Can't even visualize this problem correctly? Question was written wrong?

[bitter portal]

I've drawn the picture illustrating this problem and realize the question doesn't make sense because won't sup A be +infinity and lim inf be a finite value between the upper bound and lower bound? (Picture below)

[dire gulch]

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[azure sparrow]

this isn't intelligible

[azure sparrow]

sup A is not infinite when {s_n} is bounded

here's the main idea used in the proof

$\lim \sup s_n = \lim_{n\to\infty}\sup_{m>n}s_n=\inf_{n\in\bN}\qty{\sup_{m>n}s_n}$

[whole widget]

kaffee

[azure sparrow]

now prove this

[azure sparrow]

you should be able to derive an analogous expression for limit infimums too.

[bitter portal]

ok so BW theorem says every boudned sequence has a convergent subsequence.

so i gotta show somehow that {sup s_n} is convergent?

ok so if sup s_n is monotonically increasing, then sup s_n is all the same for all m > n

if sup s_n is monotonically decreasing, then

[azure sparrow]

@bitter portal prove this first and the rest follows

[azure sparrow]

there's no if, sup_{m>n} s_m IS monotonically decreasing

as you take away elements from the set which could be the greatest

[bitter portal]

ok so if it IS m. decreasing, then there is an L such that there is an N for all n > N: |s_m - L| < $$\epsilon$$

[whole widget]

JackieChen
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[bitter portal]

and the limit L is the infinum

by the Monotone convergence theorem

but didnt i read the problem correctly, it said set A is all real numbers a such that those a's are greater than s_n (i assume any given n).

so clearly set A can contain numbers like 100000

or infinity

[azure sparrow]

are you trying to show that the limit supremums or infimums exist

by monotone bounded sequence theorem

[bitter portal]

it just said at the last sentence

[azure sparrow]

then show that $\abs{s_n}\leq M\implies \lim \sup s_n \leq \sup s_n \leq M$

[whole widget]

kaffee

[bitter portal]

prove sup A = lim inf s_n

why lim sup? isnt it lim inf s_n <= M?

and M here is the (least) upper bound?

clearly something is not clicking for me

[bitter portal]

yeah idk, i feel like i say things but i cant connect all of them tgt

[azure sparrow]

my man do you get this? @bitter portal

[bitter portal]

i understand the statement

but i dont know how to string them together as a complete proof

[azure sparrow]

then look up the proof

[bitter portal]

the proof for? Monotone convergence theorem?

[shadow isle]

The very last subscript (index) should be m instead of n

Also the one in the middle expression

[azure sparrow]

yes that's right I missed it

[azure sparrow]

for the theorem you are asked to proof itself

[bitter portal]

ok so 1) the left side rewrite to the middle equation

then you prove that {sup s_m} is decreasing

3. then since the sequence s_n is bounded, {sup s_m} is bounded, by completeness axiom, there is an infinum for {sup s_m}

4. finally use monotone convergence theorem to prove that lim n -> infinty sup s_m = inf{sup s_m}

idk thats my best shot for reasoning

[azure sparrow]

yep that about it

[bitter portal]

ok then i can look that up in the textbook easy

so how does that lead to the questions in the problem?

[azure sparrow]

are you free on 2 hours?

[bitter portal]

yes

but can we do later today?

sorry, im busy with school so i dont have free time to sit down and be free for 2 hrs straight

[azure sparrow]

I made a typo, it'd be "in" not "on"

[brisk estuary]

I think the following elements can be helpful to you

we can start showing for the case of A, at least

1. A is convex (and also, clearly upperbounded)

so if a and a' are both in A, then anything in between is also in A

2. convex subsets of R are intervals

and finally, for the finishing touch, if a is in A and a < sup A, then sup A >= liminf sn > a

Do tell me if that helps, even remotely

sup A >= liminf sn

liminf sn > a

I'll let you find out how you can show that

if you're willing to take my lead

[bitter portal]

never talk about convex in class so i have no idea that true definition

nor should i even wrote that in the answer

all i know that its bounded, aka also upper bounded

[brisk estuary]

It doesn't matter all that much whether you know the specific term of convexity or not

The point is just that

If a' is in A, then everything less than a' is also in A

[bitter portal]

no, i will get deducted points for introducing things not in class

but yeah i get that ur trying to help

so ty

[brisk estuary]

I mean I am also trying to understand how much I can introduce without breaking the rules

[bitter portal]

oh ok, thats just basic inequality and denseness of R?

