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Help with math plz | |||||||
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Plz hlp I hav a problem in math i dun understand plz hlp me here
let M(S) denote the monotone clbum generated by a set of sets S i dun understand why ∀S S is a ring ∀A x∈M(S) ∃{A_n} (A_n⊂A_(n+1)∧S_n∈S∧(∪(S_n)=S) plz explain this to me i dun understand why this is true, its a fact used for proving the uniqueness of extention of measure |
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Posted On: 03/07/2008 5:28PM | View a dallop a daisy...'s Profile | # | ||||||
lol? |
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Posted On: 03/07/2008 5:47PM | View Ziro's Profile | # | ||||||
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acutually i made a mistake plz forgiv me the actual statement i dun understand is (corections are bolded): ∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A) thar i think i wroat it in proper formal language plz tell me why this is true a dallop a daisy edited this message on 03/07/2008 8:40PM |
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Posted On: 03/07/2008 8:36PM | View a dallop a daisy...'s Profile | # | ||||||
[This message has randomly exploded, causing the “divide by 0” effect, and thus creating a pime taradox.] Warpvader edited this message on 03/10/2008 9:52PM |
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Posted On: 03/07/2008 8:38PM | View Warpvader's Profile | # | ||||||
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Do a barrelroll. |
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Posted On: 03/07/2008 8:41PM | View Opportunity's Profile | # | ||||||
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The problem is that there are upside down “A”s in your statement… |
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Posted On: 03/07/2008 8:48PM | View Harix's Profile | # | ||||||
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the upside down A just means “for all” LOLOL NOOB |
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Posted On: 03/07/2008 8:57PM | View dreamserpent's Profile | # | ||||||
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Warpvader Posted:
o its very simple
it means “for all rings R, for all A within M(R ), there exists a sequence of sets A_n which are elements of R such that their union is A.”
to see wat im talking about, here is an example in which it is implicitly stated: http://www-maths.swan.ac.uk/staff/vl/c98.pdf go to page 18, theorem 4.2
Theorem 4.2 Let A be an algebra of sets, μ and ν two measures defined on the σ-algebra A_σ generated by A. Then μ ↾ A = ν ↾ A implies μ = ν. Proof. Choose S ∈ A_σ, then S=lim (n⇒∞) S_n, S_n ∈A, for A_σ=M(A ). Using continuity of measure, one has μ(S) = lim (n⇒∞) μ(S_n) = lim (n⇒∞) ν(S_n) = ν(S). a dallop a daisy edited this message on 03/07/2008 9:01PM |
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Posted On: 03/07/2008 8:59PM | View a dallop a daisy...'s Profile | # | ||||||
the answer is jesus |
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Posted On: 03/08/2008 4:23AM | View Miggle's Profile | # | ||||||
a dallop a daisy Posted:
(A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A) has me pretty confused. |
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Posted On: 03/08/2008 12:36PM | View Skyreal's Profile | # | ||||||
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Skyreal Posted:
it means the sequence of sets A_n is increasing and each A_n is in the ring R and the union of it is A. |
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Posted On: 03/08/2008 12:45PM | View a dallop a daisy...'s Profile | # | ||||||
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Drop the clbum before it does any more damage. |
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Posted On: 03/08/2008 9:00PM | View PhineasPoe's Profile | # | ||||||
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The game |
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Posted On: 03/08/2008 9:02PM | View VIVa-ChiCk's Profile | # | ||||||
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The only thing I remember from math was not to use my teeth. |
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Posted On: 03/08/2008 10:31PM | View Balloon's Profile | # | ||||||
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T∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A) ∀R R is a ring⇒∀A A∈M(R )⇒∃{A_n} (A_n⊂A_(n+1)∧A_n∈R∧(∪(A_n)=A)
Math Attack.
Ego -40^e-(ln(1+x^2)) Damage +50^e(x^(1/3)+x^(ln(1+mod(x^.5))))
oh.. where x is the number of times you fapped in the past 1.32 days
PWNED THIS THREAD! Log in to see images! |
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Posted On: 03/08/2008 10:43PM | View crackednut's Profile | # | ||||||
How about you stop trying to look smart on the internet there |
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Posted On: 03/08/2008 10:45PM | View Leadpipe's Profile | # | ||||||
Leadpipe Posted:
Eeee! I was found out! How?
Oh wait, were you referring to me? |
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Posted On: 03/09/2008 11:52AM | View Skyreal's Profile | # | ||||||
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do your own algebra homework and don’t expect the internets to do it for you Log in to see images!
Also, “An alternative way to show that an open subset X of R is the disjoint union of open intervals is to consider the set S of all sets of disjoint open intervals with union contained in X, with an appropriate order, apply Zorn’s lemma to S to obtain a maximal one, and then show that the maximal one in fact covers all of X.” If that helps. Abyssinth edited this message on 03/10/2008 8:25PM |
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Posted On: 03/10/2008 8:20PM | View Abyssinth's Profile | # | ||||||
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I think this Log in to see images! |
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Posted On: 03/11/2008 10:20PM | View Neurotic Ninja's Profile | # | ||||||
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Leadpipe Posted:
LETS ALL QUOTE OUR ACT SCORES!!
i got the highest score on everything possible and am currently developing a unified theory of everything. i attend super genius school and am the top in my clbum. |
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Posted On: 03/12/2008 12:55PM | View nanalatinojesus ...'s Profile | # | ||||||