Quotient of the set by equivalence relation of equality

Question

If our equivalence relation R is (x = y), then is the quotient of the set X(assume that it's only natural numbers) by R is the set of all natural numbers? Because as I understand, all of our equivalence classes are the numbers themselves:

[x] = {x}

Answer 1

If the equivalence relation is on $\Bbb N$, and we are using the relation of equality, we do not, strictly speaking, get the natural numbers back. Instead we get something that looks very much like it.

Specifically (if you allow zero to be a natural), $$ X/R = \Big\{ \{x\} \; \Big| \; x \in \Bbb N \Big\} = \Big\{ \{0\}, \{1\}, \{2\}, \cdots \Big\} $$

Notice the difference between this, and the actual natural numbers: each element of $X/R$ is a set containing exactly one of the natural numbers, as opposed to $\Bbb N$, which are the natural numbers themselves.

The reason for this is that equivalence classes (i.e. the elements of a quotient set) are, themselves, sets. Even if you have singletons, they are still just sets containing one element.

That is to say, quotient sets are, effectively, sets made of sets, regardless of what set and relation they come from.

Answer 2

Technically the quotient is $\big\{\{n\}:n\in\Bbb N\big\}$, which is not the same as $\Bbb N$, because $n\ne\{n\}$. However, the quotient map $\Bbb N\to\Bbb N/R:n\mapsto\{n\}$ is a bijection, so for most purposes we can identify $\Bbb N/R$ with $\Bbb N$, even if this is a bit sloppy.