Compact topological space

[vvaann]

Let S be a compact topological space, let T be a topological space, and let f be a function from S onto T. Of the following conditions on f, which is the weakest condition sufficient to ensure the compactness of T?

A. f is a homeomorphism.
B. f is continuous and 1 - 1.
C. f is continuous.
D. f is 1 - 1.
E. f is bounded.

Answer key:

SPOILER: C

[eastside00_99]

Theorem: Let X and Y are topological spaces and f maps X to Y and C a compact subset of X. If f is continuous then f(C) is compact.

A homeomorphism is a continuous bijective function that has a continuous inverse.

So, by the above theorem, A and B are too much. Also, we are not necessarily dealing with metric spaces and boundedness is a property of metric spaces, and so E, although in another context could be correct, it is not in this context.

Now, I am not entirely sure about D; the best thing to do is to construct a counterexample. There exists an bijective function g: [0,1] --> ]0,1[ since these two sets have the same cardinality (give these sets the subspace topology). ]0,1[ is not compact. So the existence of the gijective function g does not imply compactedness.

Therefore, it is C.