Problem and its derivative from GR8767#64

Forum for the GRE subject test in mathematics.
mhyyh
Posts: 9
Joined: Mon Oct 10, 2011 9:12 pm

Problem and its derivative from GR8767#64

Postby mhyyh » Fri Oct 28, 2011 10:13 pm

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?

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: Problem and its derivative from GR8767#64

Postby yoyobarn » Sat Oct 29, 2011 2:00 am

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?


I thought the statement I is true.

According to http://www.google.com/url?sa=t&rct=j&q=if%20x%20is%20compact%20then%20y%20is%20compact&source=web&cd=5&ved=0CDwQFjAE&url=http%3A%2F%2Fwww.math.uiowa.edu%2F~jsimon%2FCOURSES%2FM132Fall07%2FPreimageOfCompactSpace.pdf&ei=ppWrTsTNHIm3rAem7_mxDA&usg=AFQjCNHnAs6Z1XZZGFryr5A7qEoByvzHhQ&sig2=CyH28AlXVT5hgug5bEGCkg,
Theorem. If f : X → Y is a continuous surjection and X is compact, then Y is
compact.

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

Re: Problem and its derivative from GR8767#64

Postby owlpride » Sat Oct 29, 2011 4:26 am

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.

mhyyh
Posts: 9
Joined: Mon Oct 10, 2011 9:12 pm

Re: Problem and its derivative from GR8767#64

Postby mhyyh » Sat Oct 29, 2011 10:25 am

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




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