Analysis

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What is up with this definition:

1. a is an upper bound for S if for all
2. b is the least upper bound (or l.u.b or supremem or sup) for S if b is an upper bound, and moreover whenever a is any upper bound for S.


2. seems to be a bit redundant, though... if then 'b' is not an upper bound, and if , then 'b' just fits 1.

[mark_alec]

Seems to make sense to me. a is just any upper bound. b is the smallest possible upper bound.

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oh i see, i didn't notice that 'a' didn't have to be in S

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is not onto? How come? :/

Also what is ?

[humph]

Take given by .
The point is that there's a difference between the codomain and range of a function.


And is just the identity function on , that is, the function given by .

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Oh, so only 'in general' it's not onto?

Thanks humph

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Are the cardinality of all uncountable sets the same? For example, is the set of irrational numbers equivalent to the set of real numbers?

[kamil9876]

No, for example there is a famous result that for any set the power set of has strictly greater cardinality. The power set of S is the set of all subsets of S.

therefore

As for the set of irrational numbers, it's cardinality is the same as the cardinality of since is the union of the irrationals and the rationals(rationals are countable) and we have another interesting result you should prove yourself:

if is an infinite set and is a countable set then .

Let be the irrationals, be the rationals.


edit: Reminds me of a very difficult problem you shouldn't waste your time on "Is there a set with cardinality strictly in between the cardinality of integers and the cardinality of reals?".

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Thanks kamil.
I'm not completely familiar with the methods of proof yet so I'll try to prove that a bit later.

Also

edit: Reminds me of a very difficult problem you shouldn't waste your time on "Is there a set with cardinality strictly in between the cardinality of integers and the cardinality of reals?".

Isn't this the theorem that Cantor spent years on and went insane trying to prove? haha i think i'll pass

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4. How does Zorn's lemma work?
"A partially ordered set in which any chain has an upper bound has a maximal element."

Say you have: , then it doesn't have a maximal element. (or do partial orders require countable sets?)



Thanks

[kamil9876]

No definitely no restriction on countable sets, Zorn's lemma can be used to prove that every Vector space has a basis (and there are LOTS of different spanning sets for a given vector space).

your set (with the usual ordering of the reals)  does not satisfy the hypothesis, here is a chain:



It is not bounded above by any element in (so you can't say since it is not in )

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Oh so when it says it has an 'upper bound' it means the upper bound is in Y?
I thought the upper bound didn't have to be in Y.
So should have an upper bound.

[kamil9876]

Wait hold on, I think I have misunderstood your question; are you worried about the fact that has no maximal element? or are you more concerned with the existence of a maximal element in (then again you havn't told me what is)

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I'm more worried about Y... X is just any old set which you get the ordered subset Y from.

[kamil9876]

"A partially ordered set in which any chain has an upper bound has a maximal element."

let me restate it to be more clear:

"If a partially ordered set, has the property that every chain has an upper bound. Then has a maximal element."