|
|
#1 (permalink) |
|
Will power
![]() ![]() Join Date: Nov 2002
Location: Hanoi and Munich
Posts: 401
![]() |
Compact topological space
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
_ _ _ _ SIG _ _ _ _
Go to TWE section and learn writing with me. Correct my essays and I will participate in reviewing yours! |
|
|
|
|
|
#2 (permalink) |
|
I JUST got here.
Join Date: Jun 2007
Posts: 4
![]() |
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. |
|
|
|
Contact TestMagic TestMagic Forums Archive Privacy Statement
TestMagic Locations
Legal
Privacy
SEO by vBSEO 3.2.0
Copyright © 2009 TestMagic
Ad Management by RedTyger