TITLE: Are there uncountable antichains in $\mathcal{P}(\mathbb{N})$
QUESTION [6 upvotes]: As the title states, I am unsure of how to construct an uncountable antichain in the power set $\mathcal{P}(\mathbb{N})$.

Recall that an antichain is a set $\mathcal{A}\subset\mathcal{P}(\mathbb{N})$ where each element of $\mathcal{A}$ is not a subset of another element of $\mathcal{A}$. 

It seems that everything I attempt requires me to assign a property to the "$n$-th set" but that already makes it countable...

REPLY [2 votes]: Old question, but I thought I'd post my solution too since the $1-1$ correspondence felt more intuitive to me than the other solutions on this page.
Construct the antichain:
$$\mathcal{A} = \{\alpha : \alpha \text{ for each unique } \alpha = \{a : \text{ Either } a = 2n \text{ or } a = 2n - 1 \text{ for each } n \in \mathbb{N}\} \}$$
For the $1-1$ correspondence use the function:
$$f(\alpha) = \{n : n = \frac{a}{2} \text{ for } a \in \alpha \text{ if } a \text{ is even }\}$$
which gives you $1-1$ and onto function $f: \mathcal{A} \rightarrow P(\mathbb{N})$
The intuition is at each pair of even-odd numbers in an element in the antichain, we make a choice whether to include the even or odd number. This is equivalent to making a choice of whether to include an element in $\mathbb{N}$ or not when constructing a subset of $\mathbb{N}$.