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How do I prove that the complement of the closed interval $[a,b]$ is an open set.

I have a theorem that says an open set is a union of open intervals.

Can I simply say the complement of the closed interval $[a,b]$ is $(-\infty, a) \cup (b, \infty)$?

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    $\begingroup$ Yes, you are correct. $\endgroup$ – Suhail Mar 26 '15 at 12:45
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    $\begingroup$ You are indeed right, but you are not using the correct argument. You should you the reciprocal of that theorem (wish is also true), that a union of open intervals is an open set. $\endgroup$ – João Rocha Mar 26 '15 at 13:12
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I denote $A=[a,b]^c$ the complement of $[a,b]$. $$x\in A\implies x\notin[a,b]\implies \exists \delta>0: ]x-\delta,x+\delta[\cap [a,b]=\emptyset$$ and thus $]x-\delta,x+\delta[\subset A.$

Therfore, $A$ is open.

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Yes, that works. You would want to say more explicitly that $[a,b]$ is closed (in $\mathbb R$) because its complement (in $\mathbb R$) is open.

You can also prove it strictly by definition.We want to show that if a sequence $\{x_n\} \in [a,b] \subset \mathbb R$ converges to some limit $x \in \mathbb R$, then $x \in [a,b]$.

Suppose to the contrary that a closed interval $[a,b]$ is not closed. Then there exists some sequence $\{x_n\} \in [a,b]$ which converges to $x \in \mathbb R$ but $x \not \in [a,b]$. Then either $a \in (-\infty, a)$ or $a \in (b, \infty)$.

Then the distance between $x$ and either of the endpoints $a$ or $b$ must be nonzero. Let $s = min \{|x-a|, |x-b|\} > 0$ by construction. Then $B_s(x)$, the open ball around $x$ with radius $s$, contains zero points in $[a,b]$.

This contradicts the definition of a limit point. $\textit Any$ open ball around a limit point of a set must contain infinitely points of that set. (Recall the definition of a limit of a sequence in $\mathbb R$. If $\{x_n\}$ converges to $x$, then for any $\epsilon > 0$, there exists an $N$ such that $\forall n \ge N$, $|x_n - x| < \epsilon$. But here, if you set $\epsilon$ equal to $\frac s 2$, no such $N$ exists. In other words, you cannot pick a number in the interval $[a,b]$ close enough to $x$ to "bridge the distance.")

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