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.
 A: 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.
