Answer to Question #114827 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

Let's denote $A=(a,b)$ and $M=[a,b]$ are any arbitrary open and closed interval of $\mathbb{R}$ respectively.

Claim: $A$ is open set in $\mathbb{R}$

Proof:

We will show that every point of $A$ is interior point of $A$,hence $A$ will be open in $\mathbb{R}$ . Consider any arbitrary point $p \in A$ and $r_p>0$ , such that $B(p,r_p):=\{x \in \mathbb{R} |\: |x-p|<r_p\} \: (\clubs)$ is the open $r_p$ -ball of $p$ ,Clearly if we choose suitable small enough $r_p$ such $r_p$ -ball always exist then $B(p,r_p) \subset A$ ,Hence we are done.

Alternatively, observe that

$$A=\cup_{p \in A} \: B(p,r_p) \: (\spades)$$

Since, by the hypothesis,$B(p,r_p)$ is open set , therefore union of open set is open.Hence $A$ is open.

Now, consider the set $M$, we claim that $M$ is closed in $\mathbb{R}$ .

Consider the compliment of $M$ is $M'=\mathbb{R} \backslash M$ , if we show $M'$ is open in $\mathbb{R}$ we will be done.

Let, $m \in M' \implies m \notin M$ ,thus as defined in $(\clubs)$ , $B(m,r_m)$ and considering $(\spades)$ implies

$$M'=\cup_{m \in M'} B(m,r_m)$$

Hence, $M'$ is open in $\mathbb{R}$ as arbitrary union of open set is open set.Therefore, $M$ is closed in $\mathbb{R}$ .

We are done.