Answer to Question #115964 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 $a∈(c,d)$ 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 $[c,d]$ is closed interval in $R$ , so $[c,d]^c = (-\infin,c) \cup (d,\infin)$ . In part (i), we proved every open interval is open set and also we known union of two open set is open. So compliment of given closed interval is open set. So, given closed interval is closed set.