Does $\mathcal{P}(\mathbb{N})$ contain an uncountable antichain?

Question

I was given the following question on my homework:

Given a set $B$, a subset $\mathcal{A}$ of $\mathcal{P}(B)$ is called an antichain if no element of $\mathcal{A}$ is a subset of any other element of $\mathcal{A}$. Does $\mathcal{P}(\mathbb{N})$ contain an uncountable antichain?

My attempt at an approach:

Consider all subsets of $\mathbb{N}$ with $n$ elements. By intuition, none of these sets are subsets of one another. Therefore, the set containing all of these subsets (let's call it $\mathcal{A}'$) is an antichain and its cardinality is trivially infinite.

Let's say,

$$\mathcal{A}_0 = \{0, 1, 2, ..., n\}$$ $$\mathcal{A}_1 = \{1, 2, 3, ..., n+1\}$$ $$\mathcal{A}_k = \{k, k+1, k+2, ..., n+k\}$$

These sets are all contained in $\mathcal{A}'$. Now, let's show that $\mathcal{A}'$ is uncountable:

Take every set in $\mathcal{A}'$. Take the first element in that set and make it the singleton set containing that element. For example,

$$\mathcal{A}_0 = \{0, 1, 2, ..., n\} \rightarrow \{\{0\}, 1, 2, ..., n\}$$ $$\mathcal{A}_1 = \{1, 2, 3, ..., n+1\} \rightarrow \{\{1\}, 2, 3, ..., n+1\}$$ $$\mathcal{A}_k = \{k, k+1, k+2, ..., n+k\} \rightarrow \{\{k\}, k+1, k+2, ..., n+k\}$$

These are all new sets that are $\in \mathcal{P}(\mathbb{N})$, but they were not in our previously defined $\mathcal{A}'$. Thus, the antichain $\mathcal{A}'\subseteq\mathcal{P}(\mathbb{N})$ is uncountable. This is inspired by Cantor, but I'm not really sure if the proof makes much sense in the realm outside of my brain. Any ideas?

Edit: I am really trying to avoid using binary trees or solutions involving varying degrees of decimal expansions; I've seen them and they make no sense to me. I thought that this might be a simpler way of explaining it, please let me know if that's incorrect.

Answer 1

When you are asked to exhibit an uncountable collection you should be prompted to think of how you can have an infinite number of binary choices. You can partition the naturals greater than $0$ (to avoid the question of whether $0$ s a natural) into countably many subsets of the form $\{2n-1, 2n\}$. Now take all the subsets of the naturals that have exactly one member of each of these sets. There are $\mathfrak c$ of them and none is a subset of any other.

Answer 2

Since $\mathbb Q$ has the same size as $\mathbb N$, their power sets are order-isomorphic. Hence it suffices to find an uncountable antichain in $\mathcal{P}(\mathbb Q)$, which is easy:

$\{(r, r+1) \cap \mathbb{Q}: r \in \mathbb{R} \}$ is an antichain of size $\mathcal c > \aleph_0$.