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Old 09-25-2009, 08:51 AM   #1
vvxtiopmx

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Default a compact set on the reals will always have finite measure
What does this have to do with fisting or dickgirls?
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Old 09-25-2009, 08:52 AM   #2
jeockammece

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d(x,y)
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Old 09-25-2009, 09:08 AM   #3
blodwarttufla

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nvm, I didn't realize I could use without proof the fact that continuous functions on a compact set are uniformly continuous
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Old 09-25-2009, 11:05 AM   #4
zueqhbyhp

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My math homework.
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Old 09-25-2009, 12:20 PM   #5
Nupbeaupeteew

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nvm, I didn't realize I could use without proof the fact that continuous functions on a compact set are uniformly continuous
That follows straight from the definition of compactness.
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Old 09-25-2009, 04:36 PM   #6
Depolit

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I'm a math major. This is a required class.

Also, I pretty much ran out of classes in the theory department on the CS side.
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Old 09-25-2009, 05:34 PM   #7
wheettebott

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Invalid major
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Old 09-25-2009, 05:50 PM   #8
expomeHattePe

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Not being able to prove Heine-Cantor in 1 line
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Old 09-25-2009, 06:55 PM   #9
Anypeny

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You're the type of guy that's into S&M.
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Old 09-25-2009, 07:04 PM   #10
SaraKonradtt

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Whatver turns your crank.
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Old 09-25-2009, 07:11 PM   #11
sesWaipunsaws

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"William Thurston's elliptization conjecture states that a closed 3-manifold with finite fundamental group is spherical, i.e. has a Riemannian metric of constant positive sectional curvature."
"OH BABY! DON'T STOP NOW, BABY!"
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Old 09-25-2009, 07:16 PM   #12
Scukonah

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fap fap fap
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Old 09-25-2009, 09:35 PM   #13
egoldhyip

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Kuci is letting us peer into his soul. Let it be. Oooo, let it be.
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