# Is the set $A$ of anti-symmetric relations over $\mathbb{N}$ countable? [duplicate]

I think this set is uncountable, but I don't really know how to prove this. Maybe somehow find an injection from $$\mathcal{P}( \mathbb{N} \times \mathbb{N}) \rightarrow A$$?

For any $$X\subseteq \mathbb N$$, the relation $$R_X = \{(x,x) | x \in X\}$$ is anti-symmetric.
Alternatively you can count such antisymmetric relations on a set having $$n$$ elements, which are
$$2^n 3^{\frac{n^2-n}{2}}$$
Set $$n=\aleph_0$$ for set of natural numbers, so that
$$2^{\aleph_0} 3^{\frac{\aleph_0^2-\aleph_0}{2}}>2^{\aleph_0}=c(\text{cardinality of real numbers }\mathbb R)$$
For any $$S \subset \mathbb{N}$$ take $$\mathcal{R}_S = \bigcup_{s \in S} \{(s,s+1)\}$$. It is clear that $$\mathcal{R}$$ is anti-symmetric, as it's a subset of anti-symmetric relation $$\mathcal{R}=\{(x,y):x, and that this transformation is injective, so there are at least $$\# \mathcal{P}(\mathbb{N})$$ anti-symmetric relations on $$\mathbb{N}$$, thus their number is uncountable.