Real analysis Sup(A) and Inf(A)

[trim sedge]

can someone help me find the Sup and Inf for this set A = { ( E(x) + 1 ) / x , x > 1/2 } E(x) is the integer part function

[crude bramble]

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[junior crescent]

any thoughts of your own @trim sedge ?

how does E(x) / x behave for positive x, for instance?

[trim sedge]

yes

so the first thing is i got that for every Y in A 1 < y < 3

@junior crescent

[jolly dawn]

my ear die

hello

[trim sedge]

?

[trim sedge]

XD

[jolly dawn]

how does (E(x) +1) correlate to x?

what is integer par?

[trim sedge]

E(x) + 1 > x

[jolly dawn]

it the : number - (not integer par)

[jolly dawn]

yes and what can we deduce from this

[trim sedge]

that it is bigger than 1

[jolly dawn]

it might be a >=

[trim sedge]

no it cant be

equal to 1

[jolly dawn]

aight

now

there's another side to our story

[trim sedge]

this is what i got so far

[jolly dawn]

uhm

okay you didn't tell us what you got yet

[trim sedge]

E(x) + 1 / x < x+1 / x = 1 + 1/x < 1 +2 =3

[jolly dawn]

1+ what? how'd we get there

[trim sedge]

x > 1/2 therefore 1/x < 2

[jolly dawn]

okay

[trim sedge]

now i got that it is bounded

the only thing left is to use the epsilon definition to get both inf(A) and sup ( A)

[jolly dawn]

Let x = n+f where f is the fractional part, f is in [0,1).
then floor(x) = x-f = n
floor(x) + 1= n+1 > n+f as we saw
floor(x)+1/x = n+1/n+f
so our set is basically :
{n+1/n+f : f in [0,1) and n+f > 1/2}

||now you'd see that n+1/n+f > 1 but for large enough n we can get arbitrarily close to 1 from below as the 1 and f don't contribute much.||
Infimum in the spoiler

now for the supremum you'd clearly see that we can try to maximize it when preferably n=0 and we have something of form 1/f

you can figure the rest out

@trim sedge

[trim sedge]

well this is not enough to prove that 1 is the inf

we need to find an Y in A such that for every e > 0 , Y < 1 + e

and to find Y we need to find an x > 1/2 that satisfies the proposition

[jolly dawn]

it's the inf nm

[jolly dawn]

but do you see that Y < 1 readily

[trim sedge]

i know that

but we need to prove that it is the greates lowerbound

[jolly dawn]

Would finding Y < 1+e suffice

[trim sedge]

yes

[jolly dawn]

we already have Y < 1 just show that there always exists an Y > 1 - e

[trim sedge]

well thats what i am stuck on XD

[jolly dawn]

I gave you the means.

[trim sedge]

it seems that i cant find one

[jolly dawn]

fix any f
(say f = 0.5)
then we work with x of form :
0.5,1.5,2.5,...... on and on
then you'd see,
lim n-> infty (n+1)/(n+f) = 1

as f is finite

this is enough to prove that it's arbitrarily close to 1 isn't it?

@trim sedge

[trim sedge]

yes

but is it enough to prove that it is the greatest lower bound

you know the definition of a INF

[jolly dawn]

bro you tell me

I know the definition of an inf

suppose we have a set

$\qty{f(n) : n \in \mathbb{N}}$ and we have that, $\$ $f(n) < 1$ and $f(n) \rightarrow 1$

[sour thorn]

tetotetotetotetotetotetoteto

[jolly dawn]

prove that 1 is the inf

sup*

sup

[trim sedge]

INF(A) = m == { 1) for every x in A x >=m 2) for every e > 0 there exist x in A such that x < e + m

[jolly dawn]

$\qty{f(n) : n \in \mathbb{N}}$ and we have that, $\$ $f(n) > 1$ and $f(n) \rightarrow 1$

[sour thorn]

tetotetotetotetotetotetoteto

[jolly dawn]

then 1 is the inf

prove this

@trim sedge

[trim sedge]

ok

[jolly dawn]

this is very hard to read, sorry but please use a little bit of latex

as I have a headache and an ear ache and a cold so I have more sympathetical privileges rn :3

