Question #114827
Prove that an open interval in R is an open set and a closed intervals is a closed set
1
Expert's answer
2020-05-15T18:02:39-0400

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

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

Proof:

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

Alternatively, observe that

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

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


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

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

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



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

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

We are done.


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