proving supremum and infimum of bounded sequences

[plucky tinsel]

hi! im having trouble writing a proof for this. i was thinking of using that since the sequence b_n max is bounded monotonic & increasing, we know it converges to the supremum and conversely we can do the same for b_n min but im not sure if this logic is right or how to format this as a formal proof

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[proven dragon]

the logic is right if you are talking about sup a_n

or are you talking about lim sup a_n

[plucky tinsel]

just sup a_n

[proven dragon]

then your proof sketch is valid

what did you try till now?

Do you know what "completeness of R" means?

[plucky tinsel]

that the real numbers basically dont have any gaps
from my understanding we can say that every bounded monotonic sequence of real numbers converges
ive also been looking at the completeness axiom

[glacial cosmos]

That's honestly a good shot

Just beware, you do not yet know at that point that the limit of b_n is the supremum (it is the case though)

you must prove it

To be completely honest with you though I have no idea what completeness has to do here

since you seem to be using a very simple theorem to show that b_n converges

[plucky tinsel]

right now im kind of just stating the facts that since b_n max is a bounded monotonic increasingg sequence which implies it converges to the supremum that must exist because of the completeness of the real numbers

which idk if its solid enough as a proof

[glacial cosmos]

I'm sorry I think I confused completeness with something else

I thought you meant, how in a metric space, every cauchy sequence converges in it

but you're talking about the completeness of order theory right

[glacial cosmos]

That sounds about right, ok I see

Well, the proof is sound

As I said, to me it is not entirely clear without justification why b_n converges to the supremum

As in, why would it not converge to something less than the supremum

I think this part deserves a bit of elaboration

But that's just my opinion

you may ignore my opinion if it is irrelevant to you

[plucky tinsel]

so b_n converges to some value, and then i should try to show that value is the supremum

[glacial cosmos]

Yes

Although I do want to ask a question

Consider the set A = { a_1, a_2, ...}
What stops you from applying the theorem 2.6 to A ?

A is nonempty and bounded

[plucky tinsel]

im not sure, maybe just to more exhaustively show b_n max and b_n min seperately?

the theorem is just outside textbook resources, not class material

since we kind of do not have a set textbook for the class

[proven dragon]

isn't that cauchy completeness

using completeness (theorem 2.6) you can directly prove that as {a_n : n in N} forms the set in question