BG MATH [Real Analysis]

[SDSD]

I'm completely blanking here. The problem isn't that hard, but I just can't seem to think it through.

Suppose S is a non-empty set, bounded above, with no maximal element. Show that there exists an infinite sequence x_n of real numbers converging to the supremum s = sup(S).

I know this intuitively, but I'm not sure how to go about doing this. I'm thinking I use the fact the any x_n I would be working with would be bounded and convergent, but I'm not sure how to show it converges to s. Thanks.

[Woozie]

Well, not any x_n you're working with would be convergent, but it certainly would be bound. Then you know that x_n has a convergent subsequence. But we need to prove that there is a sequence that converges to supS.

Give me a few minutes to type this up.

Edit: My bad, I guess you're right, depending on how you think about it. My professor always made us start with any sequence (without assuming it's convergent) and then cite the Bolzano-Weirestrauss theorem to conclude that we can find a convergent subsequence. But I guess you could just outright say "Let x be a convergent sequence" (and we know such a sequence exist).

[Woozie]

I'm thinking use s-1/n. This is certainly a sequence converging to s. And although the entire sequence may not be in S, the "tail" end of the sequence certainly is (i.e. there exist an K such that n>K implies x_n is in S). I'm just trying to think of a formal way of writing this.

[Suiram]

I'm thinking use s-1/n. This is certainly a sequence converging to s. And although the entire sequence may not be in S, the "tail" end of the sequence certainly is (i.e. there exist an K such that n>K implies x_n is in S). I'm just trying to think of a formal way of writing this.

I don't think you're guaranteed that Sup(S)-1/n is necessarily in S for any n. But you can use the fact that S is bounded above with no maximal element to show that the open interval O(n) = (Sup(S)-1/n, Sup(S)) is non-empty, and construct a sequence from there by picking a representative from each O(n) for x_n.

[Woozie]

Hmm, maybe I'm thinking about this wrong then. If s-1/n was never in s, wouldn't that mean s-1 is greater than every element in S? This would lead to a contradiction because s is the supremum of S. What am I doing wrong here? x_x

Edit: nevermind...I think I see why this is wrong. Let me think about this for a second.

Edit 2: Yeah, my reasoning was wrong. I automatically thought of S as simply being some open interval (a,b), whereas the O.P. never said that. I can easily imagine an S such that everything the O.P. said is true, but s-1/n doesn't necessarily ever end up in S. My bad.

[Suiram]

What I mean is like, let's say his S was {x: x<2 and x irrational}, then Sup(S) - 1/n wouldn't be in S, so you couldn't exactly use them for the sequence (but you could do what I suggested above).

[Woozie]

What I mean is like, let's say his S was {x: x<2 and x irrational}, then Sup(S) - 1/n wouldn't be in S, so you couldn't exactly use them for the sequence (but you could do what I suggested above).

Yeah, my bad. I was thinking of S as simply being an open interval for some reason even though the OP never said that x_x

[Suiram]

Yeah, my bad. I was thinking of S as simply being an open interval for some reason even though the OP never said that x_x

Actually on the subject of things the OP never said, now that I look, technically you were right since the OP didn't specify that his sequence has to be contained in S, so your original answer would work (as would s, s, s, s, ... lol), although I imagine the problem probably wants it in S lol


Edit to what I said a few posts up: I meant to say that you could use that it was bounded above with no maximal element to show that S(n) = (O(n) intersect S) (where O(n) is (s - (1/n), s)) is non-empty, and then pick a representative x_n from each S(n).

[SDSD]

My friend is trying to prove this using squeezing but I disagree with all the assumptions he's making. This is the email he just sent me.

"I've defined (z_n) to be a sequence where inf (z_n) ≥ s, (y_n) where (y_n) → inf (z_n). Then if (x_n) st (y_n) ≤ (x_n) ≤ (z_n), and if (x_n) exists in S, clearly (x_n) → s by the squeezing theorem."

This seems a bit too convenient to me, and way to dependent on defining sequences to fit x_n into the squeezing theorem. Still not sure on where to go with this. I think I saw a problem like this on my homework a couple weeks ago. I'm trying to find it right now.

Edit: I'm taking the problem word for word off my review sheet. My TA still hasn't emailed me back saying if x_n has to be completely in S or just part of it. My friend is assuming it does. My problem is that I don't think it does.

Edit2: I originally wrote x_n to be s, s, s, s,... also but my TA shot me down.

[Woozie]

My friend is trying to prove this using squeezing but I disagree with all the assumptions he's making. This is the email he just sent me.

"I've defined (z_n) to be a sequence where inf (z_n) ≥ s, (y_n) where (y_n) → inf (z_n). Then if (x_n) st (y_n) ≤ (x_n) ≤ (z_n), and if (x_n) exists in S, clearly (x_n) → s by the squeezing theorem."

This seems a bit too convenient to me, and way to dependent on defining sequences to fit x_n into the squeezing theorem. Still not sure on where to go with this. I think I saw a problem like this on my homework a couple weeks ago. I'm trying to find it right now.

Edit: I'm taking the problem word for word of my review sheet. My TA still hasn't emailed me back saying if x_n has to be completely in S or just part of it. My friend is assuming it does. My problem is that I don't think it does.

I think it's safe to assume your professor wants the sequence to be in S (or else you could just use (s,s,s,s,s....) ). Also, any sequence is legit as long as the "tail" end of it is in S (because you can just toss out the beginning of the sequence and you end up with a new sequence entirely in S).

Is there any reason you can't simply do what Suriam posted?

[SDSD]

I don't see any reason I can't. It make a lot of sense actually. I'm heading out to a review session in about 30min for a different math class. I'll go see if my TA is at school too and show him.

Thanks for all the help.