Answer to Question #115731 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

A subset $E \sube \R$ is called open set if every point of $E$ is an interior point of $E.$

Let $a,b\in \R$ be any two real number .

Let $I=(a,b)$ be an open interval.

Claim: $I$ is open .

Let $y\in I$ be an arbitrary real number then $a<y<b$ .

Let $c,d$ be two real number such that $a<c<y<d<b$

$\therefore (c,d)\sube(a,b).$

Let $r=min(|c-y|,|y-d|).$

Clearly $r>0$ as $y\neq c,d.$

Define , $r_0=\frac{r}{2}$ .

$\therefore N_{r_0}(y)=\{ x:|x-y|<r_0\}$ $=-r_0<x-y<r_0$

$=y-r_0<x<y+r_0=( y-r_0,y+r_0)$

$=\text{Neighborhood of y }$

Let $x\in N_{r_0}(y)$ then $y-r_0<x<r_0+y$ .

But $r_0<r$ $\implies c<y-r_0<y \ and \ y<y+r_0<d$ Therefore , $c<x<d.$

Hence , $x\in (c,d).$

$\therefore x\in N_{r_0}(y)\sube (c,d)\sube (a,b).$

Therefore every point of $(a,b)$ is an interior point of $(a,b).$

Hence , $(a,b)$ is open.

Consider the close interval $E=[a,b]$ ,where $a,b\in \R$ .

We known that a subset $E$ of $\R$ is close iff $E^c$ is open.

Now , $E^c=\{x:x\notin [a,b]\}$ $=\{ x: x>b \ or \ x<a\}$

$=(-\infin,a)\cup(b,\infin)$

Clearly $(-\infin,a) \ and \ (b,\infin)$ are open as they are open interval.

Again we known that Union of two open set is open ,hence $E^c$ is open .

Therefore , $E$ is close.