3

Consider the partially ordered set $\langle P(\mathbb{N}), \subseteq\rangle$.

I am trying to show two things:

(a) There exists a chain $C$ such that $|C|=\mathfrak{c}$.

(b) There exists an anti-chain $A$ such that $|A|=\mathfrak{c}$.

Can you help me with this ? It seems to be fairly hard.

Caleb Stanford
  • 45,895
  • What is $\mathfrak c$? What is a chain? An anti-chain? For a), is it enough to show that there exists a chain that is not countable? – Orlando Marigliano Sep 16 '16 at 19:06
  • Instead of $\mathbb{N}$ use $\mathbb{Q}$ – Rene Schipperus Sep 16 '16 at 19:07
  • 1
    @OrlandoMarigliano This is standard terminology and notation in set theory. A chain in a partial order is a set of elements of the partial order, any two of which are comparable; similarly, an antichain is a set of elements no two of which are comparable (interestingly a different meaning is given when talking about forcing. And $\mathfrak{c}$ denotes the cardinality of $\mathbb{R}$. – Noah Schweber Sep 16 '16 at 19:08
  • oops, I accidentally assumed the continuum hypothesis in editing this. Fixed now. – Caleb Stanford Sep 16 '16 at 19:51
  • Related: http://math.stackexchange.com/questions/253192, http://math.stackexchange.com/questions/1182145 – Watson Sep 17 '16 at 08:49

1 Answers1

5

HINT: You can replace $\Bbb N$ with any other countably infinite set $S$: just use a bijection between $S$ and $\Bbb N$ to transfer subsets of $S$ to corresponding subsets of $\Bbb N$.

  • For (a) consider the subsets of $\Bbb Q$ that are the left halves of Dedekind cuts. That is, for each $x\in\Bbb R$ let $A_x=\{q\in\Bbb Q:q\le x\}$, and consider the sets $A_x$.

  • For (b) consider the almost disjoint families of this answer.

Community
  • 1
Brian M. Scott
  • 616,228