Answer to Question #115548 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

1.$(a, b)$ is the set of following elements: $x \in R , a<x<b$ . By the definition, an open set in $R$ is such set, that for every element of this set there exists some open ball with centre in this element, such that this ball also belong to the set.

Thus, for any $x \in (a, b)$,since $a<x$, $x - a = r_1 >0$, and $x<b, b- x = r_2 >0.$ Let's denote with $r = min(r_1,r_2)$ . Then the open ball with the center in x, and radius r belongs to $(a, b)$. Hence, $(a, b)$ is an open set.

If only one of $a$ or $b$ is $-\infty, +\infty$ then we will let $r$ be either $r_2,$ or $r_1$ respectively. If $a = -\infty,$ and $b= +\infty$ simultaneously, then we can let $r = 1$ , and solution will be the same.

2.$[a, b]$ is the set of following elements: $x \in R$ , $a \le x \le b$ . By the definition,a closed set in $R$ is such set, which complement is an open set. The complement for a closed interval is:

$(-\infty,a)\cup (b, +\infty)$. As we proved in 1., both of these intervals are open sets. Thus, since the union of open sets is also an open set, we obtan that $(-\infty,a)\cup (b, +\infty)$ is an open set. Therefore, [a, b] is a closed set.