Metric Space: Closure of B(x,1/2) Examined

[ehrenfest]

[SOLVED] metric space

Homework Statement

If x and y are two points in a metric space and d(x,y) = 1, is it always true that the closure of B(x,1/2) does not contain y?

In general, is $$closure( B(x,r)) = \{z | r \geq d(x,z)\}$$

Homework Equations

The Attempt at a Solution

 

[ehrenfest]

Never mind. The answer is of course because closed balls are closed in a metric space. Why are closed balls in a metric space closed. Because their complement is open. Why is their complement open? Because, for any point x in the complement, let d be the distance from x to the closed ball. Then B(x,d/2) is a nbhd of x that lies in the complement. d is always nonzero and well-defined because otherwise x would be a limit point of the closed ball.

 

[morphism]

In general, is $$closure( B(x,r)) = \{z | r \geq d(x,z)\}$$

Hold on - this is not true. In the reals with the discrete metric, closure(B(x,1))={x}, while {z : 1>=d(x,z)}=R.

But what is true is that the closure of B(x,r) will always sit inside {z : r>=d(x,z)}.

 

[ehrenfest]

I see. Thanks.