Equivalence relation on $\mathbb{N}$ with infinitely many equivalence classes

Question

Does there exist an equivalence relation, defined on $\mathbb{N}$, with infinitely many equivalence classes, all of which contain infinitely many elements?

I see no reason for such a relation not to exist, but I'm having trouble finding an example.

Answer 1

Yes. First note that $\lvert\mathbb N\rvert=\lvert\mathbb N\rvert^2$ so it would be strange if there weren't.

For instance let $\sim$ be the equivalence relation defined by $p^n\sim p^k$ for all $p$ prime and $n,k\ge 1$, and let all numbers which aren't of the form $p^n$ be equivalent.

So $$1\sim 6\sim 10\sim 12\sim 14\sim 15\sim\cdots$$ $$2\sim 4\sim 8\sim 16\sim\cdots$$ $$3\sim 9\sim 27\sim\cdots$$

and so on.

Edit: To elaborate on my first point. Let $f:\mathbb N^2\to\mathbb N$ be any bijection. Then set $f(n,k)\sim f(n,l)$ for all $n,k,l$ to get an equivalence relation as wanted.

Answer 2

An example: Let $a\equiv b$ if the highest power of $2$ that divides $a$ is the same as the highest power of $2$ that divides $b$.

So $\{1,3,5,7,9,\dots\}$ is one equivalence class.

$\{2,6,10,14,18,\dots\}$ is another.

$\{4,12,20,28,36,\dots\}$ is another.

And so on.

There are many other examples.

Answer 3

Little less constructive...

Take your favorite bijection $\mathbb{N} \stackrel{f}{\to} \mathbb{N} \times \mathbb{N}$, let $\pi_2$ be the projection in the second coordinate, and consider $h=\pi_2 \circ f$. Now, define $x \sim y \iff h(x)=h(y).$

Answer 4

Here's another example: Let $a \sim b$ if $a$ and $b$ have the same number of prime factors (up to multiplicity). We can put $1$ in with the prime numbers to meet your criterion of all classes being infinite. Then the classes are: $$\{1,2,3,5,7,11,\ldots\}\\ \{4,6,9,10,14,\ldots\}\\ \{8,12,18,20,\ldots\}\\ etc.$$