Hi, I am going to fight MathSub in Nov. And pls help me this rub ..

64. Let S be a compact topological space, let T be a t.s., 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.

The answer is C.

Can you explain the reason?

sfmathgre.blogspot.com only offered solutions of 05&97

And it is the derivative, tested in 2004:

f: X -> Y is continuous bijection.

I. if X is compact then Y is compact

II. if X is Hausdorff, then Y is Hausdorff

III. if X is compact and Y is Hausdorff, then f^(-1) (the inv of f) exist.

the answer provided by student of 04 is only III is right

Why the first statement is wrong?

## Problem and its derivative from GR8767#64

### Re: Problem and its derivative from GR8767#64

I thought the statement I is true.mhyyh wrote:And it is the derivative, tested in 2004:

f: X -> Y is continuous bijection.

I. if X is compact then Y is compact

II. if X is Hausdorff, then Y is Hausdorff

III. if X is compact and Y is Hausdorff, then f^(-1) (the inv of f) exist.

the answer provided by student of 04 is only III is right

Why the first statement is wrong?

According to http://www.google.com/url?sa=t&rct=j&q= ... gug5bEGCkg,

Theorem. If f : X → Y is a continuous surjection and X is compact, then Y is

compact.

### Re: Problem and its derivative from GR8767#64

I. is true and not hard to prove.

Let {U_alpha} be an open cover of Y. Then {f^{-1}(U_alpha)} is an open cover of X because f is continuous. Since X is compact, it has a finite subcover {f^{-1}(U_i)}. Then {U_i} is an open cover of im(f), and im(f) = Y since f is surjective.

Back to your first question: the continuous image of a compact space is compact, so continuity of f is certainly sufficient. f being 1-1 (in addition to being onto by hypothesis) doesn't help because bijections need not preserve any topological properties of a space. For example, the "identity map" from the interval [0,1] with the Euclidean topology to [0,1] with the discrete topology is a bijection but not continuous and the image is not compact. Answer E doesn't make sense at all: you need a metric to ask if a function is bounded, but a general topological space does not come with a metric.

Let {U_alpha} be an open cover of Y. Then {f^{-1}(U_alpha)} is an open cover of X because f is continuous. Since X is compact, it has a finite subcover {f^{-1}(U_i)}. Then {U_i} is an open cover of im(f), and im(f) = Y since f is surjective.

Back to your first question: the continuous image of a compact space is compact, so continuity of f is certainly sufficient. f being 1-1 (in addition to being onto by hypothesis) doesn't help because bijections need not preserve any topological properties of a space. For example, the "identity map" from the interval [0,1] with the Euclidean topology to [0,1] with the discrete topology is a bijection but not continuous and the image is not compact. Answer E doesn't make sense at all: you need a metric to ask if a function is bounded, but a general topological space does not come with a metric.

### Re: Problem and its derivative from GR8767#64

Thanks. I got it. Really appreciate of ur helps ^_^