Question #115731
Prove that an open interval in R is an open set and a closed intervals is a closed set
1
Expert's answer
2020-05-19T20:08:13-0400

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

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

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

Claim: II is open .

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

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

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

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

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

Define , r0=r2r_0=\frac{r}{2} .

Nr0(y)={x:xy<r0}\therefore N_{r_0}(y)=\{ x:|x-y|<r_0\} =r0<xy<r0=-r_0<x-y<r_0

=yr0<x<y+r0=(yr0,y+r0)=y-r_0<x<y+r_0=( y-r_0,y+r_0)

=Neighborhood of y =\text{Neighborhood of y }

Let xNr0(y)x\in N_{r_0}(y) then yr0<x<r0+yy-r_0<x<r_0+y .

But r0<rr_0<r     c<yr0<y and y<y+r0<d\implies c<y-r_0<y \ and \ y<y+r_0<d Therefore , c<x<d.c<x<d.

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

xNr0(y)(c,d)(a,b).\therefore x\in N_{r_0}(y)\sube (c,d)\sube (a,b).

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

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

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

We known that a subset EE of R\R is close iff EcE^c is open.

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

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

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

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

Therefore , EE is close.




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