1. Aug 6, 2008

### boombaby

1. The problem statement, all variables and given/known data
If f is a continuous mapping of a metric space X into a metric space Y, Let E be any subset of X. How to show, by an example, that f($$\overline{E}$$) ($$\overline{E}$$ is the closure of E) can be a proper subset of $$\overline{f(E)}$$ ? And is there something wrong with my attempt below?

2. Relevant equations

3. The attempt at a solution
If E is compact, $$\overline{E}$$ = E, f(E) is compact, $$\overline{f(E)}$$ = f(E). Hence, f($$\overline{E}$$)= $$\overline{f(E)}$$
If E is not compact, $$\overline{E}$$ is closed and hence is compact, if E is bounded in R^{k}. f($$\overline{E}$$) is compact and hence $$\overline{f(\overline{E})}$$ = f($$\overline{E}$$).
since f(E) $$\subset{f(\overline{E})}$$ , $$\overline{f(E)}$$ $$\subset{\overline{f(\oveline{\overline{E}})}}$$ = f($$\overline{E}$$). It is also true that f($$\overline{E}$$) $$\subset{\overline{f(E)}}$$. Hence f($$\overline{E}$$) = $$\overline{f(E)}$$
In both cases, f($$\overline{E}$$) is not a proper subset of $$\overline{f(E)}$$

I've no idea of other kind of function that is continuous and with E otherwise defined.
Any hint would be greatly appreciated:)

2. Aug 6, 2008

### HallsofIvy

Staff Emeritus
?? The problem was to show that f($$\overline{E}$$) can be a proper subset of $$\overline{f(E)}$$ and you have given two examples where that is NOT true. What's "wrong with that" is that you haven't answered the question!

3. Aug 6, 2008

### Dick

Exactly. You want an example. There are perfectly fine examples from R->R. You don't need anything exotic. Can you find one? Hint, think about what happens if f has a horizontal asymptote.

4. Aug 6, 2008

### boombaby

to HallsofIvy,
I gave those two examples because I could not find a disired example :(

to Dick,
O I think I get it now...Let E=[1,$$\infty$$), f(x)=1/x and everything is done.....it is strange that I didn't get it...

Thanks a lot to all :)

5. Aug 6, 2008

### Dick

Not too strange. You were only thinking of bounded sets. This is called "thinking in the box". You're welcome.