Question #115208
Prove that an open interval in R is an open set and a closed intervals is a closed set
1
Expert's answer
2020-05-18T12:00:47-0400

Lemma: Arbitrary union of open set is open set.

Proof:

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

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

is open. Consider any arbitrary pU    p \in U \implies \exist at least one AjA_j such that pAjp \in A_j which implies that there exist an open ball of radius rpr_p such that B(p,rp)AjB(p,r_p) \subset A_j (Since, by our hypothesis every AiA_i is open iI\forall i\in I ,hence, AjA_j is open for some i=jIi=j \in I .)which implies that

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

Thus, pp is interior point of UU and as pp is arbitrary, therefore all points of UU are interior points .Hence UU is open.


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

Claim: AA is open set in R\mathbb{R}

Proof:

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


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

Therefore, from the above Lemma, implies AA is open.


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

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


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