Answer to Question #115965 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

.ANSWER.

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

$a=\frac { a+a }{ 2 } <\frac { c+a }{ 2 } =\frac { 2c+a-c }{ 2 } =c+\frac { a-c }{ 2 } \ <c-\varepsilon \ <c+\varepsilon <c+\frac { b-c }{ 2 } =\frac { b+c }{ 2 } <\frac { 2b }{ 2 } =b$ .

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

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

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

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

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

$[a,b]$ is closed.