Q:

Definitions of connected space

I have seen several definitions of connected space, but I would like to discuss those from Wikipedia. I am concerned about these:
$X$ is disconnected, if it is the union of two disjoint nonempty open sets.
What about $[0,1] \cup [2,3] $ ? It is not the union of two disjoint nonempty open sets, so it is connected?
$X$ is connected, when it cannot be divided into two disjoint nonempty closed sets.
What about $(0,1) \cup (2,3) $ ? It cannot be divided into two disjoint nonempty closed sets, so it is connected? 
Maybe I don't understand what "divided" means.

A:

For the first point, note that we are considering $X = [0,1] \cup [2,3]$ as a space in its own right; a subspace of the real line $\mathbb{R}$.  In particular, the open subsets of $X$ are of the form $W \cap X$ where $W$ is an open subset of $\mathbb{R}$.  This means that $U = (-1 , 2 ) \cap X = [0,1]$ and $V = ( 1 , 4 ) \cap X = [ 2,3 ]$ are both open subsets of $X$.  They are also disjoint and their union is $X$.  Therefore they witness that $X$ is disconnected.
A similar argument can be made about your second example.
In short, it is important to understand what we mean by open (and closed) sets in subspaces of familiar topological spaces.  It may be confusing to think of $[0,1]$ as open, but this is only because in this instance we are not considering it as a subset of the familiar real line, but something a bit different.

A:

The problem is not with your understanding of divided, but rather with your understanding of closed. In the space $X=(0,1)\cup(2,3)$, the sets $(0,1)$ and $(2,3)$ are closed. This is because the topology $\tau$ on $X$ is the subspace (or relative) topology inherited from $\Bbb R$. A subset $U$ of $X$ is open in $X$ if and only if there is a $V\subseteq\Bbb R$ such that $V$ is open in $\Bbb R$ and $V\cap X=U$. Of course $(0,1)$ is open in $\Bbb R$, and $(0,1)\cap X=(0,1)$, so $(0,1)$ is open in $X$. By the definition of closed set this means that $X\setminus(0,1)$ is closed in $X$. And $X\setminus(0,1)=(2,3)$, so $(2,3)$ is closed in $X$. A similar argument shows that $(0,1)$ is also closed in $X$. Indeed, both of these sets are clopen (closed and open) as subsets of $X$, even though they are only open as subsets of $\Bbb R$. Openness and closedness depend not just on the set, but on the space in which it is considered.
You have the same problem with your first example: the sets $[0,1]$ and $[2,3]$ are clopen in the subspace $Y=[0,1]\cup[2,3]$ of $\Bbb R$, not just closed. For example, $[0,1]=\left(-\frac12,\frac32\right)\cap Y$, and $\left(-\frac12,\frac32\right)$ is open in $\Bbb R$, so $[0,1]$ is open in $Y$.

