# Cardinality of equivalence relations in $\mathbb{N}$

I came across this long proof on this site:

Cardinality of relations set

But I would like to know whether my direction can work.

Say we want to find the cardinality of all equivalence relations in $\mathbb{N}$. Since it is a subset of all relations in $\mathbb{N}$, I conclude it has a cardinality smaller or equal to $\aleph$. Now, define an injective function from $P(\mathbb{N})$ to the set of equivalence relations by matching each subset of $\mathbb{N}$ with the identity relation (which is an equivalence relation in $\mathbb{N}$.

Therefore the cardinality of all equivalence relations in $\mathbb{N}$ is greater or equal to $\aleph$ and using CSB we get the desired result.

Seems legit?

• Already wanted to upvote your previous question but was marked as duplicate. Was not too sure why to be honest. – Pedro Jun 3 '15 at 1:21
• "subgroup of $\mathbb{N}$"? – mlbaker Jun 3 '15 at 1:22
• Oops. In Hebrew the word for "group" is used to describe a "set". Quite confusing. Will edit. – Whyka Jun 3 '15 at 1:24
• I don't think you're using $\aleph$ correctly (you seem to be using it for the cardinality of the powerset of $\mathbb{N}$). You may prefer to write: "I conclude it has cardinality smaller than or equal to $2^{|\mathbb{N}|^2},$ and hence cardinality smaller than or equal to $2^{\aleph_0}$." – goblin GONE Jun 3 '15 at 1:44
• Isn't $2^{\aleph_0}$ equal to $\aleph$? – Whyka Jun 3 '15 at 1:48

Your approach is sound. I think you can choose a better injection from $$P(\mathbb{N})$$ into equivalence relations. Let us say $$0 \in \Bbb N$$ and we will inject $$P(\Bbb N \setminus\{0\})$$ into the equivalence relations on $$\Bbb N$$. I would suggest you take a subset of $$\Bbb N \setminus\{0\}$$ to the equivalence relation that groups all members of the subset and $$0$$ into an equivalence class and leaves all the rest of $$\Bbb N$$ under the identity.
If we don't do the trick with $$0$$, all singleton subsets will be mapped to the identity relation and you don't have an injection.