[brisk estuary]

Because in itself I am not using a very complicated idea

Not density no

It's just that if a' is in A, and a < a', then a is in A too

Which is a very simple idea for starters

[bitter portal]

uh idk if a is less than the infinum, then is it not in A?

a is less than a' up until a certain point tho, then its not in set A.

[brisk estuary]

Well, you czn try to visualize it

With your graph above, it's a very good one

You draw a bar at y=a', so there is a finite number of terms below a'

[bitter portal]

wait, go back to what u said here. this makes sense

[brisk estuary]

Perfect

We can resume from there

[bitter portal]

i think i lost you with this statement

but yeah back to the original one

[brisk estuary]

So take a' arbitrarily small

Close to -infinity

It is still in A

(Because the sequence is bounded)

[bitter portal]

so convex subsets of R are intervals and intervals are notated like this: (a , b)?

[brisk estuary]

Either [a, b], (a, b), [a, b) or (a, b]

[bitter portal]

oh you mean a as in a term in the series. ok, got it

yeah then the bounded definition makes sense, ok thanks for clarifying

[brisk estuary]

So the idea is actually very simple here, you effectively want to find out that A is in the form: (-infinity, c) or (-infinity, c]

With c = sup A

[bitter portal]

this makes sense except you squish in the liminf sn in the middle

and i dont know where your "work" gets you there yet

[brisk estuary]

Yep so there are 2 inequalities to prove

First sup A >= liminf sn

We will prove that, I will guide you so that you can try replicating it for the other one

We will prove it by contradiction

If liminf sn > sup A, then we can take a in the middle

[bitter portal]

does direct not work? not that it matters but direct is "straightforward"? (ignore this, just walk through the proof first and we can come back to this)

[brisk estuary]

a = (liminf sn + sup A)/2

And you will find out that a is in A

Since there will be a finite number of terms below a, given that the liminf is greater

Which contradicts that a > sup A

[bitter portal]

wait i lost you here. how did u conclude that?

[brisk estuary]

No problem, I will explain

So a is still lower than the liminf

But since the inf converges to the liminf

Then starting from a certain rank, the u_n = inf(m >= n) s_m will be arbitrarily close to the liminf

Get this so far?

In cleaner words:

let u_n = inf(m >= n) s_m. Since s_n is bounded, u_n is nondecreasing and converges (to the liminf sn).

In particular, after a certain rank n0, u_n > a, so at most n0 terms are below a.

Which means in particular that at most n0 terms of the sequence (s_n) are also below a

[brisk estuary]

Also I was mistaken, the other one can be proven directly provided the first one

A simple squeeze will do

[bitter portal]

oh sorry, my wifi lost connection there

let me read what u said

[brisk estuary]

No worries, I thought I lost you again haha

[bitter portal]

i just got connections back

[brisk estuary]

But yeah don't hesitate to ask, I will clarify whatever you need me to

[bitter portal]

yeah idk how u conclude (liminf sn + sup A)/2 < liminf sn

it could be bigger, smaller depending on supA

[brisk estuary]

Because if we supposed that sup A < liminf sn, I just took the number in the middle

[bitter portal]

oh right

yeah

[brisk estuary]

It's just a mean

[bitter portal]

ok yeah i read it as is and i forgot to reference it back to the top

my bad

[brisk estuary]

No worries

[bitter portal]

this is just the definition of lim inf?

[brisk estuary]

Yep

[bitter portal]

inf goes to infinity of subsequence will tend to lim inf

[bitter portal]

ok i got it.

[brisk estuary]

The "smallest number ahead" is u_n

And it tends to the liminf

[brisk estuary]

So if the liminf is above a, then there is only a finite number of terms below a

[bitter portal]

im rereading something. im lost somewhere.

