Post 2: inf B = lim sup s_n

[verbal gate]

Continued from the first post.

(I will write my thoughts below first)

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[verbal gate]

So we first prove that B is an interval.

Let b $\in$ in B

[terse dawn]

JackieChen

[verbal gate]

If $b' < b$, then ${n: s_{n} < b' } $ is also finite. Therefore b' $\in$ B. This shows that (-$\infty$, sup B) or (-$\infty$, sup B].

[terse dawn]

JackieChen

[verbal gate]

wait i have a question, does s_n have to be monotically nondecreasing?

can it be monotically nonincreasing?

[unique charm]

@verbal gate for this part trying making u_n negative and B negative too

it'll flip the bounds

[verbal gate]

what do u mean?

that now ${n \in N: s_{n} , is \leq than , b }$ is finite?

[unique charm]

ignore this

[terse dawn]

JackieChen

[unique charm]

how does it show that

what is that proposition

it shows an interval?

[unique charm]

changing the inequality signs won't do anything here

@verbal gate have you heard of limit points?

of a sequence

are the limit points of the sequence if they exist

maybe that's bad notation

limit points are the points to which the sequence gets arbitrarily close arbitrarily often

$\qty(\forall \epsilon >0)\qty(\forall k\in\bN)\qty(\exists m\in \bN_{>k})\text{ } \abs {s_m-l} <\epsilon$

[terse dawn]

kaffee

[unique charm]

this makes "l" a limit point

[verbal gate]

thats the definition of a limit of a sequence, yes.

[unique charm]

Now the limit supremums and infimums are just the supremums and infimums of this set of limit points

[unique charm]

it's not

a limit point need not be a limit

a sequence can only ever have one limit

but it may have countably many limit points

$\qty{\frac{1}{m}+\frac{1}{n}:m,n\in\bN}$

[terse dawn]

kaffee

[unique charm]

while being bounded too

so these are limit points

[unique charm]

it has all 1/n as a limit point

consequently

we have that

$\qty(\exists k\in\bN)\qty(\forall m\in\bN_{>k})\text{ } s_m \in \qty[\lim\inf s_n,\lim\sup s_n]$

[terse dawn]

kaffee

[verbal gate]

so limit points are just the points inside a given epsilon bound?

once it is greater than big N

[unique charm]

i wrote the definition down

make of it what you will

if you want to propose "so it's like [some alternate definition]" do it via a logical statement it's easier to understand and use

[verbal gate]

i just never see it written in the ross textbook or define it like that

[unique charm]

now you know

[verbal gate]

(kenneth ross: real analysis, class use that textbook)

[unique charm]

do you get this?

from the definition of limit points and limit extremums

you should be able to prove it so do so

[unique charm]

this instantly proves both your claims

as if x>lim sup then eventually all the terms in the sequence are < x.

try proving it for now

[verbal gate]

ok so this is saying that all points with a bigger index than a given index k is in the set from lim inf sn to lim sup sn

[unique charm]

yep

but prove this

it's necessary to the understanding of limit extremums and why they are important

because every bounded sequence has limit extremums even if it doesn't have a limit!

hint : everything can be proved from the 3 alternate definitions of the limit extremums

[verbal gate]

hang tight, im reviewing lim extremums again.

so theres 2 definition of limsup

limsup $s_{n}$ = $inf {m \in N: \sup s_m }$

[terse dawn]

JackieChen

[verbal gate]

the limsup is the "lowest" ceiling out of all the sup's.

[unique charm]

prove it with whatever you have