Is con't fn maps compact sets to compact sets converse true?

1. Oct 2, 2008

pantin

Is "con't fn maps compact sets to compact sets" converse true?

The question is here,
Suppose that the image of the set S under the continuous map f: s belongs to R^n ->R is compact, does it follow that the set S is compact? Justify your ans.

I already know how to prove the original thm, it requires us using another thm: Given S belongs to R^n, a belongs to S, and f: S->R^m, the following are equivalent:
a. f is con't at a.
b. For any {x_k}sequence in S that converges to a, the sequence {f(x_k)} converges to f(a).

If I need to prove the question on the top, I have to get the converse of this thm first.

And I see someone post a similar question before, please take a look as well:

"That f is continuous and that there is a continuous inverse, g, say.

So all we're doing is using the more basic fact that the continuous image of a compact set is compact.

Ie K compact implies f(K) compact, and f(K) compact imples gf(K)=K is compact."

Here, I agree this method, but I doubt this is not enough to prove my question, isn't it?

2. Oct 3, 2008

Moo Of Doom

Re: Is "con't fn maps compact sets to compact sets" converse true?

Indeed it is not enough to prove your assertion, as f may not have a continuous inverse. It may not have an inverse at all!

Do you think the converse is true or false? Have you tried looking for counter-examples?

3. Oct 3, 2008

HallsofIvy

Staff Emeritus
Re: Is "con't fn maps compact sets to compact sets" converse true?

Think about the function f(x)= 0 for x any member of Rn.

4. Oct 3, 2008

pantin

Re: Is "con't fn maps compact sets to compact sets" converse true?

you are right, but i still dont 100% get it.

my professor gave me a similar example, f(x)=c.

as you said, x can be any number, assume S={x_k}, f(S)=c, right?

but if a set is made up by the points on a line, say y=c, then this is not compact because it's not closed?
i am not sure here.
if the set of points on a line is counted to be a compact set, then you are right cuz {x_k} is not compact.