2007 February 13th, 06:14 AM
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#41 (permalink) |
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a ghost
 Join Date: Dec 2005
Posts: 335
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ugh im not ready! i am reading over piont set topology stuff.. accumulation point - i'm not sure if i understand this clearly. so x is an accumulation pt of S if the deleted nbhd contains a pt in S, but doesn't it really depend on the magnitude of your neighbourhood?? obviously i am going to like, do badly on this test
 also the whole open cover subcover nonsense, i can figure out that a set is compact if it's closed and bounded, but without using that theorem, how do i know that, say a set S=(0,2) and for a sequence A_n=(1/n,3), that this is not compact? erm..yeah. im screwed Edit: sorry i just realized you asked for a "funny" question - well i have none of those lol i guess never mind |
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2007 February 13th, 06:55 AM
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#44 (permalink) | ||
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Within my grasp!
Join Date: Oct 2006
Location: NYC, NY
Posts: 469
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Quote:
Quote:
To show that S is not compact, take an open cover, such as the set A = {(1/n,3) | n in N}. Clearly, the union of this family of open sets covers S, since for every x in S we can find a n sufficiently large, such that x is in A_n. (Just choose n so that 1/n < x). But there is no finite subcover, for suppose there were one. Then there is a set of indices M = {m1, m2, ..., mk} such that Union_{m1,m2,...mk} A_n contains S. Let mMax = max M. Then A_mMax contains all other sets in this subcover. But we can find a real number between 0 and 1/mMax that is in A and not in A_mMax. Hence, this subcollection does not cover the set. Therefore, there exist no finite open subcover for this open cover, whence A is not compact. |
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2007 February 13th, 04:42 PM
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#45 (permalink) |
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Within my grasp!
Join Date: Feb 2007
Location: Baltimore, MD
Posts: 217
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It is always easier to prove something is not compact than proving that it is, since you must show for EVERY open cover there exists a finite subcover. It can help to use rules, like every closed subset of a comact set is compact, and so forth. Good luck!
PS: Nice example, Promo! |
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2007 February 13th, 05:12 PM
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#46 (permalink) | |
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Eager!
Join Date: Oct 2006
Posts: 47
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i think before understanding all this shxt, u need to have a solid comprehension of completeness axiom and archemedian property. those are like what gives rigid explanation for all this.
F={(1/n),3)} is an open cover for (0,2), because for any e>0, we can always find N, s.t. for all n>N, 1/n<e (archemidian  ), so that each of those (1/n,3) contains (0,2)but F does not have a finite subcover for (0,2). since if there is one, we can pick m=max{n}, so that the union of all sets in this subcover is (1/m,3), which certainly does not cover (0,1/m] Quote:
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2007 February 14th, 01:25 AM
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#48 (permalink) | |
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TestMagic Guru-in-Training
Join Date: Feb 2005
Posts: 690
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Quote:
_ _ _ _ SIG _ _ _ _
University of Wisconsin - Madison: Took my Masters and ran. |
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