A subset E⊆R is called open set if every point of E is an interior point of E.
Let a,b∈R be any two real number .
Let I=(a,b) be an open interval.
Claim: I is open .
Let y∈I be an arbitrary real number then a<y<b .
Let c,d be two real number such that a<c<y<d<b
∴(c,d)⊆(a,b).
Let r=min(∣c−y∣,∣y−d∣).
Clearly r>0 as y=c,d.
Define , r0=2r .
∴Nr0(y)={x:∣x−y∣<r0} =−r0<x−y<r0
=y−r0<x<y+r0=(y−r0,y+r0)
=Neighborhood of y
Let x∈Nr0(y) then y−r0<x<r0+y .
But r0<r ⟹c<y−r0<y and y<y+r0<d Therefore , c<x<d.
Hence , x∈(c,d).
∴x∈Nr0(y)⊆(c,d)⊆(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∈R .
We known that a subset E of R is close iff Ec is open.
Now , Ec={x:x∈/[a,b]} ={x:x>b or x<a}
=(−∞,a)∪(b,∞)
Clearly (−∞,a) and (b,∞) are open as they are open interval.
Again we known that Union of two open set is open ,hence Ec is open .
Therefore , E is close.
Comments