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I feel that this is probably really obvious, but I don't know how to get started. Is it true that every set in a metric space is the union of connected, pairwise-separated sets? And does this generalize to topologies easily?

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Try to read about connected components. – Sigur Aug 8 '12 at 1:36
Thank you for the terminology, it was very helpful :) – Eric Stucky Aug 8 '12 at 2:32

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It does indeed generalize to arbitrary topologies, although it's a slightly weird definition, because it's defined a little backwards, in terms of what it isn't rather than in terms of what it is.

A separation of a topological space $\langle X, {\mathcal T}\rangle$ is two nonempty sets, $U$ and $V$, whose union is $X$, such that each is disjoint from the closure of the other. And we say that the space $X$ is disconnected if there is a separation of it, and connected if not.

So for example, $\Bbb R$ with the usual topology is the prototypical example of a connected space, but ${\Bbb R}\setminus \{0\}$ is disconnected: a separation of it is $U=(-\infty,0), V=(0, \infty)$. Similarly, $\Bbb Q$ with the usual topology is disconnected, since a separation is $U=(-\infty,\sqrt 2)\cap{\Bbb Q}, V=(\sqrt 2, \infty)\cap{\Bbb Q}$.

It's easy to show that a space $X$ is disconnected exactly when there is a nonempty subset of $X$ that is both open and closed. For example, in $\Bbb R$ with the usual topology, the only sets both open and closed are $\Bbb R$ and $\emptyset$, so once again we have the $\Bbb R$ is connected when it has the usual topology. On the other hand, when $\Bbb R$ is given the half-open interval topology (open sets are unions of half-open intervals $[a, b)$), the set $[0, 1)$ is both open and closed, and so $U=[0,1), V=(-\infty, 1)\cup[1,\infty)$ forms a separation and the space is disconnected.

Subsets of a topological space $X$ can be identified as connected or disconnected, if they are connected spaces themselves in the subspace topology inherited from $X$.

Every subset $S$ of a space can be considered to be a disjoint union of connected components, which are just the maximal connected subsets $S$, and it's exactly the connected components you are asking about. A connected subset $S$ has exactly one component, and a disconnected subset has more.

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Minor quibble: this definition implies that the empty space is connected, which I don't think is "right." The "right" definition is that a connected topological space is one with exactly one connected component (where we say that two points are in the same connected component if they cannot be separated in the sense of your definition). The empty space has zero connected components. – Qiaochu Yuan Aug 8 '12 at 2:25
I just double-checked in Kelley General Topology, and although he doesn't discuss the empty space specifically, I think I reproduced his definition accurately. He does say "…it follows at once that any indiscrete space is connected," which would include the empty space. It's on pp.52–53 if you want to look yourself. But perhaps the fashion of how to regard this case has changed since 1955 when the book was written. I don't think it is very important. – MJD Aug 8 '12 at 2:29
Thank you, this was very helpful. A small note: I was looking for unions of (pairwise) separated connected sets. But if they weren't separated, then they're probably not maximal, right? – Eric Stucky Aug 8 '12 at 2:32
Exactly right. If you have two connected subsets of $S$ that cannot be separated, then their union is a connected subset of $S$, and the two original sets were not maximal. Theorem 1.22 on page 55 of Kelley includes "If $A$ and $B$ are distinct [connected] components of a space, then $A$ and $B$ are separated". You might want to check it out. – MJD Aug 8 '12 at 2:34
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@Mark: I think it is natural to require that the decomposition of a space into its connected components is unique. This is false if you allow the empty space to be connected in the same way that unique prime factorization is false if you allow $1$ to be prime. There is an entry in the nLab about this general phenomenon: ncatlab.org/nlab/show/too+simple+to+be+simple – Qiaochu Yuan Aug 8 '12 at 3:54

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