[trim sedge]

$\text{inf}(A) = m \quad \equiv \quad \left{
\begin{aligned}
&1) \text{ For every } x \text{ in } A, , x \geq m \
&2) \text{ For every } \varepsilon > 0, \text{ there exists } x \text{ in } A \text{ such that } x < \varepsilon + m
\end{aligned}
\right.
$

[sour thorn]

I5Pirate

[trim sedge]

this is it

i understand what are you trying to say

but the professor didnt like that answer and said it is not complete

[jolly dawn]

[trim sedge]

XD

[jolly dawn]

it's wrong lol

[trim sedge]

no

[jolly dawn]

oh wait it the Infimum

[trim sedge]

bruh

[jolly dawn]

ye

[jolly dawn]

this is it yes

[jolly dawn]

now prove this @trim sedge

that 1 is the Infimum

[trim sedge]

well if i proved this i wouldnt need help in the first place lol

there is little to no information about the function

[jolly dawn]

I already gave you all needed information

1 is the sup

[jolly dawn]

in this one it's the inf

[trim sedge]

and how did you know its the Inf

[jolly dawn]

why?
f(n) >= 1 given
f(n) tends to 1
Thus, for all e>0 we can always find some n such that,

|f(n) - 1| < e
=> f(n) - 1 < e
=> f(n) < 1 + e
which proves that 1 is the inf doesn't it?

[trim sedge]

bruv you need to prove that the limit is 1

you cannot just say the limit is one

[trim sedge]

you said that u can always find an n

can you give me that n

[jolly dawn]

yeah why not we can always find such a finite n

by the definition of a limit

[trim sedge]

i know we can find an n

[jolly dawn]

hah

$\lim_{n\to \infty} \frac{n+1}{n+f}=1$

[sour thorn]

tetotetotetotetotetotetoteto

[trim sedge]

ok let me give you an example

[jolly dawn]

there

[trim sedge]

well our professor dont like that u stated that the limit is 1 without a proof

i know that the limit is 1

[jolly dawn]

okay why the fuck should I care?

prove it yourself

[jolly dawn]

I don't mean it in an offensive way

just, you can prove it yourself :3

and add it on :3

[trim sedge]

💀 bro u said it the most offensive way possible

[jolly dawn]

I don't mean it

anyways

[jolly dawn]

you can add the proof of this

if you remember, we aren't giving you the whole proof written in gold

just the sketch

I ended up Givin you the whole proof nonetheless

[trim sedge]

bruv and i told you that u r using a circular reasoning

💀

with all due respect that aint a proof

[jolly dawn]

what.

where did I use circular reasoning.

[trim sedge]

you said that the limit is one

[jolly dawn]

and HOW is that circular

@trim sedge

[trim sedge]

well if you state that its limit is one u just said that its sup is 1

[jolly dawn]

boom, you just proved it's sup is 1

that's how proves work

[trim sedge]

bro i give up

[jolly dawn]

did you read this once?

coffee and oreos go together :3

anyways

[trim sedge]

again u just used the fact that n/n = 1 to get the limit

the actually prove that 1 is the limit you gotta find that x dude

[jolly dawn]

but the limit being 1 is the proof of there existing such x.

in our case

you can prove the limit without using like anything from the problem

[jolly dawn]

"the actual prove"

thats a proof too

you want me to prove it with the epsilon delta definition?

[trim sedge]

yessss thats it

[jolly dawn]

do it yourself.

why can't you do that yourself.

why do I have to prove that the limit of (n+1)/(n+f) is 1 as n goes to infinity

[trim sedge]

well tbh i didnt try this

let me give it a try

[jolly dawn]

want a hint

[trim sedge]

yes plz

well i would have multiple cases for epsilon

[jolly dawn]

bro what.

just use long division.

@trim sedge just use long division

$\frac{n+a}{n+b} = 1 - \frac{b-a}{n+b}$

[sour thorn]

tetotetotetotetotetotetoteto

[jolly dawn]

actually with the polynomial division thing

you can derive this generally

[trim sedge]

+close