According to R Engelking - General Topology:

A topological space $X$ is called a compact space if $X$ is a Hausdorff space and every open cover of $X$ has a finite subcover, i.e., if for every open cover ${\{U_s}\}_{s\in S}$ of the space $X$ there exists a finite set ${\{s_1,s_2, \dots , s_k}\}\subset S$ such that $X = U_{s_1}\cup U_{s_2}\cup \dots U_{s_k}$.*

And $^*$ means its footnote:

The reader should be warned that some authors do not include the assumption that $X$ is a Hausdorff space in the definition of compactness.

Actually, I have seen non-inclusion of Hausdorffness in definition of compactness in C Adams & R Franzosa's Introduction to Topology. Pure and Applied!

My questions are: If Engelking's book is right,

  1. Why Hausdorffness of a space as a requirement for compactness is omitted in (especially undergraduate) texts even though Hausdorffness is a concept that is taught in those books?!

  2. Would someone please give an example of a space that is compact according to the definition without Hausdorffness, and that is not a Hausdorff space?

Thank you.

EDIT - One definition must be correct; because compactness is equivalent to many other statements e.g. closeness+boundedness (?) ; Does any equivalence of compactness still hold if it is not Hausdorff?

EDIT 2 - I think Engelking's definition is wrong in the last sentence since it must be $X \subset U_{s_1}\cup U_{s_2}\cup \dots U_{s_k}$ (?)

  • $\begingroup$ See this for an example for 2. $\endgroup$ – David Mitra Jul 13 '15 at 8:19
  • $\begingroup$ My scope in not very broad. It is for me the first time that I see the extra condition that $X$ should be Hausdorff to be compact. I tend to hold my shoulders for it. Indiscrete spaces (with more than one element) are not Hausdorff and compact according to the definition that does not demand the space to be Hausdorff. $\endgroup$ – drhab Jul 13 '15 at 8:22
  • 4
    $\begingroup$ Boundedness does not even make sense if the space is not a metric space. $\endgroup$ – ajotatxe Jul 13 '15 at 8:39
  • $\begingroup$ Equivalence between compactness and closedness+boundedness holds in a very restrictive environment, namely the reals with the standard Euclidean metric (and products thereof). $\endgroup$ – egreg Jul 13 '15 at 8:40
  • 2
    $\begingroup$ Compact = bounded and closed is only true in certain spaces. Bounded is not even defined except for metric spaces, and metric spaces are all Hausdorff. $\endgroup$ – Thomas Andrews Jul 13 '15 at 12:45

There are many examples. The simplest is any space $X$ with more than one point that has the indiscrete topology, $\{\varnothing,X\}$. The simplest $T_1$ examples are any infinite set with the cofinite topology. The line with two origins is another $T_1$ example.

Many of us topologists feel that there is no good reason to include Hausdorffness in the definition of compactness and prefer simply to add the requirement and then talk about compact Hausdorff spaces when Hausdorffness is actually required for something. Those who prefer to include Hausdorffness in the definition of compactness seem to be mostly influenced either by Bourbaki or by category theory.

  • $\begingroup$ Category theory? Really? I never met it there. Again I feel forced to say that my scope is a small one, though. $\endgroup$ – drhab Jul 13 '15 at 8:27
  • $\begingroup$ @drhab: Not in category theory, so far as I know, but by people who tend to think in category-theoretic terms. Then again, I may just be overgeneralizing from exposure to algebraic geometers. $\endgroup$ – Brian M. Scott Jul 13 '15 at 8:59

I would argue that, overall in modern mathematics, the accepted meaning of "compact" is "every open cover has a finite subcover", i.e., it does not include the requirement of being Hausdorff.

The approach of including Hausdorff-ness in the definition of compact, and instead using the word "quasi-compact" for the less restrictive condition, is traditionally associated with French mathematicians and algebraic geometry. See for example Compact and quasi-compact, and quasi-compact and compact in algebraic geometry, and French notational differences.

  • $\begingroup$ Thank you also for the collection of quotes in here $\endgroup$ – L.G. Jul 13 '15 at 8:52

There is no right or wrong in these cases. I believe that Bourbaki has “quasi compact” for the non Hausdorff case, but the terminology can be seen elsewhere.

The book where I learned topology is Kelley’s, where the Hausdorff property has its importance, but is not assumed throughout. Perhaps Engelking has his reasons for including the Hausdorff property when talking about compact spaces: indeed, it simplifies certain results because a compact subspace of a Hausdorff space is closed.

The prime example of a compact non Hausdorff space is any infinite set with the cofinite topology. Such a space is obviously compact, because the non empty open subsets have finite complement, so any open cover admits a finite subcover.

The simplest proof I remember of the equivalence between Zorn's lemma and Tychonov's theorem uses the cofinite topology, so non Hausdorff compact spaces do have their relevance.

Other important non Hausdorff compact spaces arise in algebraic geometry, when the Zariski topology is considered.


For 2, take for example:

$$X=\{a,b\}$$ $$\tau=\{\emptyset,X\}$$

Every finite space is compact, so $X$ is compact (according to your second definition), but $X$ is not Hausdorff because there is no open set that contains $a$ but not $b$.

To include a less trivial example, let $(X,\tau)$ be any compact and Hausdorff space with more than one point. Now select any $x\in X$, and remove from the topology every neighbourhood of $x$, except $X$ itself. That is, define $$\tau'=\{U\in\tau: x\notin U\}\cup\{X\}$$

This is indeed a topology, because arbitrary unions and finite intersections of sets from $\tau'$ are certainly in $\tau'$.

The identity $i:(X,\tau)\to (X,\tau')$ is continuous, because the second topology is croaser than the first, so $(X,\tau')$ is compact. But the second topology is not Hausdorff, since $x$ can not be separated from any other point.

  • 1
    $\begingroup$ In fact, $X$ can be any space containing at least two points; it need not be finite. In this case, if $X$ is endowed with the indiscrete topology $\tau=\{\varnothing, X\}$, then $X$ is compact but not Hausdorff. $\endgroup$ – triple_sec Jul 13 '15 at 8:16

Others have already explained the differences in usage of various conventions. My first course in topology was taught by Engelking so perhaps I can say something about his motivations for this choice. He had very specific standards for notation and terminology. In particular, if I recall correctly, one guideline was that notions that are used more often deserve shorter names. If Engelking decided to exclude hausdorffness from the definition of a compact space, then the phrase "compact Hausdorff" would occur much more often in his book than "compact" on its own and that would be a waste of words. Hence he made a more economical choice of including this condition in the definition.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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