[bitter portal]

yeah i was rereading back this part

[brisk estuary]

Yeah

So basically, since u_n converges to the liminf, starting from a certain rank n0, u_n > a

[bitter portal]

only a finite # of terms below a? thats because a can grow above liminf?

[brisk estuary]

(Provided that a < liminf)

So at most n0 terms of u_n's are below a

So, also at most n0 terms of s_n's

[bitter portal]

wait ur applying the defintion of divergence here right? that all values are greater than a value M.

this is a subtle definition of divergence

[brisk estuary]

Here? No not really

I mean we are exploiting convergence for sure

But not divergence as far as I know

It's just that our chosen a, which was supposed go be greater than sup A, actually turns out to fit the bill to be in A due to the convergent nature of u_n

Which is a problem

[bitter portal]

ok i read this and this makes much sense here

i see clear application of montone convergence theorem

[brisk estuary]

That is very good then

In that case, is it clear in your mind that sup A >= liminf sn ?

[bitter portal]

i forgot, is capital A the sequence of sm's?

[brisk estuary]

A is the set defined in your exercise

[bitter portal]

i get what ur saying here mathematically, but i still cant connect why a is in A

[shadow isle]

I think we began by choosing an a from A, didn't we?

[brisk estuary]

A is the set of a in R such that {n : s_n < a} is finite

[brisk estuary]

No

[bitter portal]

a is defined as the mean

[shadow isle]

Ah okay, let me review

[brisk estuary]

We supposed that sup A < liminf sn, and we took a as the middle point

[bitter portal]

probably bad word choice (causes confusion as an element)

[brisk estuary]

And it turns out that, for the reason explained above, a fits the bill to be in A

[shadow isle]

Gotcha

[brisk estuary]

Since essentially, your candiates of s_n < a are in {s0, s1, ..., s_{n0-1}} (the other ones are automatically eliminated because u_n > a for any n >= n0)

Whichever they are, there is a finite headcount

Not sure if that clears your doubts as for why a fits the bill to be in A

[shadow isle]

Yeah, a slightly more direct way to write this part

First sup A >= liminf sn
We will prove that, I will guide you so that you can try replicating it for the other one
We will prove it by contradiction
If liminf sn > sup A, then we can take a in the middle
a = (liminf sn + sup A)/2
And you will find out that a is in A
Since there will be a finite number of terms below a, given that the liminf is greater
Which contradicts that a > sup A
might be to take arbitrary element x of A and establish liminf sn to be no less than x, thus making liminf sn an upper bound of A
Scratch that, that only makes sup A <= liminf sn

[brisk estuary]

The thing is

Provided sup A >= liminf sn, ghis is relevant

Since for any a in A, sup A >= liminf sn > a

And given that A is an interval, you can squeeze

[shadow isle]

Yeah, the desired result was not established

[brisk estuary]

But we want the first one, which is tricky to prove without contradiction

[shadow isle]

True

We are basically using the fact that any real number x less than liminf s_n is an element of A

[brisk estuary]

Quick illustration

[shadow isle]

which makes (-inf, liminf s_n) a subset of (-inf, sup A)

[brisk estuary]

As you can see onthe note it is almost 4am so I must go rest a bit

I entrust you to Rafain's capable hands

[shadow isle]

Let x < liminf s_n; denote e := liminf s_n - x > 0.
liminf s_n = lim(n->+inf) inf{s_m: m>= n}.
By the definition of limit (and monotonicity of the sequence inf{s_m: m>= n}), there exists natural M such that liminf s_n - inf{s_m: m >= n} <= e for any m >= M.
Then {n in N: inf{s_m: m >= n} <= x} = {n in N: liminf s_n - inf{s_m: m >= n} >= e} has at most M elements, making it finite.
Since s_n >= inf{s_m: m >= n}, {n in N: s_n <= x} is a subset of {n in N: inf{s_m: m >= n} <= x} and is also finite.
Hence x is an element of A.

Since the choice of x in (-inf, liminf s_n) is arbitrary, (-inf, liminf s_n) is a subset of A.
Hence sup A >= liminf s_n.

