In Munkres (Chapter 3, Section 23, p. 148), Munkres shows that if a subspace $Y$ of a space $X$ is not connected, then there are two disjoint open subsets $A,B$ such that the union of $A$ and $B$ contains $Y$. What doesn't make sense is that a separation of $Y$ only requires two open subsets of $Y$ which are disjoint, as subsets of $Y$. That is, $A$ and $B$ certainly can't intersect anywhere in $Y$, but who says they have to be disjoint? Or is asking whether a subset $Y$ of $X$ is connected as a subset of $X$ a different question than saying "okay, here is the topology on $Y$ as a subset of $X$  now is $Y$ connected under that topology?"

$\begingroup$ There are examples of topological spaces $X$ and a nonconnected subspace $Y$ such that there are no disjoint subspaces $A$ and $B$ of $X$ satisfying your conditions. Are you sure that's what Munkres actually says? $\endgroup$ – Robin Chapman Jul 4 '10 at 12:24

$\begingroup$ Aren't you missing the condition that the closure of $A$ should not intersect $B$ and viceversa? $\endgroup$ – José FigueroaO'Farrill Jul 4 '10 at 12:31

$\begingroup$ For the few of us that don't have Munkres handy, can you provide the relevant statements and definitions for your first sentence? In particular, I am thinking about the space X = {a,b,c} and Y = {a,c} with topology on X generated by {a,b} and {b,c}. So as you stated it, that can't possibly be a theorem (but you said "Munkres showed"). $\endgroup$ – Willie Wong Jul 4 '10 at 12:32

$\begingroup$ Your statement is exactly why I'm confused. Here is what Munkres writes, wordforword: "If Y is subspace of X, a separation of Y is a pair of disjoint nonempty sets A and B whose union is Y, neither of which contains a limit point of the other. The space Y is connected if there exists no separation of Y." I assume he means "disjoint" in X. $\endgroup$ – David Corwin Jul 4 '10 at 12:34

3$\begingroup$ A subspace $Y$ of a space $X$ is connected if it is connected in the subspace topology, i.e., if there do not exist disjoint open subsets $U$ and $V$ of $Y$ such that $Y=U\cup V$. Munkres is giving an alternative characterization of this definition. $\endgroup$ – Keenan Kidwell Jul 4 '10 at 13:09
Per your comment, I think you misunderstood what Munkres is trying to say.
If Y is subspace of X, a separation of Y is a pair of disjoint nonempty sets A and B whose union is Y, neither of which contains a limit point of the other. The space Y is connected if there exists no separation of Y.
I read it to mean two definitions are given. Firstly, that he defines what it means to have a separation of a subspace $Y$ inside of the space $X$. Then he defines a space $Y$ as connected if it cannot be separated within itself. ($Y$ is trivially a subspace of itself, so the first definition of a separation can be used.)
Now, in the example I gave where $X = \{a,b,c\}$, $Y = \{a,c\}$, with topology on $X$ generated by $\{a,b\}, \{b,c\}$, the subspace $Y$ is not a connected set in $X$, as it is not a connected space in its subspace topology. But the space $X$ is connected, so the connected component of $\{a\}$ in $X$ is the whole space. (Whereas the connected component in $Y$ is itself.)
This shouldn't be so strange if you consider a more intuitive example: Let $X$ be the open interval $(0,1)$, and $Y$ be the subset $(0,1/4)\cup (3/4,1)$. Then $Y$ is not connected. The connected component containing the point $1/8$ in $X$ is the whole space, whereas the connected component when considered in $Y$ is just the interval $(0,1/4)$.
In other words, the connected component of a point in $X$ is a subspace $Y$ such that $Y$ is connected in the subspace topology and such that $Y$ and $X\setminus Y$ are both open. (And $Y$ of course contains the point in question.)
Two sets are called disjoint if they have no element in common. Thus for two subsets of $Y\subset X$ there is no distinction to be made between "disjoint in $Y$ and "disjoint in $X$".
Call two sets in a space separated if (they are disjoint and) neither one contains a limit point of the other. By the nature of the subspace topology, for two subsets of the subspace $Y\subset X$ there is no distinction between "separated in $Y$" and "separated in $X$". That is, a point of $Y$ in in the closure of a subset of $Y$ from the point of view of $X$ if and only if this is true from the point of view of $Y$. So there is no ambiguity in asking whether $Y$ can be expressed as the union of two nonempty separated sets.
But note that "having disjoint closures" is a stronger condition on two subsets of $Y$, for which there would be ambiguity.
(My parenthesis above was to ward off a real ambiguity of language: some people use "limit point of $A$" to mean any point in the closure of $A$; others do not include isolated points of $A$ as limit points.)
I wonder if the questioner was giving "disjoint" some topological meaning. I have noticed that topology students sometimes get the idea that it has such a meaning. This may be suggested by topologists' habit of using the expression "the disjoint union of spaces" for what is sometimes called the topological disjoint union or topological sum, i.e. the coproduct of objects in the category of spaces.

$\begingroup$ But your definition of separated sets is weaker than the one Munkres is using. While yours is by all means a good definition, I don't think it will help the original poster with understanding what he read in Munkres' book. $\endgroup$ – Willie Wong Jul 4 '10 at 14:47

1$\begingroup$ I have not defined the term "separation" and Munkres has not defined the term "separated". After defining what I mean by a pair of sets in a space being separated, as I did above, I could go on to say that a "separation" of a set $Y$ in a space $X$ means a separated pair of nonempty sets $A$ and $B$ in $X$ whose union is $Y$. This agrees with Munkres's definition. $\endgroup$ – Tom Goodwillie Jul 4 '10 at 19:42

$\begingroup$ Oh, oops. I misunderstood your post. Mea culpa. $\endgroup$ – Willie Wong Jul 5 '10 at 10:51

$\begingroup$ Ok. I understand everything now. Just to point out, my main issue was that if we have two subsets of $X$, and their intersections with $Y$ are disjoint in $Y$, that doesn't mean they are disjoint in $X$. $\endgroup$ – David Corwin Jul 6 '10 at 9:23
 A subset of a topological space is naturally endowed with a topology, namely, the subspace topology.
NOTE: Whenever we speak of a topology on a subspace, unless specified otherwise, we mean this subspace topology. It seems natural to assume the following definition of a connected topological space (and Munkres does so) :
 A topological space X is connected if whenever any two nonempty open sets A and B of X cover X, then A \cap B is nonempty.
It follows that any subspace of X is connected if it is connected with respect to the induced (subspace) topology on it.
 A connected subspace is a subset which is a connected space wrt the induced topology. (A connected component is a maximal (wrt to inclusion) connected subset of X. )
Now, Munkres gives a characterization of connectedness of a subspace:
If Y is subspace of X, a separation of Y is a pair of disjoint nonempty sets A and B whose union is Y, neither of which contains a limit point of the other. The space Y is connected if there exists no separation of Y.
The proof given there is clear. One point to note is that the following are equivalent for subsets A and B of X:
 ...A and B whose union is Y and neither of which contains a limit point of other.
 A and B are both closed and open in Y and their union (in Y) is Y.

$\begingroup$ Your first bullet point is wrong. You need to add the condition such that $A\cup B = X$. Else by your definition almost no space is connected. $\endgroup$ – Willie Wong Jul 4 '10 at 13:25