Not understanding sequences question from 2020 Miklos Schweitzer Competition The Hungarian Miklos Schweitzer 2020 competition just ended and while they do post the problems in English they haven't done so for the 2020 one yet. I got an unofficial English translation online but the first question is confusing me and I'm not sure if I'm at fault or if there is a translation error (although it is likely the former).
The question is as follows:
We say that two sequences $x, y: \mathbb{N} \to \mathbb{N}$ are completely different if $x(n) \neq y(n)$ holds for all $n \in \mathbb{N}$. Let $F$ be a function assigning a natural number to every sequence of natural numbers such that $F(x) \neq F(y)$ for any pair of completely different sequences $x, y$, and for constant sequences we have $F(k,k,...) = k$. Prove there exists an $n \in \mathbb{N}$ such that $F(x) = x(n)$ for all sequences $x$.
It seems to me that any $n$ would work rather obviously. If I choose and $n$, for constant sequences it doesn't matter and for completely different sequences it also doesn't matter which  because the constraint that they aren't equal is immediate. And for all other sequences just use the same $n$. There doesn't seem to be any restrictions on $F$ being injective. I really think I'm missing something obvious here so I'd appreciate any help.
 A: Let $X$ be the set of sequences $x:\Bbb N\to\Bbb N$, and let $F:X\to\Bbb N$ be as stated in the problem. For $x,y\in X$ write $x\perp y$ iff $x(k)\ne y(k)$ for each $k\in\Bbb N$.
Observe first that $F(x)\in\operatorname{ran}x$ for each $x\in X$: if $n\in\Bbb N\setminus\operatorname{ran}x$, and $c_n\in X$ is the constant function $c_n(k)=n$ for all $k\in\Bbb N$, then $x\perp c_n$, so $F(x)\ne F(c_n)=n$. We’ll use this fact a number of times.
Let $X_0=\{x\in X:x\text{ is injective}\}$; for each $x\in X_0$ there is a unique $k_x\in\Bbb N$ such that $x(k_x)=F(x)$. We will show first that $k_x=k_y$ for all $x,y\in X_0$ and hence that there is an $\ell\in\Bbb N$ such that $k_x=\ell$ for each $x\in X_0$, i.e., such that $F(x)=x(\ell)$ for each $x\in X_0$.
Let $x,y\in X_0$, and suppose first that $F(x)\ne F(y)$. Let
$$z:\Bbb N\to\Bbb N:k\mapsto\begin{cases}
F(y),&\text{if }F(x)\in\{x(k),y(k)\}\\
F(x),&\text{otherwise.}\\
\end{cases}$$
Then $z\in X$, and $z\perp x$ (since $x$ is injective), so $F(z)\ne F(x)$. However, $$F(z)\in\operatorname{ran}z\subseteq\{F(x),F(y)\}\,,$$ so $F(z)=F(y)$, $z\not\perp y$, and there must be some $k_0\in\Bbb N$ such that $z(k_0)=y(k_0)$.
If $y(k_0)=F(x)$, then $z(k_0)=F(y)\ne F(x)=y(k_0)$, so we must have $y(k_0)\ne F(x)$. But then $z(k_0)=y(k_0)\ne F(x)$, and $z(k_0)\in\{F(x),F(y)\}$, so $y(k_0)=z(k_0)=F(y)$, and hence $k_0=k_y$. Moreover, $z(k_y)=F(y)$ implies that $F(x)\in\{x(k_y),y(k_y)\}=\{x(k_y),F(y)\}$, and $F(x)\ne F(y)$, so $x(k_x)=F(x)=x(k_y)$, and therefore $k_x=k_y$.
Now suppose that $F(x)=F(y)=n$, say, and to get a contradiction suppose that $k_x\ne k_y$. For $m\in\Bbb N\setminus\{n\}$ let
$$z_m:\Bbb N\to\Bbb N:k\mapsto\begin{cases}
m,&\text{if }k=k_x\\
n,&\text{otherwise.}
\end{cases}$$
Then $z_m\in X$, and $z_m\perp x$, so $F(z_m)\ne F(x)=n$, but $F(z_m)\in\operatorname{ran}z_m=\{m,n\}$, so $F(z_m)=m$. Fix $x'\in X_0$ such that $F(x')\ne n$, and let $n'=F(x')$. (E.g., let $$x':\Bbb N\to\Bbb N:k\mapsto n+1+k\,,$$ or take $x'$ to be any enumeration of $\Bbb N\setminus\{n\}$.) At least one of $k_x$ and $k_y$ is different from $k_{x'}$; without loss of generality $k_x\ne k_{x'}$.
Fix $m\in\Bbb N\setminus\{n,n'\}$, and let
$$z':\Bbb N\to\Bbb N:k\mapsto\begin{cases}
m,&\text{if }k=k_{x'}\\
n',&\text{otherwise;}
\end{cases}$$
then $F(z')=m=F(z_m)$, but $z'\perp z_m$, so $F(z')\ne F(z_m)$, and we have the desired contradiction. Thus, we must have $k_x=k_y$ in this case as well, and it follows that $k_x=k_y$ for all $x,y\in X_0$.
Let $\ell\in\Bbb N$ be such that $k_x=\ell$ (and hence $x(\ell)=F(x)$) for all $x\in X_0$; we want to show that $x(\ell)=F(x)$ for all $x\in X$, so let $x\in X$. For each $n\in\Bbb N$ define $y_n\in X$ by recursion as follows:
$$y_n(k)=\begin{cases}
n,&\text{if }k=\ell\\
\min\Big(\Bbb N\setminus\big(\{y_n(j):j<k\}\cup\{n,x(k)\}\big)\Big),&\text{otherwise.}
\end{cases}$$
In other words, $y_n(\ell)=n$, and for $k\in\Bbb N\setminus\{n\}$, $y_n(k)$ is the smallest natural number different from $x(k)$ and from all $y_n(j)$ with $j<k$. Then $y_n\in X_0$, so $F(y_n)=y_n(\ell)=n$. If $n\ne x(\ell)$, then $y_n\perp x$, and hence $F(x)\ne F(y_n)=n$. Thus, $F(x)$ must be $x(\ell)$, as desired.

I was led to consider the set $X_0$ by the observation that if we replace $X$ by $Y$, the set of bounded sequences of natural numbers, the result is false. To see this, let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. Let $y\in Y$; $\operatorname{ran}y$ is finite, so there is a unique $m_y\in\Bbb N$ such that $y^{-1}[\{m_y\}]\in\mathscr{U}$. Let
$$F:Y\to\Bbb N:y\mapsto m_y\,,$$
and for each $y\in Y$ let $U_y=y^{-1}[\{m_y\}]\in\mathscr{U}$.
Let $n\in\Bbb N$. If $c_n$ is the constant sequence defined above, then clearly $F(c_n)=m_{c_n}=n$. Now suppose that $x,y\in Y$; then $U_x\cap U_y\ne\varnothing$, so fix $k\in U_x\cap U_y$. If $x\perp y$, then $F(x)=x(k)\ne y(k)=F(y)$, so $F$ has the desired properties. But $\mathscr{U}$ is free, so $\bigcap\mathscr{U}=\varnothing$, and there is therefore no $\ell\in\Bbb N$ such that $y(\ell)=m_y=F(y)$ for all $y\in Y$.
Thus, the original result must depend in some way on the unbounded sequences, and the simplest unbounded sequences are the injective ones.
