There's no way this isn't a trick

[timber cape]

I need help finding the supremum of these set

[potent gyro]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[sonic yacht]

Miguel

Miguel

Miguel

[timber cape]

Why did they give me the same 3 times

Is there any trick

Isn't the proof the same for them all?

[deft quiver]

they want you to suffer

[iron pivot]

Miguel hi

@timber cape do yk what a sup is?

[timber cape]

Yes

[iron pivot]

describe it to me

[timber cape]

For all elements of the set, it is bigger than them all

[iron pivot]

are you sure?

[timber cape]

Bigger or equal

And if there is another number in the real numbers that also meet that criteria, then the sup is smaller or equal than it

Which can also be described in a fancier and more convenient way using epsilon

[iron pivot]

mkay

first one is straight forward

2-eps

2nd one is the same but eps in Q

3rd one kinda tricky

[timber cape]

Epsilon in Q?

Hmmm

[iron pivot]

it'd be between sup Q and sup R

but as R-Q is dense in R

denser than Q

it's greater than sup Q

[timber cape]

I don't think we've learnt that I is dender than Q

[iron pivot]

in a way sup R and sup R-Q are same

[timber cape]

And sup Q is smaller

Because there is an irrational between 2 and the biggest element of the set?

But the sup of the second one is 2

Isn't it

[iron pivot]

yes

biggest rational element*

[timber cape]

Yes

[timber cape]

.

Like even if you can find an irrational, you can also find another rational between that irrational and 2

I guess if you repeat that long enough, it would be 2

[empty patio]

@timber cape what seems to be the issue

yes theyre all 2

[timber cape]

Ok

[timber cape]

Now I have the issue to prove they're two

For example for the first one

I've been trying to show that the sup cannot be bigger than two or smaller than 2, so that it has to be 2

Showing that it cannot be bigger than 2 is easy because 2 is already an upper bound so the sup cannot be higher by it

I have problems showing that the sup cannot be smaller

[empty patio]

The first one is the R one correct?

[timber cape]

Yes

So I want to show that the supreme cannot be less than 2

If it was less than two, that means that

[empty patio]

Well, you start by supposing the sup is less than 2, then find a contradiction

[timber cape]

$\forall \epsilon > 0 \exists a \in (-\infty, 2)=A|\text{sup}(A)-\epsilon < a ≤ \text{sup}(A)<2$

So at first I tried using some values for epsilon, such as 2 - sup(A)

But none of them seemed to give a contradiction

[empty patio]

This statement is the definition of the sup

[timber cape]

Yes

But with the assumption that sup(A) < 2

You could write it there like

[empty patio]

wait, this will not give you a contradiction because it is not the full definition

[sonic yacht]

Miguel

[timber cape]

There it is

[empty patio]

this can be true, but it doesn't mean necessarily that a number is a sup

let me explain what i mean

Assume there is a $m$ such that $\forall, \epsilon > 0, ; \exists a \in A=\qty(-\infty, 2) : : ; m-\epsilon < a \leq m$

[sonic yacht]

Pyro #Pyro4manager

[timber cape]

Yes

[empty patio]

this doesn't mean m is the sup

[timber cape]

Hmmm

[empty patio]

in fact this only means that $m \leq 2$

[sonic yacht]

Pyro #Pyro4manager

[empty patio]

any number less than or equal to m will satisfy this property

[timber cape]

Well not

Imagine m = 1

Oh

Ok

Yes

[empty patio]

so what do you think the issue is

[timber cape]

I have to use the first condition

[empty patio]

Yes

It has to satisfy this, but also be an upper bound

[timber cape]

$\forall a \in A 2 > \text{ sup}(A) ≥ a$

[empty patio]

it's the other way

[timber cape]

Right

[empty patio]

so you're trying to show that a number less than 2 cannot be a sup

[timber cape]

And I have that sup(A) < 2, yes

[empty patio]

But I just showed you that a number less than 2 does satisfy the second property

so in order to find a contradiction, you need to prove that a number less than 2 cannot satisfy the first property

which is being an upper bound

[sonic yacht]

Miguel

[empty patio]

so how would you show that $\forall, m < 2$, $m$ is not an upper bound

[sonic yacht]

Pyro #Pyro4manager

[timber cape]

Finding a number in A greater than m

[empty patio]

Yep

try it

[timber cape]

If m < 2, then m + x = 2 for some x in the reals

So m + x/2 is in the set

Woooo

[empty patio]

yes!

[timber cape]

Ok this is great thank you so much

[empty patio]

alternatively you can just take $(m+2)/2$ (the number between m and 2)

[sonic yacht]

Pyro #Pyro4manager

[empty patio]

you're welcome

[timber cape]

Yeah and for the other two cases I just have to show that those numbers exist

Because they're dense in R

[empty patio]

however it will be a little different for the other cases, because this is in R and you can use any number you want to finda contradiction

[empty patio]

Yeah

for example in Q you can't use $\frac{m+2}{2}$ as a contradiction, because that is not necessarily rational

[sonic yacht]

Pyro #Pyro4manager

[timber cape]

Ok ok I'll try to see if I can do it

I've been stuck all day with this, you've taken me out of this pit of depression

Thanks you so much again

I'll delete the post once I do it