Homeomorphism, the idea and definitions are not too hard to memorize but specific construction/proving it doesn't exist is almost like a millenium question to me.

Say in this case; I'm sure my professor made it easy by restricting the set to a three-point set.

A tweak on the possible topologies lets me determine which ones are Hausdorff and which ones are not; it's simple, I don't need to think about "is there, out of trillions of possible maps $f$, that are homeomorphic or not?"

I just need to pick two distinct elements from the set and look at the topology, check if there is at least one disjoint set in the topology that contains each elements. Otherwise, game over.

But checking for homeomorphisms...well, I mean, there just seems no simple way of determining if there exists or not a homeomorphism. Nothing obvious.

Like, there are statements that say "if one is compact and other is not, then they are not homeomorphic" and I go "but, then I have to go through the process of proving one is compact and one is not, that's just translating the problem and doesn't help!"

There are no simple ways, like "if there are different numbers of sets in the topologies, then they cannot be homeomorphic" or something like that. I mean, that'll be easy, as long as I can count. There just seems no convenient way of showing these.

Which topologies are non-homeomorphic in $X=\{a,b,c\}$ and how do I show it, and most importantly, how should I find them from scratch

I mean, does something like $id_X:(X,\tau_{disc}) \to (X,\tau_{indisc})$ for discrete and indiscrete work as a nonexample of homeomorphism between them? Sending $id(a)=a$ for instance has an obvious inverse $id^{-1}(a)=a$ but $a$ is open in the discrete toplogy, but not in the indiscrete topology. So the space with discrete and indiscrete topologies cannot be homeomorphic?

Does anyone have a good explanation?

  • $\begingroup$ Yup, the discrete and indiscrete topologies are non-homeomorphic. However, the discrete topology is Hausdorff, so you're not done yet. $\endgroup$ – Noah Schweber May 5 '16 at 21:32
  • $\begingroup$ Hi, yeah I kinda realized....are these types of problems only solvable by essentially guesstimating+some trial and error methods? $\endgroup$ – John Trail May 5 '16 at 21:35

Using the indiscrete topology and a non indiscrete one is sound. Now consider the topology $\{\emptyset,\{a\},\{b,c\},X\}$.

Note that if all singletons are closed, you get the discrete topology, so at least one singleton must be non closed, in order the topology is not Hausdorff. If its closure contains all three points, you have the indiscrete topology. So you need a singleton whose closure contains exactly two points.


Since your space is finite, there is only a finite number of possible topologies on it (the upper bound is, I think, $2^{2^{n}-2}$ for a set of size $n$). In fact, I can tell you that there are exactly 29 possible topologies on $\{a,b,c\}$; you can list them.

The next step is to look at your list and find two topologies that satisfy the required conditions.

  • $\begingroup$ Hi Will, so you'd say that to tackle this problem....you'd list all the topologies($29$) and try to work out those that satisfy the conditions...using the fact that the topologies must have same cardinality? So $\tau_1,\tau_2$ can be homeomorphic iff they have the same number or sets? $\endgroup$ – John Trail May 5 '16 at 21:34
  • $\begingroup$ I'm not actually sure the stuff about cardinality is correct. But certainly a good approach to this problem is to start by writing down a lot (maybe not all) of the possible topologies, and seeing what you can figure out from there. The space given is is finite, so use this to your advantage. $\endgroup$ – Will R May 5 '16 at 21:51
  • $\begingroup$ Homeomorphic spaces must have topologies with the same cardinality; it's quite easy to see. $\endgroup$ – egreg May 6 '16 at 10:11
  • $\begingroup$ @egreg: I thought so originally, but I was pressed for time and couldn't think it through. Looking back now, I think I see it: homeomorphisms map sets in $\tau_{1}$, and only those sets, to sets in $\tau_{2}$. Therefore a homeomorphism suggests a natural surjection from $\tau_{1}$ to $\tau_{2}$, and its inverse suggests a natural surjection from $\tau_{2}$ to $\tau_{1}$. Does it sound like I'm on the right lines? $\endgroup$ – Will R May 6 '16 at 13:31
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    $\begingroup$ @WillR Yes, Your idea is good $\endgroup$ – egreg May 6 '16 at 13:32

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