The cardinality of the set all symmetric relations on the set of natural numbers is $\mathfrak{c} = | \mathbb{R} | = 2^{\aleph_0}$

Question

Prove that a set of all symmetric relations on the set of natural numbers has cardinality $\mathfrak{c} = | \mathbb{R} | = 2^{\aleph_0}$.

Here I think that the $(a,b)b$ - will be every number there is in $\mathbb N$ and hence we will never get pass ordered pair that contain $a$ as first number. If it's true can you please tell me how to write it?

Answer 1

Hint: symmetric relations on $\Bbb N$ are in bijection with subsets of $\{(a,b)\in\Bbb N^2:a\le b\}$.

Alt route: any subset of $\{(c,c)\in\Bbb N^2:c\in\Bbb N\}$ is a symmetric relation.