can anyone teach me this(intro to real analysis)

[proven nimbus]

what does sup A $\not \in$ A tell you about A itself, specifically the presence, or lack of, a certain type of element?

[wary halo]

Céline

[torpid current]

maybe its something like 0<A<1,then sup A gonna be 1

but i dont know the genally case

[proven nimbus]

what's the definition of a supremum?

[torpid current]

the least upper bound

[proven nimbus]

correct. so if the least upper bound is already out of A, what does that tell you about A itself?

[torpid current]

order S cant be finite?

[proven nimbus]

uh, seems like we're jumping quite a few steps there

plus, I asked about A

think about this - if a bound, known to be the lowest maximum, is out of A, can A itself contain any maximal element?

[torpid current]

no

[proven nimbus]

excellent. that is our key observation.

since A cannot contain any maximal elements by virtue of sup A not being in A, what can you say about every element in A?

[torpid current]

its countable infinite?

[proven nimbus]

jumped steps again

we want to prove that

actually that's not what we want to prove, and neither can we confirm A is countably infinite

A could be uncountable, but we don't care about A's size

if A has no maximum, then for every element in A, you cannot say it is the biggest element in A. and we show that this element (call it a) cannot be the biggest element in A by saying that...?

[torpid current]

sup A is not in A?

[proven nimbus]

we concluded that earlier. not used here

well, if a is not the biggest element, that means there will always be elements that are ___ than a

[torpid current]

bigger

[proven nimbus]

there we go

[torpid current]

oh so u can say its infinite?

[proven nimbus]

not... that fast

[torpid current]

oh sor

[proven nimbus]

oh harpz is here, I'll probably take a back seat then

ok nvm let's continue

so we know there will always be elements greater than a. pick one of those elements and call it b.

then, we can say that for every element a in A, there is some b that is also in A, where a < b. agreed?

keyword: every element in A

[torpid current]

agree

[proven nimbus]

cool

this is our key observation

now, we want to show that A has a countably infinite subset

our observation showed that every element in A has at least one other element larger than itself

so we will need to construct a subset of A that has this property

now, start by picking some a_1 in A

this is our initial element

now, from our observation, if I pick an a_2, what do you want the relationship between a_1 and a_2 to be so that it can help us, and how can you justify this relationship?

[torpid current]

a2 is bigger than a1?

[proven nimbus]

and how can you justify this?

[torpid current]

maybe use induction

[proven nimbus]

how do we know such an a_2 exists?

not yet

you need to prove the base case first

[proven nimbus]

btw I haven't said this yet, but you must also show why you're allowed to just pick some random element in A

(hint: given by the question)

[torpid current]

coz its nonemply

[proven nimbus]

as for this, consider our observation

[proven nimbus]

correct.

remember to write it in your proof

[proven nimbus]

what about this?

[torpid current]

because A is in an order set?

[proven nimbus]

not exactly

if A was given as ordered but nothing else, then it can be assumed to have a maximum, and thus there will be an a in A that is greater than all other elements

[proven nimbus]

^^^

[torpid current]

a_2 can always be greater a_1 because of this?

[proven nimbus]

the message you replied to isn't our observation

[proven nimbus]

^^

[torpid current]

so it always exist a2 that is greater than a1 because of this

[proven nimbus]

yes

[torpid current]

so we can treat it as the basd case for the induction

[proven nimbus]

yes

wanna continue the rest of the proof yourself? you're quite close to the end

[torpid current]

then there also exist a_m+1 that is greater then a_m with the logic we used in a_1

[proven nimbus]

what can we conclude from this?

[torpid current]

but do i need to map the subset to a countable infinite set?

[proven nimbus]

not yet

actually yes

but you'll first need to conclude something else from this

you seem to be very keen on getting the proof done and over with lol

[torpid current]

by induction ,there is a subset (call it M) form A where there is always an element that can be greater than other element

[proven nimbus]

I don't know if that's accepted

I would have formed a sequence consisting of all the elements a_n first

if you think that's accepted or if your prof tells you it's accepted

by all means directly form this subset

you're still not done

you're one step away from concluding that M is countably infinite

[torpid current]

mapping it?

[proven nimbus]

show the mapping then

(this mapping is actually rather easy - don't overthink it)

[torpid current]

let there be a function f that map M to k(k belongs to the natual number)

first have to show its 1-1

[proven nimbus]

ok overkill

but let's hear it

[torpid current]

Let f(x1) = f(x2)

then x1 = x2

[proven nimbus]

what is the justification?

[torpid current]

Let f(a1) map to 1 from the natural number

and f(a2) to 2 and so on

so if f(x1)=f(x2)that can only mean x1 = x2

[proven nimbus]

I would be very cautious of my wording here.
you have not declared the sequence a_n to be strictly increasing. you only mentioned that B is a subset formed from a_n where there is always some element that can be greater than other elements. to some people, this will not seem like you have stated that this sequence will never have duplicates

[torpid current]

so i should mention it right after the induction that a_r+1 will always be greater than a_r

[proven nimbus]

you should conclude that the sequence of a_n is strictly increasing, which means that yes, a_(n+1) is always greater than a_n for all n

[proven nimbus]

then, you can make this conclusion

after that, you are only one step away from your conclusion

[torpid current]

then prove it is onto

[proven nimbus]

well, if you want to do that, two steps

either way, I gtg now, so you can ping helpers to check your answer after this

all the best