Find cardinality of $[f_\equiv]$, decide whether $g(n)$ exists

Question

Let us define an equivalence relation $\equiv$ in $\mathbb{N}^\mathbb{N}$ such that for all $f,g \in \mathbb{N}^\mathbb{N}$ we have

$$f \equiv g \iff \forall n \ge101 \left(f(n) = g(n) \right)$$

I want to:

(a) Find the cardinality of $[f_\equiv]$, where $[f_\equiv]$ is the equivalence class of $f$. We let $f(n)=2n^2+1$.

(b) Decide whether there is a function $g(n) \in \mathbb{N}^\mathbb{N}$ such that $\left \lvert [g_\equiv] \right \rvert = \left \lvert \mathbb{N}^\mathbb{N}\backslash \equiv \right \rvert $, where $\mathbb{N}^\mathbb{N}\backslash \equiv$ is the quotient set.

For (a), I start off with noticing that for $f \equiv g$, we must have

$$f(n) = \{ 1,3,9, \dots, 101^2+1, 102^2+1, \dots \}$$ $$g(n) = \{ *,*,*, \dots, 101^2+1, 102^2+1, \dots \}$$ where $*$ can be anything from $\mathbb{N}$.

This means that for the given $f$ we only can have $g$s given by $\mathbb{N}^{101}$. It is a relatively known fact that $\lvert \mathbb{N} \times \mathbb{N} \rvert \sim \lvert \mathbb{N} \rvert$ and this can be extended to show $\lvert \mathbb{N}^{101} \rvert \sim \lvert \mathbb{N}\rvert$, so we have $$[f_\equiv] \sim \lvert \mathbb{N}\rvert$$

For (b), we notice that for $f \equiv g$, we must have $$f(n) = \{ *,*,*, \dots, x_{101}, x_{102}, \dots \}$$ $$g(n) = \{ *,*,*, \dots, x_{101}, x_{102}, \dots \}$$ and the possible $f$s are given by $\{x_{101}, x_{102}, \dots \}$, which is $\mathbb{N}^\mathbb{N}$. We know $\vert\mathbb{N}^\mathbb{N}\rvert = \mathfrak{c}$, so $\left \lvert \mathbb{N}^\mathbb{N}\backslash \equiv \right \rvert = \mathfrak{c}$. At the same time, $[g_\equiv] \sim \lvert \mathbb{N}\rvert$ (from the argument in (a)), so there is no such function $g$.

I would appreciate any comments on this solution.

Answer 1

Both solutions look OK. The second part could use a little more explanation, but it's clear you get the idea.

I would actually provide the bijection $$F:\mathbb N^\mathbb N\setminus \equiv\longrightarrow\mathbb N^\mathbb N$$

as $F$ is fairly easy to define and the proof becomes more clear. So I would say $F$ is defined as $$F([f])(n) = f(100+n)$$ and then show

1. $F$ is well defined
2. $F$ is bijective

to conclude that $|\mathbb N^\mathbb N\setminus\equiv|=\mathfrak c$