[bitter portal]

i'll check with with @brisk estuary later. but i know this whole process is all over the place.

so i wanna write the proof neatly again into one chunk

[bitter portal]

is this the direct proof?

[shadow isle]

That's what I tried to do, yes, but I can't guarantee it is valid - a lil too rusty to be sure

[bitter portal]

do u know if u_n here is a sequence of infinums or one infinum? im confused with the U_n defintion?

u_n is a subsequence?

[shadow isle]

For each n, u_n is one infimum
{u_n: n in N} form a sequence of infima
Rion didn't distinguish these two in their notation, let me help distinguish

[bitter portal]

so this is what Rion means {u_n: n in N} for u_n

ok so its a sequence of infima

[shadow isle]

let u_n = inf(m >= n) s_m.
This defines a single infimum u_n for each n
Since s_n is bounded, u_n is nondecreasing and converges (to the liminf sn).
This one is {u_n: n in N}

[bitter portal]

so the second "u_n" is actually meant as {u_n: n in N}?

yeah i read it and i kept being confused

[shadow isle]

👍

[bitter portal]

how did Rion conclude that after a certain rank n0, u_n will always be greater than a?

is that the consequence of non-decreasing?

[azure sparrow]

it's the question condition

"a" are such that only finite ranks exist where u_n =< a

[brisk estuary]

Of convergence, to a limit that is greater than a

[brisk estuary]

No I legit meant it as a sequence

Not as a set

$u_n = \inf_{m \geq n} s_m$

[whole widget]

Rion

[brisk estuary]

This is a number sequence like any other

We established its convergence with monotonic convergence theorem

[bitter portal]

is that read the infinum of sequence s_m or the sequence of infinums of s_m subsequences?

[brisk estuary]

The infimum of the sequence numbers ahead of rank n

[bitter portal]

so its 1 number, the infinum of that sequence

[brisk estuary]

$u_n = \inf { s_m | m \geq n}$

[whole widget]

Rion

[brisk estuary]

u_n is a number yes

The sequence (u_n) converges (easy to check with monotonic convergence theorem)

And the limit is exactly, by definition, liminf s_n

[brisk estuary]

For the record this is correct

[shadow isle]

Yeah I meant the same, bad habits kicked in
I write sequences in that abused notation since the indexing usually suffices to communicate to the reader when sequence properties are called

[bitter portal]

hmm, i thought limit definiton is for all epsilon, i can find N such that all n > N: |x_n - L| < epsilon? nowhere does it have greater than (for greater than a??)?

[shadow isle]

The limit is compared to a
The tolerance is not comparable to a

[bitter portal]

tolerance here is the |xn - L|?

oh no, thats the epsilon

oh ok

[brisk estuary]

Take the tolerance as |L - a|/2

[shadow isle]

Epsilon is the tolerance

[bitter portal]

so what is this other limit we're talking about, so a is the line in the drawing but then this limit is for s_m?

[shadow isle]

Afaik yes for the part after the comma, @brisk estuary please confirm

[brisk estuary]

The limit is for u_n

[shadow isle]

Oh right, I am dizzy

[brisk estuary]

It is not know whether s_n converges

[brisk estuary]

In an attempt to clarify all the doubts, I will rewrite the earlier proof in tex

Let $u_n = \inf_{m \geq n} s_m = \inf { s_m | m \geq n}$.

Since $(s_n)$ is bounded, $(u_n)$ is nondecreasing and converges (by definition, to $L =\lim \inf s_n$).

Due to the fact that $L > a$, there exists a certain rank $n_0$ such that for any $n \geq n_0, \lvert u_n - L \rvert \leq \frac{L - a}{2}$. In particular, for any $n \geq n_0, u_n > a$.

Therefore, for any $n \in \mathbb{N}$, if $u_n \leq a$, then necessarily $n \in { 0, ..., n_0}$. Moreover, given that $\forall n, s_n \geq u_n$, the same holds: for any $n \in \mathbb{N}$, if $s_n \leq a$, then necessarily $n \in { 0, ..., n_0}$.

