(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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

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