\tau\in\mathcal{T}(X)/\sim$$

(?) Is this function well defined?

I'm interested in functions where "coarsest Hausdorff" can be switched to other properties the topology must satisfy. For example "Finest connected" as long as the maps are well defined.

(?) What would be suitable topologies to give $\mathcal P \left({X}\right)$ and $\mathcal{T}(X)/ \sim$?

I'd like to have continuous functions to be able to take a convergent limit of a sequence of subsets and assert that the corresponding sequence of topologies converge.

Basically I'd like to know if this structure has been studied since I sense it could deepen my understanding.

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Marginally related is some work done by Edgar Raymond Lorch and his student Hing Tong. You can find a nicely written overview of it (with references) in Lorch's paper Continuity and Baire functions, American Mathematical Monthly 78 #7 (Aug-Sept. 1971), 748-762. Near the bottom of p. 756 Lorch makes this colorful statement: This circumstance, that a collection of topologies is topologized, may seem a bit incestuous. –  Dave L. Renfro Dec 6 '13 at 19:40
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That is the most horrible notation I have EVER witnessed. Using $\mathcal P(\bullet)$ for the power set, then using $\mathcal{P(P(\bullet))}$ for the set of topologies?! Why not $\mathcal T(\bullet)$ for that? –  Asaf Karagila Dec 8 '13 at 12:33
    
@AsafKaragila Thanks for the help! Is it better now? –  Saal Hardali Dec 8 '13 at 12:42
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Yeah, much better. –  Asaf Karagila Dec 8 '13 at 12:48
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Maybe this nice 1975 survey article on the lattice of topologies can be of interest: projecteuclid.org/euclid.rmjm/1250130634 –  johndoe Dec 12 '13 at 10:26

3 Answers 3

up vote 10 down vote accepted
+300

In general you probably won’t be able to carry out such a program. For example, your function $f$ is not well-defined: there is no coarsest Hausdorff topology on an infinite set.

Let $X$ be infinite, and fix distinct points $p,q\in X$. Let

$$\tau_p=\big\{\{x\}:x\in X\setminus\{p\}\big\}\cup\{U\subseteq X:p\in U\text{ and }X\setminus U\text{ is finite}\}$$

and

$$\tau_q=\big\{\{x\}:x\in X\setminus\{q\}\big\}\cup\{U\subseteq X:q\in U\text{ and }X\setminus U\text{ is finite}\}\;;$$

$\tau_p$ and $\tau_q$ are compact Hausdorff topologies on $X$ and are therefore minimal Hausdorff topologies on $X$. However,

$$\tau_p\cap\tau_q=\big\{\{x\}:x\in X\setminus\{p,q\}\big\}\cup\{U\subseteq X:\{p,q\}\subseteq U\text{ and }X\setminus U\text{ is finite}\}\;,$$

which does not contain a Hausdorff topology: no topology $\tau\subseteq\tau_p\cap\tau_q$ contains disjoint nbhds of $p$ and $q$.

Of course $\tau_p$ and $\tau_q$ are homeomorphic, so this isn’t quite a counterexample, but it is indicative of the potential problems. So is the fact that there are Hausdorff topologies that do not contain minimal Hausdorff topologies: the usual topology on $\Bbb Q$ is a well-known example.

Added from Comments: Let $X$ be an infinite set of cardinality $\kappa$. There are $2^{2^\kappa}$ distinct ultrafilters on $X$, and if $\mathscr{U}$ is an ultrafilter on $X$, then $\mathscr{U}\cup\{\varnothing\}$ is a topology on $X$. There are only $2^\kappa$ permutations of $X$, so there are $2^{2^\kappa}$ pairwise non-isomorphic ultrafilters, which yield $2^{2^\kappa}$ pairwise non-homeomorphic topologies. Plainly there are at most $2^{2^\kappa}$ pairwise non-homeomorphic topologies on $X$, so there must be exactly that many.

Added: The lattice of topologies on a set is something that has been studied quite a bit. In this answer I mention several topologies on lattices and partial orders that have been studied; I don’t know whether any of them is useful for your purposes. However, you should note that for many topological properties there simply isn’t a finest or coarsest topology with that property on a given set. A couple of papers that may be of interest on that score are Manuel P. Berri, ‘Minimal topological spaces’, Trans. Amer. Math. Soc. $108$ $(1963)$, $97$-$105$, and M.P. Berri, J.R. Porter, and R.M. Stephenson, ‘A survey of minimal topological spaces’, General Topology and Its Relations to Modern Analysis and Algebra, Praha: Academia Publishing House of the Czechoslovak Academy of Sciences, $1971$, $93$-$114$

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Thanks for clarifying this. I totally forgot to mention that I will be satisfied with finite sets alone for a start. –  Saal Hardali Dec 4 '13 at 19:38
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@Shaul: On a finite set the only Hausdorff (indeed, the only $T_1$) topology is the discrete topology. I suspect that for most properties of interest the coarsest topology on a finite set with that property is going to be either the discrete or the indiscrete topology. –  Brian M. Scott Dec 4 '13 at 19:42
    
I see. What is the cardinality of $\mathcal P \left({\mathcal P \left({X}\right)}\right)/ \sim$. in Particular can you define injective functions from it to $\mathcal P(X)$? –  Saal Hardali Dec 4 '13 at 19:50
    
@Shaul: First you need to remove from $\wp(\wp(X))$ the collections that aren’t topologies. After that it gets messy, and I don’t know much. I’d start here, here, and here. –  Brian M. Scott Dec 4 '13 at 20:00
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@Shaul: Let $X$ be an infinite set of cardinality $\kappa$. There are $2^{2^\kappa}$ distinct ultrafilters on $X$, and if $\mathscr{U}$ is an ultrafilter on $X$, then $\mathscr{U}\cup\{\varnothing\}$ is a topology on $X$. There are only $2^\kappa$ permutations of $X$, so there are $2^{2^\kappa}$ pairwise non-isomorphic ultrafilters, which yield $2^{2^\kappa}$ pairwise non-homeomorphic topologies. Plainly there are at most $2^{2^\kappa}$ pairwise non-homeomorphic topologies on $X$, so there must be exactly that many. –  Brian M. Scott Dec 4 '13 at 20:14

This is not at all what you're looking for, but why not look at non-Hausdorff topologies? Did you know that every finite CW complex is homotopy equivalent to a non-Haussdorff space with finitely many points? And for every subset of a space, there is a unique topology given by that set, the empty set, and the whole set. This map is certainly continuous under any natural topology on either set. Then one could look at 'neighborhoods' of embedded points in the space of topologies. One natural topology in the set of topologies is to take as a sub basis all sets consisting if topologies where a fixed subset is open.

So, for instance, a sub basis element for the topology on the topologies on the reals might consist of all topologies where the rationale are open.

It could be fun!

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In the case of $X$ finite ($|X|=n$), the sought cardinality is equal to the number of ways to split number $n$ as a sum of positive integers. Only one of the is Hausdorff.

For example, for $n=4$, $$ n=4, \quad n=3+1, \quad n=2+2, \quad n=2+1+1, \quad n=1+1+1+1, $$ i.e., five different topologies, up to homeomorphism.

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