[whole widget]

Rion

[brisk estuary]

This means, in particular, that ${ n \in \mathbb{N} | s_n < a}$ is finite, hence $a \in A$

[whole widget]

Rion

[bitter portal]

oh ok, now i get it, ur just restating the defintion of limit and epsilon is now L - a/2

yeah i see it now

cuz the limit is L

[brisk estuary]

Ok that's very good

Is there any other point troubling you about the statement that $a \in A$ now? (and hence the contradiction)

[whole widget]

Rion

[bitter portal]

the next sentence doesnt follow the previous one?

ok so u define the limit and then you conclude u_n > a

[brisk estuary]

Starting from a rank n0 yes

[bitter portal]

thats an assumption?

[brisk estuary]

What is?

[bitter portal]

u_n > a?

[brisk estuary]

No, that is proven

$\forall n \geq n_0, -\frac{L-a}{2} \leq u_n - L \leq \frac{L-a}{2}$

[whole widget]

Rion

Rion

[brisk estuary]

$\forall n \geq n_0, \frac{L-a}{2} \leq u_n$

Ah wait I did a mistake

[brisk estuary]

$\forall n \geq n_0, \frac{a}{2} + \frac{L}{2} \leq u_n$

[whole widget]

Rion

[brisk estuary]

Given that $L > a$, it also holds:
$$\forall n \geq n_0, a < \frac{a}{2} + \frac{L}{2} \leq u_n$$

[whole widget]

Rion

[brisk estuary]

There

Ignore the one with a skull reaction on it, I couldn't delete it

[bitter portal]

ok now i see it u_n > a

btw did u derive all of this in ur head? u dont even show work for these steps and its not really intuitive for me

[brisk estuary]

Just a drawing is enough

you see that u_n converges to L, which is not a

so at some point, it will always be in [L - epsilon, L + epsilon]

so you take epsilon so that this interval does not contain a

The above wall of text is just a long winded way to convince you formally that with a well chosen epsilon, it is the case

[brisk estuary]

If it reassures you, I did think it thoroughly last night before I started messaging in this thread

So I already had a full solution in mind when I sent my first message here

[bitter portal]

ur right, i got it. half of the distance between L and a is always greater than a, thats really intuitive

last sentence s_n >= u_n, that is true because s_n = u_n (s_n happens to be the infinum in u_n) or u_n is a smaller value looking ahead

[brisk estuary]

Since $u_n = \inf { s_m | m \geq n}$, it is the inf of a set that contains $s_n$

[whole widget]

Rion

[brisk estuary]

So in particular, it has to be lower or equal to s_n

[bitter portal]

right cuz thats by def of inf

u < element for all elements in set

[brisk estuary]

less than, or equal to, but yes

[bitter portal]

yeah i follow thru everything and then since a is in set A, then it violates defintion of sup A

[brisk estuary]

Yes, exactly.

[bitter portal]

im just amazed how u come up with "scratchwork" (doing the mean, finding these inequalities and of course knowing what to contradict in the 1st place)

[brisk estuary]

I just did that with a drawing to be fair, all the downstream work is just formalizing

[azure sparrow]

There can be a simple proof

using the fact that $\qty{u_n}\subseteq A$

[whole widget]

kaffee

[azure sparrow]

okay maybe not that

but some of them are in A

so we can trace a subsequence of that through A

which tends to inf A?

[brisk estuary]

A direct proof can use this, if it is known

but that in itself is quite tedious to prove formally, well just as much as a proof by contradiction

[azure sparrow]

constructivists disagree, don't know about the OP though

[brisk estuary]

Well for someone used to it, it's not hard

but someone who just started getting into the topic can find this very challenging

[azure sparrow]

True but using constructive arguments are sometimes better for understanding

[brisk estuary]

It's definitely good to learn for future reference

So in either case I strongly suggest the OP to get used to constructivist arguments

[bitter portal]

wait did we ever prove sup A = liminf sn?

we prove sup A >= liminf sn, not strictly equal to liminf sn?

