Answer to Question #116446 in Real Analysis for Sheela John

Question

Prove that an open interval in R is an open set and a closed intervals is a closed set

Answer 1

For definiteness, let us consider any open interval in $R$ is $(c,d)=\{x∈R∣c<x<d\}$

Furthermore, let $a \in (c,d)$ and $ϵ≤a−c\ and\ ϵ≤d−a$ .

Now $a \in (c,d)$ and recall that the ϵ-neighborhood of $a$ in the set $\implies x\in V_\epsilon (a)$ so $a-\epsilon <x<a+\epsilon$ .

If we take $ϵ=min\{a−c,d−a\}$, then $\epsilon ≤a−c\ and\ \epsilon≤d−a$ .

So, $c=a−(a−c)<a−ϵ<x<a+ϵ<a+(d−a)=d$

$\implies x∈(c,d) ⟹ V_ϵ(a)⊆(c,d)$ .

Hence, every open interval in $R$ is an open set.

Now, let $[c,d]$ be a closed interval in $R$ , so $[c,d]^c = (-\infin,c) \cup (d,\infin)$ . In part (i), we proved every open interval is an open set and we also know the union of two open sets is open. So complement of the given closed interval is open set. So, the given closed interval is a closed set.