# What does a compact set look like?

So for a writing assignment in one of my classes we are asked to discuss and prove some basic results about compact sets in general topological spaces. I like proving these things, but they dont help me understand what a compact( locally compact, paracompact,...) set in a topological space "looks like." That said I'm asking for some examples of compact (locally compact, ...) sets in a variety of topological spaces. I'm also interested in some explanation of what extra "benefits" are brought about by singling out these compact sets. For example the p-adic numbers are locally compact and locally compact things (abelian groups to be precise) are a good setting in which to carry out fourier analyis.

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As you probably know: compact is an important notion because it singles out those spaces which share the basic properties of closed intervals of real numbers: continuous functions achieve a minimum and maximum, sequences have countable subsequences (here for simplicity of exposition I am ignoring the difference between compactness and sequential compactness), the intersection of a nested sequence of closed subsets is non-empty. Locally compact spaces are singled out because they share the property of the real numbers (or more generally of Euclidean spaces) that every point has a ... – Matt E Oct 20 '10 at 2:19
... neighbourhood which is compact, i.e. which has the basic analytic properties of closed intervals. – Matt E Oct 20 '10 at 2:21

You could try reading Terence Tao's notes on the subject; I found them very informative.

As for examples, here is an example and a non-example which I think are informative. The example is that, by Tychonoff's theorem, $[0, 1]^I$ is compact for any index set $I$. The non-example is that the closed unit ball in any infinite-dimensional Banach space, say $\ell_1(\mathbb{Z})$, is not compact.

Edit: Here is another perspective from which to think about compact Hausdorff spaces (and LCH spaces). A compact Hausdorff space $X$ is completely determined by the ring of continuous functions $C(X, \mathbb{R})$; this is a standard exercise which is worked out, for example, here. One gets the points of $X$ as the maximal ideals of $C(X, \mathbb{R})$ and the topology as the initial topology making every function in $C(X, \mathbb{R})$ continuous. Slightly generalized, this is the commutative Gelfand-Naimark theorem, and it says that studying commutative C*-algebras with unit is the same thing as studing compact Hausdorff spaces. Removing the requirement that we have a unit is the same thing as studying locally compact Hausdorff spaces.

So one can write down compact Hausdorff spaces by writing down commutative C*-algebras with unit. One choice is to take the space $C_b(X, \mathbb{R})$, the space of bounded continuous functions on an arbitrary topological space $X$, with the sup norm. If $X$ is completely regular Hausdorff, this gives the Stone-Cech compactification of $X$. (Which is hopeless to think about in general; even for $X = \mathbb{N}$ it's very weird, as the Wikipedia article describes.)

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Wow! What a notes! Thank you, Qiaochu. – a.r. Oct 20 '10 at 15:37

I don't know if I understand your question, but here are some elementary examples of compact and non compact spaces. You can find more examples in Wikipedia.

1. $[0,1]$ is compact with the subspace topology of the usual, Euclidean topology on $\mathbb{R}$.
2. Neither $(0,1)$, nor $\mathbb{R}$ are compact with the same topology.
3. A set with a finite number of points $A \subset \mathbb{R}$ is always compact. In fact, it is always compact, with no matter which topology.
4. With the usual topology on $\mathbb{R}^n$, $A \subset \mathbb{R}^n$ is compact if and only if it is bounded and closed. Beware: this characterization of compactness is no longer true for arbitrary metric spaces: for instance, with the bounded metric $d(x,y) = \mathrm{min} \left\{ \| x-y\|,1 \right\}$ (which is equivalent to the Euclidean one), every subset $A \subset \mathbb{R}^n$, even the whole $\mathbb{R}^n$, is bounded. So, if the former characterization was still true with this metric, every closed subset of $\mathbb{R}^n$ ($\mathbb{R}^n$ itself included) would be compact.
5. This countable set $\left\{\frac{1}{n}\ \vert \ n \in \mathbb{N} \right\} \cup \left\{ 0\right\} \subset \mathbb{R}$ is compact (with the usual topology).
6. This countable set $\left\{\frac{1}{n}\ \vert \ n \in \mathbb{N} \right\} \subset \mathbb{R}$ is not compact (with the usual topology).
7. This countable set $\mathbb{N} \subset \mathbb{R}$ is not compact (with the usual topology).
8. With the finite complement topology, $\mathbb{R}$ is compact. In fact, with this topology on the real line, every subset $A \subset \mathbb{R}$ is compact. In fact, every topological space with the finite complement topology is compact.
9. $\mathbb{Q}$ is not compact with the subspace topology of the usual, Euclidean topology on $\mathbb{R}$.
10. Btw, the empty set is always compact.
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