[brisk estuary]

Yes, we did not yet

I asked you to prove that, additionally, for any $a \in A$, $\lim \inf s_n \geq a$

[whole widget]

Rion

[brisk estuary]

By using a similar reasoning

[bitter portal]

so can i assume that since sup A >= liminf sn, this would imply sup A = liminf sn and be done?

[brisk estuary]

No, not yet

You need to prove what I asked you to prove first

[bitter portal]

hmm, i dont know where to start

[brisk estuary]

Try to follow my model of proof by contradiction

[bitter portal]

so let lim inf sn < a

[brisk estuary]

Suppose that liminf sn < a, yes

So how many terms of (s_n) are below a?

[bitter portal]

finite terms?

thats the definition of set A

[brisk estuary]

Yes, that should be the case

but is it really the case?

[bitter portal]

Suppose that liminf sn < a, then let c be (liminf sn - a)/2 so that liminf sn < c < a?

[brisk estuary]

Sure, but the other way around

c = (a + liminf sn)/2

the midpoint is the mean

not the half-difference

But yes, continue

you're on the right track, I think

[bitter portal]

im not sure what u mean here. if liminf is less than a then there are points less than a?

ok and by the definition of limit L, there exists n0 > N such that |sn - L| < (a + liminf sn)/2

[brisk estuary]

No, not s_n

again, use $u_n = \inf_{m \geq n} s_n$

[whole widget]

Rion

[brisk estuary]

this one is sure to converge to liminf s_n

[brisk estuary]

What is L?

[bitter portal]

liminf s_n = L

but i probably meant to say |u_n - L| then, my bad

[brisk estuary]

Yes, then now I agree with you

You're on the right track, what's next?

[bitter portal]

(a + L)/2 - L < sn < (a + L)/2 + L

a/2 - L/2 < sn < a/2 + 3L/2

[brisk estuary]

More accurately:

$\forall \epsilon, \exists n_0, \forall n \geq n_0, \lvert u_n - L \rvert \leq \epsilon$

[whole widget]

Rion

[brisk estuary]

So here we can indeed pick $\epsilon = \frac{c - L}{2}$

[whole widget]

Rion

[bitter portal]

yes

wait c - L/2

[brisk estuary]

yeah the c you picked earlier

as the midpoint of a and L

[bitter portal]

its not midpoint of a and L?

[brisk estuary]

So to clear your mind, here is what our current scheme looks like

$a > c > L + \epsilon > L$

[whole widget]

Rion

[brisk estuary]

It's true however we can pick your choice of epsilon too

So $a > L + \epsilon > L$

[whole widget]

Rion

[brisk estuary]

So what now?

[bitter portal]

so our midpoint can be between c and L

or a and L

bc it fits the epsilon, but ur choosing between c and L out of convenience later on?

[brisk estuary]

No, to be fair there isn't any particular reason we can't go with your choice.

[brisk estuary]

This is the most important part

The fact that we can pick epsilon so that a is not in [L - epsilon, L + epsilon]

And this will suffice to proceed

[bitter portal]

ok so we said that there is an n0 for which n > n0 a> u_n

because its just a> L + epsilon > L > L - epsilon

which means this:

[bitter portal]

here

which means that u_n < a means there is an n from {0, 1, 2, 3...n0}

and so s_n > u_n (because of monotonically increasing) then s_n < a if n from {0, 1, 2, 3...n0}

[brisk estuary]

No

$\forall n \geq n_0, u_n < a$

[whole widget]

Rion

[brisk estuary]

So you have a LOT of terms u_n that are below a

[bitter portal]

oh ok

which means that u_n > a for finite n from {0, 1, ,2 ,3 ...n0}

[brisk estuary]

Which means that $u_n < a$ for an infinite number of $n$

[whole widget]

Rion

[brisk estuary]

this is more important

[bitter portal]

ok ill give it some thought

ok so then can i conclud that liminf s_n < a?

