Continuity characterization (metric spaces)

[Carl140]

1. The problem statement, all variables and given/known data

Let (X,d) and (Y,d') be metric spaces and f: X-> Y a continuous map.
Suppose that for each a>0 there exists b>0 such that for all x in X
we have:

B(f(x), b) is contained in closure( f(B(x,a))).

Here B(f(x),b) represents the open ball with centre f(x) and radius b.
Similarly B(x,a) represents the open ball with centre x and radius a.

Prove that for all x in X and for every c > a :

B(f(x), b) is contained in f(B(x,c)).

3. The attempt at a solution

No clue here, I took an y element in B(f(x),b) so d(f(x),y) < b.
Then by assumption B(f(x),b) is contained in closure(f(B(x,a)) so y
is in closure(f(B(x,a)), and then?

[Hao]

If you don't get an answer, this forum

http://www.mathhelpforum.com/math-help/

should help you immensely.

Questions in their section on analysis are typically very well answered.

[matt grime]

This is ripe for a proof by contradiction approach (which may even yield a direct proof).

If there is a c such that B(f(x),b) is not contained in f(B(x,c)) what does that mean?

[Carl140]

Hi Matt, thanks for your reply. Assume there exists a point q in B(f(x),b) such
that q is not in f(B(x,c)).

Now since c>a it follows that B(x,a) is contained in B(x,c).
Hence f(B(x,a)) is contained in f(B(x,c)).

Since q is not in f(B(x,c)) then q is not in f(B(x,a)).

But f is continuous so f(closure(B(x,a))) is contained in closure(f(B(x,a)).

I'm stuck here. What else should I do?