I know this looks like an obvious question, but I'm not exactly sure at the method of proof for this question and suspect it involves some topology (which I've never taken a formal course on).
Suppose $I = (a, b] \cup [c, d) \subset \mathbb{R}$ satisfy $a < b < c < d$. I wish to show that it cannot be written as a union of open intervals.
Consider $\mathbb{R}$ under the standard topology. Then $I$ is disconnected because it is not an interval (is there perhaps an easier way to see this without using this result?). Because it is disconnected, it cannot be written as a union of open intervals in $\mathbb{R}$.
Is my proof correct?
Best Answer
Being disconnected is not enough: $(0,1)\cup (2,3)$ is a union of open intervals $(0,1)$ and $(2,3)$. But it is disconnected.
In order to show that your $I$ is not a union of open intervals, first you need to know that open intervals are in fact open subsets and thus a union of open intervals is open as well. And so if $I$ is a union of open intervals, then $I$ is open.
We can then utilize the definition of an open subset in $\mathbb{R}$: a subset $U\subseteq\mathbb{R}$ is open if and only if for any point $x\in U$ there is $\epsilon >0$ such that $(x-\epsilon,x+\epsilon)\subseteq U$. In other words: a subset of $U\subseteq \mathbb{R}$ is open if for any point $x\in U$, our subset $U$ contains a small neighbourhood of $x$ as well.
So given $I=(a,b]\cup [c,d)$ with $a<b<c<d$ can you see why $I$ is not open? For example: why $I$ does not contain any open neighbourhood of $b$?