# show union of two intervals is not connected

Let $X$ be a topological space. $X$ is connected if $X\neq U\cup V$ with open sets $U,V$ and $U\cap V=\emptyset$.

If you consider $A:=(0,1]\cup(2,3)\subset\mathbb R$, $A$ is not connected.

But how can you prove it? Clearly I have to find those open sets like above but how? Thanks!

• $A$ is already written as the union of two intervals... And each interval is open in $A$. – N. S. Jun 5 '13 at 18:23
• Another useful result is: A set $E \subset \mathbb{R}$ is connected if and only if it has the following property: If $x \in E, y \in E$ and $x < z < y \implies z \in E.$ So an alternate proof (though completely unnecessary in this case) would be to note that $1 \in A$, $5 \over 2$$\in A and 1 <$$3\over 2$$< \frac{5}{2}$, but $\frac{3}{2} \not \in A$. So $A$ is not connected. – user70962 Jun 5 '13 at 18:45
• This is fine unless we're looking at the order topology on $A$, in which case it is connected. :) But I'm guessing that the OP meant the question as stated. – Ted Shifrin Jun 5 '13 at 19:07

Just take $U=(0,1]$ and $V=(2,3)$: those sets are both open in $A$. To see that $U$ is open in $A$, observe that it’s equal to $A\cap(0,2)$, where $(0,2)$ is open in $\Bbb R$.
• @BrianM.Scott thanks. But I don't get it why $U$ is open in $A$. Can you say a little bit more about it? – user81109 Jun 5 '13 at 18:43
• @psipi: It’s just the definition of the relative topology on a subset of a space. If $X$ is a topological space, and $A\subseteq X$, the relative (or subspace) topology on $A$ is $\{U\cap A:U\text{ is open in }X\}$. In other words, a set $V\subseteq A$ is (relatively) open in $A$ if and only if there is an open set $U$ in $X$ such that $V=U\cap A$. – Brian M. Scott Jun 5 '13 at 18:47
Note that in $A$ the interval $(0,1]$ is open. It is open because $(0,1]=(0,1.1)\cap A$, so it is relatively open.
Therefore $A$ is the union of two nontrivial open sets, and therefore not connected.