- #1
pantin
- 20
- 0
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?
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?