Antichain of P(N): Determining Uncountability with Known Sets

[R.P.F.]

Homework Statement

P(B) refers to the collection of all subsets of B.
Given a set B, a subset A of P(B) is called an antichain if no element of A is a subset of any other element of A. Does P(N) contain an uncountable antichain？

Homework Equations

The Attempt at a Solution

If I can build a bijective map between an antichain of P(N) and another set of known cardinality, then I will be able to know if the antichain is uncountable. Sets I know that are uncountable: R, sequences of 0's and 1's. Is it one of these two?

Sorry I'm not able to do much with this problem. Any help is appreciated.

 

[Raskolnikov]

hmm...you've got me interested! but I can't put together a full proof at the moment, maybe in the morning after some sleep haha. If it's of any help, here's one more uncountable set to add to your list: P(N). If this is new to you, read through an explanation of Cantor's Theorem.

 

[Dick]

Hint: divide N into pairs, {1,2}, {3,4}, {5,6}... Now form sets by choosing exactly one element of each pair. What properties does the resulting collection of sets have?

 

[R.P.F.]

Hint: divide N into pairs, {1,2}, {3,4}, {5,6}... Now form sets by choosing exactly one element of each pair. What properties does the resulting collection of sets have?

hmm...you've got me interested! but I can't put together a full proof at the moment, maybe in the morning after some sleep haha. If it's of any help, here's one more uncountable set to add to your list: P(N). If this is new to you, read through an explanation of Cantor's Theorem.

I think I kinda got it. So this has to do with the cardinality of P(N) right? Because P(N) is uncountable and we could build an antichain of the same cardinality, so the antichain is also uncountable.
Thank you both!

 

[R.P.F.]

Oh sorry I think I got it wrong. I still need to use Cantor's diagonal Thm.

Thanks!

 

[Dick]

I think I kinda got it. So this has to do with the cardinality of P(N) right? Because P(N) is uncountable and we could build an antichain of the same cardinality, so the antichain is also uncountable.
Thank you both!

Yes. You need to show i) the resulting sets are an antichain and ii) the cardinality of the antichain is the same as P(N).