# 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.

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 $Y$.

• Oh, I see! So, in the definition of disconnected set, instead of "two disjoint nonempty open sets", there should be "two disjoint nonempty subsets of X open in X". Commented Jun 20, 2013 at 23:10
• @Ivan: Yes, that’s exactly right. The same goes for the closed set definition: they’re closed in $X$. Commented Jun 20, 2013 at 23:16
• @BrianM.Scott How can the two definitions of connectedness using the word "open" and also "closed" be the same? Wikipedia says the two disjoint sets must be closed for example- and my lecture notes use "open, disjoint, nonempty"- why do both definitions work out? I am asking this for arbitrary sets and not just for this particular example. Commented Feb 8, 2016 at 17:44
• @Arcane1729: (Apparently I didn’t see this comment seven years ago, but I’ll answer it now for the benefit of future readers.) The definitions are equivalent. Suppose that $X=A\cup B$, where $A$ and $B$ are disjoint and non-empty. If $A$ and $B$ are open, then $B=X\setminus A$ and $A=X\setminus B$ are also closed, and if $A$ and $B$ are closed, then $B=X\setminus A$ and $A=X\setminus B$ are also open. Commented May 3, 2023 at 18:41

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.

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.
[0,1] is open because $X=[0,1] \cup [2,3]$ has the subspace topology from $\mathbb R$, and say $[0,1]= (-1,\frac{3}{2}) \cap X$. For the same reason, $[2,3]$ is open as well. So $X$ is visibly the disjoint union of nonempty open sets.