Question #115965
Prove that an open interval in R is an open set and a closed intervals is a closed set
1
Expert's answer
2020-05-21T16:29:48-0400

.ANSWER.

1)An open interval is a set (a,b)={xR:a<x<b}(a,b)=\left\{ x\in R:a<x<b \right\} .Let c(a,b)c\in (a,b) and ε=min{bc2,ca2}\varepsilon =min\left\{ \frac { b-c }{ 2 } ,\frac { c-a }{ 2 } \right\} .ε<bc2\varepsilon <\frac { b-c }{ 2 } and ε>c+a2-ε>\frac { -c+a }{ 2 } . Therefore

a=a+a2<c+a2=2c+ac2=c+ac2 <cε <c+ε<c+bc2=b+c2<2b2=ba=\frac { a+a }{ 2 } <\frac { c+a }{ 2 } =\frac { 2c+a-c }{ 2 } =c+\frac { a-c }{ 2 } \ <c-\varepsilon \ <c+\varepsilon <c+\frac { b-c }{ 2 } =\frac { b+c }{ 2 } <\frac { 2b }{ 2 } =b .

So, for all c(a,b)c\in (a,b)\quad exist ε>0 , such that (cε ,c+ε)(a,b)(c-\varepsilon \ ,c+\varepsilon )\subset (a,b) .By the definition of an open set , any open interval is an open set.

2)Similarly, the sets (,a)={xR:<x<a}(-\infty ,a)=\left\{ x\in R:-\infty <x<a \right\} and (b,+)=={xR:b<x<+}(b,+\infty )==\left\{ x\in R:b<x<+\infty \right\} are open sets . Because , for all c (,a)c\in \ (-\infty ,a) exist ε=c+a2\varepsilon =\frac { -c+a }{ 2 },(c+ε=c+a2<a+a2=a)(c+\varepsilon =\frac { c+a }{ 2 } <\frac { a+a }{ 2 } =a) , such that (cε ,c+ε)(,a)(c-\varepsilon \ ,c+\varepsilon )\subset (-\infty ,a) .

If d(b,+)ε=db2(dε=d+b2>b)d\in (b,+\infty )\quad \varepsilon =\frac { d-b }{ 2 } \quad (d-\varepsilon =\frac { d+b\quad }{ 2 } >b) , then (dε,d+ε)(b,+)(d-\varepsilon \quad ,d+\varepsilon )\subset (b,+\infty \quad ) .

3) A closed interval is a set [a,b]={xR:ax b}[a,b]=\left\{ x\in R:a\le x\ \le b \right\} =R((,a)(b,+))=R\diagdown \left( (-\infty ,a)\cup (b,+\infty ) \right)

That is, the closed interval is the complement of the set I=(,a)(b,+)I=(-\infty ,a)\cup (b,+\infty ) .From 2) it follows that the set II is open. Therefore, by the definition of the closed set, the set

[a,b][a,b] is closed.


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