Example of a connected set

Question

Can someone please give an example of a connected set? The formal definition is that if the set $X$ cannot be written as the union of two disjoint sets, $A$ and $B$, both open in $X$, then $X$ is connected. I cannot visualize what it means. For example, a convex set is connected. Why is it so? I looked up in the internet, it was of no help. Any help would be appreciated.

Thanks in advance!

Answer 1

The first step is to think about the definition in the usual metric topology on the real line $\Bbb{R}^1$. Here, for example, any finite open interval is connected, because if you try to separate it into two disjoint smaller open intervals, and say interval $A$ contains some point that is less than a point in interval $B$, then $x= \inf \tilde{A}$ is not in $A$ because arbitrary reals greater than $\inf \tilde{A}$ are all in $\tilde{A}$ hence not in $A$. And $x$ is not in $B$ since if it were, for a sufficiently small by non-zero $\epsilon$, the $\epsilon$-neighborhood of $x$ would be in $B$; but the part of that neigborhood on the lesser side of $x$ is also in $A$ so the sets would not be disjoint.

Let's move on to $\Bbb{R}^2$, which is much more interesting.

For a convex set $K$, let's say we divide $K$ into two disjoint open sets $A$ and $B$. Now choose any arbitrary points $a \in A$ and $b \in B$. By the definition of convexity, every point on the straight line $L$ between $a$ and $b$ is in $K$. Now let's look at $C = L \cap A$ and $D = L \cap B$. Since $L$ is topologically equivalent to a line segment in $\Bbb{R}^1$, it is connected, so for the disjoint open sets $C$ and $D$ there is a point $x$ that is neither in $C$ nor in $D$. $x$ demonstrates that the disjoint open sets $A$ and $B$ cannot cover $K$. So the convex set $K$ is connected. (This proof holds in any $\Bbb{R}^n$.)

But of course convexity is a much stronger condition than is needed for a set to be connected.

Answer 2

By the way, the reason $[1,2] \cup [3,4]$ is disconnected is that the definition really says that when considered as a subspace topology, the set cannot be written as the union of disjoint subsets.

Now in the subspace topology, the interval $[1,2]$ is in fact an open set, since any sufficiently small neighborhood in that subspace of any point $p$ in that interval lies entirely in that interval. In particular, if we take $p = 2$, then any neighborhood with $\epsilon < 1$ lies entirely in $[1,2]$ because the points between 2 and 3 are not in the subspace at all. So the decomposition into two disjoint open sets $[1,2] \cup [3,4]$ shows that this set is in fact not connected.