Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $\tau_1$ and $\tau_2$ are two topologies on a set $X$ with the property that $K\subset X$ is compact with respect to $\tau_1$ if and only if $K$ is compact with respect to $\tau_2$. Then is this enough information to determine whether $\tau_1 = \tau_2$?

If not (which is suspect to be the case) is there a nice counterexample, and what is the minimum amount of extra information required for $\tau_1 = \tau_2$?

I feel like there would be an issue regarding 'points at infinity.'

share|cite|improve this question
up vote 12 down vote accepted

It is not enough information. A simple (and perhaps frightening) counterexample is as follows: if $X$ is an uncountable set, then in both the discrete and co-countable topologies the compact sets are exactly the finite sets. It's frightening because these two topologies share virtually no properties: one is metrizable (hence perfectly normal), the other isn't even Hausdorff; one is Lindelöf, and the other has maximal Lindelöf number.

I'm not certain what additional information would be required to conclude that two topologies are the same. (I'll think about this some more.)

Addition 1

Hausdorffness is not sufficient. The discrete space on $\mathbb{N}$ and the Arens–Fort space are both Hausdorff, and the compact subsets are exactly the finite subsets, but these spaces are not homeomorphic. (Since they are both countable, we can transfer the topology of one onto the underlying set of the other, and the compact sets will be the same.)

Actually, this example shows that perfect normality, Lindelöfness, and paracompactness are not sufficient.

share|cite|improve this answer
In the same spirit, the cofinite and trivial topology are both such that every set is compact, but if the underlying set is not a singleton they are very different. (But your example is better.) – Asaf Karagila Mar 21 '14 at 10:08

Consider a finite set $X$,then every subset $A$ of $X$ is compact in the trivial topology $\{\emptyset,X\}$, but every subset $A$ of $X$ is also compact in the discrete topology on X (the topology where all subsets of X are considered open). If $|X|>1$, then these topologies are obviously not equal.

I will return to your second question if it happens that I have anything useful to say about it.

share|cite|improve this answer
Your first paragraph doesn't answer the question: you need to consider all the subsets of $X$, not just $X$ itself. – Najib Idrissi Mar 21 '14 at 8:22
@nik thank you, I was stupid and read the question as "there is a" but this would of course have rendered the question trivial, as the one point sets are always compact independent of the topology on the set. – Henrik Mar 21 '14 at 8:37
If you additionally require that the spaces $( X , \tau_1 )$, $( X , \tau_2 )$ are compact Hausdorff, then the result follows because in compact Hausdorff spaces the compact subsets are exactly the closed subsets. – arjafi Mar 21 '14 at 8:48
@Arthur Fischer Yes that is right, it suffices to assume that $X$ is compact in a finer topology and hausdorff in a coarser topology, this would imply that the topologies are equal. However this condition seems a bit ad hoc, to the question at hand, so I deleted that part of my question. Thank you for your comment. – Henrik Mar 21 '14 at 9:45
In fact, when your $X$ is a finite set, there are many topologies possible and clearly knowledge of the compact subsets gives absolutely no information here. – Jeppe Stig Nielsen Mar 21 '14 at 12:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.