Connectedness of Disjoint Union of Connected Sets

The definition of connected sets is:

A topological space $X$ is connected iff there do not exist sets $U, V \subset X$ such that: $U, V \neq \varnothing$, $U \cap V = \varnothing$ and $U \cup V = X$, with both $U$ and $V$ both open and closed.

I am having trouble applying this definition to certain cases--for example, the union of two intervals in the real number line with the usual topology.

Intuitively, $C=(0,1) \cup (2,3)$ should be disconnected (and I found a special definition of connectedness for open sets that allows me to prove that), but I don't see how to apply the actual definition of connectedness to prove that (or to prove, for example, the same problem with closed sets).

$C$ being disconnected should imply the existence of $U$ and $V$ satisfying the above property, but I can't find any.

• You do not have to check for both sets that they are each open and closed. You can check for one set that it is open and closed (then for the other set you don't have to check anything at all), or you check for both sets that they are both open, or that they are both closed. For if $V$ is open, then in the situation you describe, $U$ is the complement of $V$, so it must be closed. – Stefan Hamcke Oct 14 '13 at 14:04

Take $U=(0,1)$ and $V=(2,3)$: these sets are both open and closed in the space $C$. $(0,1)$ are open in $C$ because each is the intersection with $C$ of a set open in $\Bbb R$, and each is closed in $C$ because it’s the complement in $C$ of an open subset of $C$. The fact that neither is closed in $\Bbb R$ is irrelevant.
Choose $U=(0,1), V=(2,3)$. Then $C=U\cup V$, $U,V$ are open, and $U \cap V = \emptyset$. Since $X \setminus U = V$ and $X \setminus V = U$ we see that $U,V$ are closed as well. Hence $C$ is not connected.
If you want to see whether a subset $Y$ of a topological space $X$ is connected, and if you insist on using what you call the "actual" definition of connectedness, you have to apply it to the space $Y$ with its relative topology as a subspace of $X$; your so-called "actual" definition applies only to a topological space, not to a subset of a topological space.
Here is an equivalent definition which is easier to use: $X$ is connected iff there do not exist sets $U,V$ such that $U,V\ne\emptyset$, $U\cup V=X$, $U\cap\mathrm{Cl}(V)=\emptyset$, and $\mathrm{Cl}(U)\cap V=\emptyset$. In words, $X$ is connected iff it is not the union of two nonempty sets, each of which is disjoint from the closure of the other. The advantage of this definition is that you can apply it directly to a subset of a topological space without bothering with a relativized topology.