Why the set $\{f \in \omega^\omega : \exists_{m}\forall_{m \leq n}^{}(f(n)\neq g(n))\}$ is $F_{\sigma}$ and meager?

Question

For any $g \in \omega^\omega$,

Why the set $\{f \in \omega^\omega : \exists_{m}\forall_{m \leq n}^{}(f(n)\neq g(n))\}$ is $F_{\sigma}$ and meager?.

I do not know why this set is $F_{\sigma}$ and meager.

A suggestion please.

Answer 1

If you want to prove directly that the set is $F_\sigma$, let the set in question be $F$, and for each $m\in\omega$ let

$$\begin{align*}F_m&=\left\{f\in{^\omega\omega}:\forall n\ge m\big(f(n)\ne g(n)\big)\right\}\\ &=\bigcap_{n\ge m}\left\{f\in{^\omega\omega}:f(n)\ne g(n)\right\}\;; \end{align*}$$

clearly each $F_m$, being an intersection of closed sets, is closed, and $F=\bigcup_{m\in\omega}F_m$, so $F$ is an $F_\sigma$.

It remains to show that each $F_m$ is nowhere dense; try showing that each basic open set in ${^\omega\omega}$ contains a point not in $F_m$. I’ve completed the argument in the spoiler-protected block below.

Now let $U$ be a non-empty open subset of ${^\omega\omega}$. Fix $f\in U$ arbitrarily; there is an $\ell\in\omega$ such that $B=\{h\in{^\omega\omega}:h\upharpoonright\ell=f\upharpoonright\ell\}\subseteq U$. Given $m\in\omega$, let $n=\max\{\ell,m\}$, and define $h\in{^\omega\omega}$ by making $h\upharpoonright n=f\upharpoonright n$, $h(n)=g(n)$, and $h(k)=0$ for $k>n$. Then $h\in U\setminus F_m$, and since $F_m$ is closed, it is not dense in $U$. Thus, each $F_m$ is closed and nowhere dense, and $F$ is $F_\sigma$ and meagre.