Continuous functions on dense subsets

[Eulogy]

Hi, can someone give me pointers on this question

Homework Statement

Prove or provide a counterexample: If f : E -> Y is continuous on a
dense subset E of a metric space X, then there is a continuous function
g: X -> Y such that g(z) = f(z) for all z element of E.

The Attempt at a Solution

I'm not sure if the statement is true or not. I have tried to find counter-examples using continuous functions on the rationals or irrationals. For example f: Q -> R , f(x) = x. This is continuous for every x in Q. However it is easy to find a mapping g: R -> R which is continuous and g(z) = f(z) for all z element of Q. ie given by g(x) = x. I am yet to find a counter-example (in R anyway). However if the statement holds I'm not to sure how I would begin to prove it.

Thanks!

 

[HallsofIvy]

There is a reason you can't find a counter-example!

Let a be a point in X that is not in E. Since E is dense in X, there exist a sequence of points, $\{x_n\}$ in E that converges to a. Show that, since f is continuous in E, the sequence $\{f(x_n}\}$ converges and define that limit to be f(a).

 

[micromass]

There is a counterexample.

Take E=]0,1] and X=[0,1].

Can you find a function on E whose limit to 0 does not exist?

 

[Eulogy]

Thanks guys, makes a lot more sense now!