Let X d be a metric space and F X Show that diamF diamF Sol

Let (X, d) be a metric space, and F X. Show that diam(F) = diam(F).

Solution

We have given that (X,d) is a metric Space, and F X.

Let us see the definition of the diam(F).

Definition :- Let (X,d) be a metric space and FX. The diameter 0 diam F of F is

diam F = sum{d(x,y) : x,y F}.

It follows from the definitions that F is bounded if and only if diam F < .

If F is open, it does not contain any of the (common) boundary points. Hence they all belong to F^c..

And So F^c must be closed.

Conversely, if F^c is closed, it contains all boundary points, and hence F can not have any.

This means that F is Open.

Assume that F is closed and that \"a\" does not belong to F. We must show that a sequence from F cannot converge to a. Since F is closed and contains all its boundary points,

a has to be an exterior point, and hence there is a ball B(a; )

around a which only contains points from the complement of F.

But then a sequence from F can never get inside B(a, ), and hence cannot converge to a.

This shows that diam(F) =diam (F^c).