Answer to Question #115962 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

An open set $S$ in $R$ is such set, that if $x \in S$ then there exist such $\varepsilon >0$ , that $(x-\varepsilon, x+\varepsilon)$ $\in S$. Since for every $x \in (a, b)$ , $a < x<b$, then (if either $a\neq -\infin$ or $b \neq + \infin$) we will define $\varepsilon$ as $min(|x-a|, |x-b|)>0.$

If, for instance, $a = - \infin$ and $b \neq +\infin$, then $\varepsilon := |x-b|>0$, same for $a \neq - \infin, b = +\infin$. Notice, that $a$ can not be equal to $+\infin$, as well as $b$ can not be equal to $+ \infin$, since $a<b$. If $a= -\infin, b = + \infin$ simultaneously, we let $\varepsilon = 1$.

With such $\varepsilon$, by the definition, $(x- \varepsilon, x+\varepsilon) \in (a, b),$ if $x \in (a, b)$. Thus, $(a, b)$ is an open set.

A closed set $S$ in $R$ is such set, that its complement is open. Let's show that $[a, b]^{C}$ is open. Indeed, $[a, b]^{C} = (-\infin,a) \cup (b, +\infin)$. Notice, that If either $a = -\infin$, or $b = + \infin$, then there is only one interval in the complement.

$[a, b]$ is the set of following elements: $x \in R$, $a \le x \le b$ . We have already proved that open interval is an open set. Thus, since the union of open sets is also an open set, we obtan that $(-\infty,a)\cup (b, +\infty)$ is an open set too. (if there was only one interval in the complement, then the last sentence can be ommitted).

So, $[a, b]$  is a closed set.