[brisk estuary]

With a bit more elaboration, you can also conclude that $s_n < a$ for an infinite number of $n$ too

[whole widget]

Rion

[bitter portal]

ok but no contradiction yet

[brisk estuary]

When this is proven, then you have a contradiction

because then $a$ cannot be in $A$

[whole widget]

Rion

[bitter portal]

no i think it already contradicts with what the definiton of A is

in the problem

[brisk estuary]

Yes, indeed, but you first need to show that s_n < a for an infinite number of n

After that, it won't be hard to conclude

but we're one step short

[brisk estuary]

To show this, you can fall back on the definition of u_n

[bitter portal]

ok u_n is a term in a subsequence by the mct theorem that its bounded and convergent

(u_n) = liminf{u_n: n > N}

[brisk estuary]

in particular, with a small enough $\delta > 0$:
$$u_n + \delta > \inf_{m \geq n} s_m$$
So you can find a $s_{\phi(n)} \in [u_n, u_n + \delta]$ with $\phi(n) \geq n$.

So in fact, the set ${\phi(n) | n \in \mathbb{N}}$ is infinite

[whole widget]

Rion

[bitter portal]

i have no idea whats going on here

[brisk estuary]

Sorry, let me break down step by step

[bitter portal]

so u_n > inf s_M originally?

[brisk estuary]

So you know that $u_n = \inf_{m \geq n} s_m$

[whole widget]

Rion

[brisk estuary]

So in particular, $u_n + \delta > \inf_{m \geq n} s_m$

[whole widget]

Rion

[bitter portal]

ok and by archimedean principle, you can find phi that makes it bigger

yeah

[brisk estuary]

So in particular, u_n + delta is NOT the inf

so you can find an element below it

which we call $s_{\phi(n)}$

[whole widget]

Rion

[brisk estuary]

where $\phi(n)$ is the $m \geq n$ that verifies: $u_n < s_{m} < u_n + \delta$

[whole widget]

Rion

[bitter portal]

right i see that.

[brisk estuary]

So now, you collect those $\phi(n)$ in one set

[whole widget]

Rion

[brisk estuary]

And in particular, since $\phi(n) \geq n$, then $\phi$ cannot be stationary (i.e. stay constant after a certain rank)

[whole widget]

Rion

[brisk estuary]

So the set ${\phi(n) : n \in \mathbb{N}}$ gotta be infinite

[whole widget]

Rion

[bitter portal]

ok i see

so a is not in the set A because...

[brisk estuary]

Well, remember that u_n converges to L

So u_n + delta isn't too far off L + delta

Or more formally, apply the property using $\delta_n = \frac{a - u_n}{2}$

[whole widget]

Rion

[brisk estuary]

So you can find your $\phi(n)$'s

[whole widget]

Rion

[brisk estuary]

So now, the set of ${n : s_n < a}$ is proven to be infinite

[whole widget]

Rion

[brisk estuary]

Since for any $n \geq n_0$, it holds: $s_{\phi(n)} < a$

[whole widget]

Rion

[brisk estuary]

And ${ \phi(n) : n \geq n_0 } \subset { n : s_n < a }$

[whole widget]

Rion

[brisk estuary]

The former being infinite for reasons stated above

[bitter portal]

dang, why is it this long to prove the =?

[brisk estuary]

No offense but I am detailing to the finest detail every single step, hence it is so long

But the full proof should fit within 10 lines at most

[bitter portal]

no no. you're good. im just wondering why its so long but like its only 1 problem ( didnt prove the second half: inf B = lim sup s_n)

[brisk estuary]

You don't need to prove it

You can say we use the same reasoning

Or just apply our already proven results for the sequence (-s_n), which lim inf is -B

Both are clean ways of finishing

[bitter portal]

so from here, on the left side, is that representing for liminf and so liminf is an element of A so there are elements of A that are less than liminf, and that contradicts???

[brisk estuary]

The liminf is not necessarily an element of A, but this subset inclusion shows that the right set is infinite

Which will contradict that a is in A

[bitter portal]

so liminf s_n > a and sup A > liminf s_n and therefore sup A = liminf s_n?

