1. ## Cardinality...last one

This one is purely for fun. Tell me what you think of this.

Problem: Call a mapping $f:\mathbb{N}\mapsto\mathbb{N}$ eventually constant if there exists some $N\in\mathbb{N}$ such that if $n>N$ then $f(n)=c$ for some fixed $c\in\mathbb{N}$. Let $\mathfrak{M}$ be the set of all eventually constant function.

Find the cardinality of $\mathfrak{M}$

Claim: $\mathfrak{M}\cong\mathbb{N}$ ( $\cong$ is used here to denote equipotence)

Proof: Let $F_N\left(c\right)=\left\{f|f:\mathbb{N}\mapsto\mat hbb{N}\text{ and }f\left(n\right)=c,\text{ }n>N\right\}$, and construct a mapping $\tau:F_N(c)\mapsto\mathbb{N}^N\times\{c\}$ by $f\mapsto\left(f\left(1\right),\cdots,f\left(N\righ t),c\right)$. To see that this mapping is injective assume that

$\tau\left(f\right)=\tau\left(\bar{f}\right)\implie s\left(f(1),\cdots,f\left(N\right),c\right)=\left( \bar{f}\left(1\right),\cdots,\bar{f}\left(N\right) ,c\right)$ where it clearly follows that $f(n)=\bar{f}(n),\text{ }\forall n\in\mathbb{N}$ where it follows that $f=\bar{f}$. To see that it's surjective one merely needs to

know that for any $\left(n_1,\cdots,n_N,c\right)$ the function $f\left(k\right)=\begin{cases} n_k & \mbox{if}\quad 1\leqslant k\leqslant N \\ c & \mbox{if}\quad k>N\end{cases}$ is in $F_N$ and $f\mapsto\left(n_1,\cdots,n_N,c\right)$. The conclusion follows, so $F_N(c)\cong \mathbb{N}^N\times\{c\}\cong\mathbb{N}^N$ where the last equipotence is

obvious. Thus, for a fixed $c$ all the possible functions that are eventually $c$ is $\bigcup_{c\in\mathbb{N}}F_N(c)$. Now, since each $F_N(c)$ is countable it follows (since the union of countably many countable sets is

countable) that for a fixed $c$ that $\bigcup_{N\in\mathbb{N}}F_N(c)$ is countable. Lastly, noticing then that $\mathfrak{M}=\bigcup_{c\in\mathbb{N}}\bigcup_{N\in \mathbb{N}}F_N(c)$ and reapplying the theorem that the union of countably many countably sets is countable

finishes the argument. $\blacksquare$

What do you think? It's late, and maybe I'm paranoid but I feel I am overlooking something.

Thank you very much for taking the time to read it.

2. I think you are quite right.
If I am supposed to prove this, I will bigin with proving that the collection of all finite subset of N is countable, and then the collection of eventually constant mapping is embeded in the cartesian product of the collection of all finite subset of N, and N.

3. Originally Posted by Drexel28
This one is purely for fun. Tell me what you think of this.

Problem: Call a mapping $f:\mathbb{N}\mapsto\mathbb{N}$ eventually constant if there exists some $N\in\mathbb{N}$ such that if $n>N$ then $f(n)=c$ for some fixed $c\in\mathbb{N}$. Let $\mathfrak{M}$ be the set of all eventually constant function.

Find the cardinality of $\mathfrak{M}$

Claim: $\mathfrak{M}\cong\mathbb{N}$ ( $\cong$ is used here to denote equipotence)

Proof: Let $F_N\left(c\right)=\left\{f|f:\mathbb{N}\mapsto\mat hbb{N}\text{ and }f\left(n\right)=c,\text{ }n>N\right\}$, and construct a mapping $\tau:F_N(c)\mapsto\mathbb{N}^N\times\{c\}$ by $f\mapsto\left(f\left(1\right),\cdots,f\left(N\righ t),c\right)$. To see that this mapping is injective assume that

$\tau\left(f\right)=\tau\left(\bar{f}\right)\implie s\left(f(1),\cdots,f\left(N\right),c\right)=\left( \bar{f}\left(1\right),\cdots,\bar{f}\left(N\right) ,c\right)$ where it clearly follows that $f(n)=\bar{f}(n),\text{ }\forall n\in\mathbb{N}$ where it follows that $f=\bar{f}$. To see that it's surjective one merely needs to

know that for any $\left(n_1,\cdots,n_N,c\right)$ the function $f\left(k\right)=\begin{cases} n_k & \mbox{if}\quad 1\leqslant k\leqslant N \\ c & \mbox{if}\quad k>N\end{cases}$ is in $F_N$ and $f\mapsto\left(n_1,\cdots,n_N,c\right)$. The conclusion follows, so $F_N(c)\cong \mathbb{N}^N\times\{c\}\cong\mathbb{N}^N$ where the last equipotence is

obvious. Thus, for a fixed $c$ all the possible functions that are eventually $c$ is $\bigcup_{c\in\mathbb{N}}F_N(c)$. Now, since each $F_N(c)$ is countable it follows (since the union of countably many countable sets is

countable) that for a fixed $c$ that $\bigcup_{N\in\mathbb{N}}F_N(c)$ is countable. Lastly, noticing then that $\mathfrak{M}=\bigcup_{c\in\mathbb{N}}\bigcup_{N\in \mathbb{N}}F_N(c)$ and reapplying the theorem that the union of countably many countably sets is countable

finishes the argument. $\blacksquare$

What do you think? It's late, and maybe I'm paranoid but I feel I am overlooking something.

Thank you very much for taking the time to read it.
Your proof looks fine. It is, essentially, what I would do. In fact, I spend a wee while writing up here what I would do, posted it, then re-read your post to discover what I had written was, essentially, the same as what you had written. Essentially.

That said, what I did was instead of defining the function $F_N(c)$ I wasn't explicit with the $c$. When you are defining your bijection near the top of your proof you can just forget about the $c$'s as taking a cross product with the natural numbers will preserve your countability. This just makes defining everything neater and should, I think, remove a few lines afterwards.