Answer to Question #115791 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

open real interval is an open set:

proof:

Let (a,b) be an open interval in R.

Let z∈ R be such that a<z<b

define δ such that δ< min{b-z, z-a} , then δ >0

Let B($\delta$  , z) be an open ball in R

now, since z+ < b and a< z-$\delta$  , so,

B($\delta$  ,z) lies in (a,b)

hence, there exists a neighborhood in (a,b)

thus, each point of (a,b) is an interior point

Hence,(a,b) is an open set.

Therefore, an open real interval is an open set in R.

closed real interval is a closed set:

proof:

Let [a,b] be a closed real interval.

[a,b]^c = ( -$\infty,....a) \cup (b,...\infty)$

Since $(-\infty,...a) and (b,.....\infty)$ are open intervals, they are open sets in R.

Since the union of open sets in R is an open set in R.

Therefore $(-\infty,...a) \cup (b,...\infty)$ is an open set in R

That is [a,b]^c is an open set in R.

As the complement of the open set is a closed set in R.

([a,b]^c)^c = [a,b] is a closed set in R.

Hence, a closed real interval is a closed set in R.