# Does there exist a topology for a set $X$ which is compact and Hausdorff?

For every set $X$ and every topology $\tau$ over $X$ we have that $\tau$ contains the trivial topology $\{ X, \emptyset\}$, which is compact, and is contained in the discrete topology $\{ S: S \subseteq X\}$, which is Hausdorff. I was wondering if there is any topology on X "between" the trivial and the discrete such that it has both properties.

It seems that there is such a topology for specific sets, such as the natural numbers, but I haven't found any result for arbitrary $X$. I don't know if any additional condition must be established on $X$ for the result to hold, or if it isn't possible in general.

I would deeply appreciate some help on this.

• Just a little thought: do you know that if $X$ has topology $\tau$ making $X$ compact and Hausdorff, then there is not finer topology than $\tau$ making $X$ compact? – user39280 Apr 11 '13 at 18:05
• ... and no coarser topology making it Hausdorff :-) – Stefan Hamcke Apr 11 '13 at 18:11
• By the way, if there is a such topology for a set $X$, then there is a topology for any set which is bijective to $X$, since we can just "carry over" the topology from one set to the other. For example, a topology on $\mathbb R$ making it compact and Hausdorff is difficult to imagine, but if we add a point $\infty$, then it suddenly becomes very easy. But since there is a bijection between $\mathbb R$ and $\mathbb R\cup\{\infty\}$ we could also define this topology on $\mathbb R$ directly. It would look a bit ugly, however. – Stefan Hamcke Apr 11 '13 at 18:19
• Crossposted at MO. – Zhen Lin Apr 11 '13 at 18:28

If $X$ is finite, then the discrete topology is Hausdorff compact and we're done.

If $X$ is infinite, fix some $x\in X$, and let $X'=X\setminus\{x\}$ have the discrete topology which is a locally compact topology, and take $X$ to be the one-point compactification of $X'$. That is $x$ is "the point in $\infty$".

Then clearly $X$ is compact, and to see it is Hausdorff, take $u,v\in X$ to be distinct. At most one of them is $x$ itself, if it is the case assume $v=x$. Taking $\{u\}$ and $X\setminus\{u\}$ we have two disjoint open sets and we are done.

Another, in fact much simpler (but using the axiom of choice, whereas the previous one doesn't) example would be the following:

Let $\alpha$ be an ordinal such that $|X|=|\alpha|$, then $\alpha+1$ as an ordinal space is both Hausdorff and compact. To see that it is Hausdorff take $x<y\in\alpha+1$ then $(0,x+1),(x,\alpha)$ as two open intervals witnessing that.

The compactness of successor ordinals follows from the fact that given a cover by open intervals we can index the endpoints of these intervals, and form a decreasing sequence of ordinals. Since every decreasing sequence of ordinals is finite we can find a finite subcover.

From there to the general case of open sets the switch is simple, replace each open set by all the intervals it contains; find a finite subcover; replace each interval in the finite subcover by a particular open set which contains it.

• My first thought upon reading the question was that it might have been dependent on AC. Then I saw that you answered and that got my hopes way up. Then I read your answer and was a bit disappointed ;-). Nice answer, +1 from me! – Jason DeVito Apr 11 '13 at 18:25
• @Jason: Some time ago when I was working on an answer to some question here about compact spaces and the axiom of choice I was exploring all sort of compact constructions in choiceless contexts. One-points never failed to disappoint me (although they were useless in that context anyhow). – Asaf Karagila Apr 11 '13 at 18:32
• In what sense is the order topology "in fact much simpler" than the one-point compactification? – Martin Apr 11 '13 at 18:51
• @Martin: I don't know really. At the time it seemed like a much simpler solution. – Asaf Karagila Apr 11 '13 at 18:53
• Ah, set-theorists (to be) :-) +1 have your nice answer – Martin Apr 11 '13 at 18:54