For instance, I know that for a general topological space ($X, \mathcal{T}$), a subset $K \subset X$ is compact by definition if every open cover of $K$ has a finite subcover.

I guess what I am trying to ask is the following: We know that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded. But for general topological spaces ($X, \mathcal{T}$) and a subset $K \subset X$, this may not necessarily hold, correct?

In other words, am I wondering if for general topological spaces a subset of our space can be compact but may not necessarily be closed?

My book states the following proposition: A subset of a topological space is compact $\iff$ it is compact in itself (with respect to the relative topology). Is this the best we can do as a characterization of compactness for subsets of general topological spaces??

Thank you.


To answer the first question, a compact subset of a Hausdorff space must be closed, but a compact subset of a non-Hausdorff space need not be. Let $X$ be any infinite set, and give $X$ the cofinite topology; then every subset of $X$ is compact, but the only closed subsets besides $X$ itself are the finite subsets.

There are some characterizations of compactness in general topological spaces, however.

Theorem. Let $X$ be a topological space. The following are equivalent:

  • $X$ is compact.
  • Every centred family of closed subsets of $X$ has non-empty intersection. (A centred family is one in which every finite subfamily has non-empty intersection.)
  • Every filter on $X$ has a cluster point.
  • Every filter on $X$ is contained in a finer convergent filter.
  • Every ultrafilter on $X$ converges.
  • Every net in $X$ has a cluster point.
  • Every net in $X$ has a convergent subnet.
  • Every universal net in $X$ converges.
  • Every infinite subset of $X$ has a complete accumulation point. (This is less familiar; you’ll find a proof here.)
  • $\begingroup$ And every infinite subset $A$ of $X$ has a complete point of accumulation. (where for every neighbourhood $O$ of that point $|O \cap A| = |A|$ in cardinality). $\endgroup$ – Henno Brandsma Dec 11 '16 at 13:41
  • $\begingroup$ @Henno: Thanks; I’d forgotten about that one. $\endgroup$ – Brian M. Scott Dec 11 '16 at 20:29
  • $\begingroup$ Nitpicking; Centered family: Every non-empty finite family has non-empty intersection, if you define $\cap \emptyset=\emptyset.$ $\endgroup$ – DanielWainfleet Dec 13 '16 at 17:24
  • $\begingroup$ @user254665: But you don’t: since you’re working entirely with subsets of $X$, $\bigcap\varnothing=X$. $\endgroup$ – Brian M. Scott Dec 13 '16 at 20:24

Any finite topological space is compact, but there exist topological spaces in which not every finite subset is closed.

Take, for instance, the Sierpinski space $\Sigma_2$: the set of points of $\Sigma_2$ is $\{0,1\}$ and the open sets are $\emptyset$, $\{0\}$ and $\{0,1\}$. Then $\{0\}$ is finite - hence compact - but is not closed.


In addition to the other answers, if $d$ is a metric on $X$ then the metric $e(x,y)=\min (1,d(x,y)$ generates the same topology that $d$ does. And every subset of $X$ is bounded with respect to $e.$ So boundedness is not a topological property: It cannot generally be determined, by examining the topology generated by $d,$ whether a subset of $X$ is bounded or not. But compactness is a topological property.

Consider the case $X=\mathbb R$ with $d(x,y)=|x-y|$ and $e(x,y)=\min (1,|x-y|).$ With respect to $e,$ every subset of $X$ is bounded.

And also: When $(X,d)$ is a metric space, the topology on $X$ is generated by some unbounded metric $f$ iff $X$ is non-compact.


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