Any math fags out there that could help a Sup Forumsro with some hw?

Any math fags out there that could help a Sup Forumsro with some hw?

bump

shameless self bump

bumpity bumping

Don't ask just fuckin post

lay it on me cunt

a+b = b+a
a*b = b*a
(a + b) + c = a + (b + c)

Oh boy, if only there was a board for people who need help with things. One where anons help fulfill others' requests maybe? There should be one of these that is safe for work though. That idea would be so fucking nifty!

Post your fucking problem faggot

Ok heres the question.

Let x be any bounded sequence and S be the set defined by S={z\inR | finitely many elements of x are less than z}. Suppose the Sup(S)=u and is not an element of S. Prove that there exists a subsequence of x that converges to u.

I wish!

confused. What are a,b, and c?

bumping 4 halp

z/inR
Im not familiar with this writing explain short pls

z is an element of the real numbers

hahaha underage b&

hmm are you saying not understanding a poorly defined statement implies being under the age of 18? Prove it

Bumping 4 halp on this problem

For any \epsilon > 0, there is an element s of S with u - s < \epsilon. That means there are infinitely many elements of x in [s, u). Pick one, call it x_1. Replace \epsilon by \epsilon/2 and repeat the argument. You obtain the desired convergent sequence. QED.

When it comes to set theory, you start with what you are given, and deduce what you can from it.

>x is a bounded sequence
>S=z in R
Therefore, the bounded sequence x is a subset of z in R = S.
>superset of S = u, not element of S.
Therefore, S is a subset of u.
Prove there exists a subsequence of x that converges to u.
There isn't a subsequence of x that converges to u, because x is sequence within S, and u is not an element of S.
QED

It is precisely because x is a bounded sequence, and also finite elements less than z that the infinite R which exists in S is a subset of u, meaning u is also infinite set greater than S = z in R that x as a bounded finite set less than z cannot converge on u.

Cause u is not in S, there are infinitly many x

Sup(s) is the supremum of S

Can you further explain why for any /epsilon>0 there is an element s of S with u-s < /epsilon