[brisk estuary]

Almost there

sup A >= liminf sn >= a

Now remembering that A is an interval in the form (-infinity, sup A) or (-infinity, sup A]

You can make a go to sup A (in the sense of limits)

And still maintain the inequality

[bitter portal]

actually isnt this just the monotone convergence theorem?

inf increases, then the lim {inf} reaches the upper bound at sup A

[brisk estuary]

No it is not

It is indeed known that the sequence (u_n) converges

But it is clearly not trivial that the limit itself is sup A

We spent a lot of effort proving that

[bitter portal]

right, but we prove sup A >= liminf sn , not strictly equals.

[brisk estuary]

Correct

We also proved that for any a in A, liminf sn >= a

And for the reasons stated above it gives you grounds to perform some kind of squeeze

As in: sup A >= liminf sn >= sup A

Hence, sup A = liminf sn

[bitter portal]

by squeeze theorem

[bitter portal]

ok, is it possible if u can combine the main parts all in 10 lines or less?

i know there's a lot of work in between and I did got lost along the way 😆

but thank you for ur effort

[bitter portal]

u said it could be rewritten w 10 lines or less rite?

[azure sparrow]

this is like saying

"Can you combine a 45 minute lecture into a 30 second youtube short"

try to backread and note stuff down on your own it'll help you a ton

[bitter portal]

ok

but there are a lot of side details here that is to the side so i might double check with Rion again

but thank you

[azure sparrow]

sure double-check after noting them down

if you want I could write you a full proof too if that helps

[brisk estuary]

Sure I will try to make condensed notes

Firstly, let us show that A is an interval. Let $a \in A$. If $a' < a$, then ${n : s_n < a'}$ is also finite, hence $a' \in A$. This shows that $A = (-\infty, \sup A)$ or $(-\infty, \sup A]$.

[whole widget]

Rion

[hoary juniper]

rion___ has been timed out for 5m
Reason: Similar-Messages Spam

rion___ | 276638393923010560

[whole widget]

Rion

[brisk estuary]

Part 2: Let us show that $\forall a \in A, L \geq a$. Let $a \in A$, and suppose that $a > L$. Let $c \in (L, a)$. Likewise, there exists $n_0 \in \mathbb{N}$ such that $forall n \geq n_0, u_n \geq c < a$. Let $\delta_n = c - u_n > 0$. With any $n \geq n_0$, since $\inf m \geq n s_m < u_n + \delta_n = c < a$, we can pick $\phi(n) \geq n$ such that $s_{\phi(n)} < u_n + \delta_n$. Since $\phi$ is not stationary, ${ \phi(n) : n \ in \mathbb{N}}$ is infinite, which contradicts $a \in A$.

[whole widget]

Rion

[brisk estuary]

Finally: $\forall n > 0, a_n = \sup A - \frac{1}{n} \in A$, since $A$ is an interval. The squeeze theorem imposes that as we take the limit,$\sup A \geq L \geq \lim_{n \rightarrow \infty} a_n = \sup A$.

[whole widget]

Rion

[bitter portal]

wow, thank u @brisk estuary . I dont know how I can give back with ur help. I know im not the strongest in Real Analysis but i appreciate ur help

[finite gorge]

@bitter portal has given 1 rep to @brisk estuary

[bitter portal]

i'll ask u for help on the second part if thats ok w u; i dont want to bother @brisk estuary too much, Rion helped me out a lot this time.

i'll actually make another post bc this one is too cluttered with the first part already.

[bitter portal]

https://discord.com/channels/624314920158232616/1239414520376463371

[brisk estuary]

As said before, saying that an analogous reasoning yields the result is perfectly fine

Alternatively, by flipping the sequence, B fulfills the same role as A and the results are applicable.

[bitter portal]

wdym flipping the sequence, like reversing the last term to become the first and so on?

[azure sparrow]

he meant multiplying everything by -1 I think

[brisk estuary]

Multiplying by -1

$B = \left{ b : {n : s_n > b } \text{ is finite} \right}$

[whole widget]

Rion

[brisk estuary]

$B = \left{ b : {n : (-s_n) < (-b) } \text{ is finite} \right}$

[whole widget]

Rion

[brisk estuary]

Which, to no one's surprise, has the same form as A here

we just replaced s_n by -s_n

and a by -b