Answer to Question #117246 in Real Analysis for Salik

Question

Prove that
a) Any Open ball is an open set
b) Any closed ball is closed set.

Answer 1

1. ANY OPEN BALL IS AN OPEN SET

proof: Let B(x₀,r) be an open ball with center x₀ and radius r in a metric space (X,d)

Let y€B(x₀,r)

Define r₁ =r-d(x₀,y)

We claim that B(y,r₁) belongs to B(x₀,r)

To see this let z€B(y,r₁)

Then

d(z,x₀) < d(z,y) +d(y,x₀) < r₁+(r-r₁)

So z€B(x₀,r)

So B(y,r₁) belongs to B(x₀,r)

Hence B(x₀,r) is an open set.

2. ANY CLOSE BALL IS A CLOSED SET

proof: Let B̅(x,r) be a closed ball. We prove that B̅'(x,r)= C(say) is an open ball.

Let y€C then d(x,y)>r

Let r₁=d(x,y) then r₁>r and take r₂=r₁-r

Consider the open ball B(y,r₂/2) we prove that B(y,r₂/2) belongs to C.

For this let z€B(y,r₂/2) then d(z,y)<r₂/2

By triangular inequality

d(x,y)<d(x,z)+d(z,y)

d(x,y)<d(z,x)+d(z,y)

d(z,x)>d(x,y)-d(z,y)

d(z,x)>r₁-r₂/2 = (2r₁-r₂)/2

d(z,x)> (2r₁-r₁+r)/2 , r₂=r₁-r

d(z,x)>(r₁+r)/2

d(z,x)>(r+r)/2 , r₁>r

d(z,x)>r

z does not belong to B̅(x,r)

This shows that z€C

B(y,r₂/2) belongs to C.

Hence C is an open set. So B̅(x,r) is closed.