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

[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?

 

[Moo Of Doom]

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?

 

[HallsofIvy]

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

 

[pantin]

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

you are right, but i still don't 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 because {x_k} is not compact.