$(a, b] cup [c, d)$ cannot be written as a union of open intervals

Question

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?

freakish · Accepted 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 $Usubseteqmathbb{R}$ is open if and only if for any point $xin U$ there is $epsilon >0$ such that $(x-epsilon,x+epsilon)subseteq U$. In other words: a subset of $Usubseteq mathbb{R}$ is open if for any point $xin U$, 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 $b$?

Kavi Rama Murthy · Answer

$(0,1) cup (2,3)$ is a disconnected but it is a union of open intervals. So your argument fails.
Any union of open intervals is open. But the given set is not open. Hence you cannot express it as  union of open sets.

mathcounterexamples.net · Answer

Your proof is not correct. For example $J = (a,b) cup (c,d)$ is the union of $(a,b)$ and $(c,d)$. $J$ is open and disconnected.
However, the union of open intervals is an open subset of the real. As it is not the case of $I$, $I$ is not the union of open intervals.

$(a, b] cup [c, d)$ cannot be written as a union of open intervals

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