# Discrepancy in terminology regarding the notion of compactness of topological spaces

Last semester we encountered the following definition of compactness in our first course on topology:

A topological space $X$ is said to be compact if every open cover of $X$ has a finite subcover.

This is also the definition found in Lee and Munkres. Now we have a first course on complex analysis, where we talked a bit about the topology on $\mathbb{C}$. The professor mentioned that he looked through the notes of our topology course and found a major mistake: The terminology of compactness is only established (or makes sense) for Hausdorff spaces, i.e. his definition would be (I also checked this in the literature, for example the definition occurs in general topology by Ryszard Engelking):

A topological space $X$ is said to be compact if $X$ is a Hausdorff space and every open cover of $X$ has a finite subcover.

This awoke my curiosity, so I looked up on wikipedia and found the following:

Some branches of mathematics such as algebraic geometry, typically influenced by the French school of Bourbaki, use the term quasi-compact for the general notion, and reserve the term compact for topological spaces that are both Hausdorff and quasi-compact.

So my question is:

• Why is this difference regarding the terminology of compactness?
• Is it really that bad to not demand Hausdorffness of a compact space?
• Part 1: the distinction is used only by Bourbaki and followers. Part 2: Kelley and followers don't distinguish. The main difference is that a compact subspace need not be closed in a non Hausdorff space. – egreg Feb 21 '17 at 21:02

## 3 Answers

The reason for the difference is the same as the reason for any difference in terminology in math: different people made slightly different definitions at different times in different places. There's not any deep reason.

There is also a linguistic divide in this case--in English, "compact" usually does not require Hausdorffness, but in French it usually does. If your professor works in French, that would explain why he called it a "mistake" to not include Hausdorffness.

So no, it's not really that bad to not demand Hausdorffness. Just be careful if you are working in a language other than English where the convention may differ, or if you are working in a field like algebraic geometry where it is common to require Hausdorffness even in English. Basically, if you ever see people saying "quasicompact", that means that they require compact spaces to be Hausdorff.

To provide some context for the situation in algebraic geometry, there are two different kinds of topologies that are used in algebraic geometry. First, there are topologies that are like the Euclidean topology, which are Hausdorff and for which compactness is a restrictive and powerful property. Second, there are Zariski topologies, which are almost always compact but rarely Hausdorff. So the compactness of a space in the Zariski topology does not mean that the geometric object you're thinking about is "compact" in a strong geometric sense (e.g., that it behaves similar to a space that is compact in the Euclidean topology). For this reason it is appealing to use the weaker term "quasicompact" for such topologies.

I would note also that compact Hausdorff spaces have a much more powerful theory than general compact spaces, due to results such as Urysohn's lemma that guarantee there are lots of continuous real-valued functions on any compact Hausdorff space. So for a lot of applications, compact Hausdorff spaces have a much nicer and conceptually different feel from general compact spaces, similar to how abelian groups differ from general groups.

It's actually quite rare for non-Hausdorff topological spaces to appear in analysis. Algebraic geometers, it seems, use them more often: the Zariski topology is generally not Hausdorff.

Many of the nice properties of compact sets that we are used to are only true in the Hausdorff case. Thus in a non-Hausdorff space, compact sets need not be closed. A continuous bijection from one compact space onto another need not be a homeomorphism if the second space is not Hausdorff.

• Ah thanks. Yes, in the first exercise sheet we do have a look at the Zariski topology. – TheGeekGreek Feb 22 '17 at 7:23

For what it's worth, I'll tell you why I chose not to include the Hausdorff property in the definition of compactness in my books. The first reason, of course, is the desirability of following the usual convention in English-language mathematical writing. But there's also a more substantive reason why I think this is the right convention: The most important theorem about compactness, in my opinion, is the fact that continuous images of compact sets are compact. If the definition of compactness included the Hausdorff property, this would not be true -- for example, let $X=\{0,1\}$ with the trivial topology, and define $f\colon[0,1]\to X$ by $f(1)=1$ and $f(x)=0$ if $x<1$. Then $f$ is continuous (when $[0,1]$ is given the standard topology), but its image would not be compact simply because it's not Hausdorff.

Even in realistic cases, such as quotient spaces, where the image is compact, in order to prove it's compact you'd also have to prove that it's Hausdorff. Who needs that?

There are plenty of theorems that apply only to compact Hausdorff spaces, or to locally compact Hausdorff spaces, but it's no big hassle just to add the Hausdorff hypothesis whenever it's needed.