Answer to Question #115208 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

Lemma: Arbitrary union of open set is open set.

Proof:

Let fix $I=\{1,2,3...\}$ and denote an open set $A_i$ for each $i \in I$ . Now we will show that

$$U=\cup_{i=1}^{\infty} A_i$$

is open. Consider any arbitrary $p \in U \implies \exist$ at least one $A_j$ such that $p \in A_j$ which implies that there exist an open ball of radius $r_p$ such that $B(p,r_p) \subset A_j$ (Since, by our hypothesis every $A_i$ is open $\forall i\in I$ ,hence, $A_j$ is open for some $i=j \in I$ .)which implies that

$$B(p,r_p) \subset \cup_{i=1}^{\infty}A_i \\ \implies B(p,r_p) \subset U$$

Thus, $p$ is interior point of $U$ and as $p$ is arbitrary, therefore all points of $U$ are interior points .Hence $U$ is open.

Let us denote $A=(a,b) \: \& S=[a,b]$ open and closed intervals respectively.

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

Proof:

Let, $\forall q \in A$ , $B(q,r_q)$ is a open ball centered at $q$ of radius $r_q >0$ , Thus, clearly

$$A= \cup_{q \in A} B(q,r_q)$$

Therefore, from the above Lemma, implies $A$ is open.

Now, consider $B'=\mathbb{R} \backslash B$ is the complement of $B$ ,thus if we prove that $B'$ is open then $B$ is a closed set in $\mathbb{R}$ .

Clearly, $B'=(- \infty,a) \cup (b,\infty)$ ,therefore again from the above Lemma $B'$ is open (since, $(-\infty,a) \: \& (b,\infty)$ are both open intervals).Hence, we are done.