Meetability of $\pm 1$-functions on $\omega$

Question

If ${\cal S}$ is a collection of functions $f:\omega\to\omega$ we say that ${\cal S}$ is meetable if there is a "global function" $g:\omega\to \omega$ such that for every $f\in {\cal S}$ there is $n\in\omega$ such that $g(n) = f(n)$.

Let ${\cal S}^{\pm 1}$ denote collection of functions $f:\omega\to\omega$ such that $|f(n) - f(n+1)| = 1$ for all $n\in\omega$ and for $k\in\omega \setminus \{0\}$, let ${\cal S}^{\pm 1}_{\leq k} = \{ f\in {\cal S}^{\pm 1}: f(0) \leq k\}$.

Is ${\cal S}^{\pm 1}$ meetable? If not, is ${\cal S}^{\pm 1}_{\leq k}$ meetable for all $k\in\omega\setminus\{0\}$?

(User Wojowu pointed out in a comment below that ${\cal S}^{\pm 1}_{\leq 1}$ is meetable.)

Answer 1

$\newcommand\om{\omega}$ The set $S:={\cal S}^{\pm 1}$ is meetable by any function $g$ of the following form:

Let $l_0,l_1,\dots$ be odd numbers in $\om=\{0,1,\dots\}$ such that $l_j/n_j\to\infty$ (as $j\to\infty$), where $n_0=-1$ and $n_{j+1}=n_j+l_j$ for $j\in\om$. (E.g., take $l_j=1+j!$ for $j\ge2$.) For each $j\in\om$ and all $n\in\om$ such that $n_j<n\le n_{j+1}$, let $g(n)=n_{j+1}-n$.

Indeed, take any $f\in S$ and suppose that $f$ is not met by $g$. Since $l_j/n_j\to\infty$, we have $f(n_j+1)\le f(0)+n_j+1\le l_j-1=g(n_j+1)$ for all large enough $j$. For all such $j$, there exists $$m_j:=\max\{n\in\{n_j+1,\dots,n_{j+1}\}\colon f(n)\le g(n)\}.$$ Since $f$ is not met by $g$ and $|f(n+1)-f(n)|=1$, we see that $f(m_j)=g(m_j)-1=n_{j+1}-m_j-1$, whence $n_{j+1}-1=f(m_j)+m_j=f(0)\text{ mod }2$ for all large enough $j$. This is a contradiction, because the $l_j$'s are all odd and hence the $n_j$'s alternate $\text{ mod }2$.

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Here are the list plots $\{(n,g(n))\colon n\in\{1,\dots,n_5\}\}$ (blue) for $l_j\equiv1+j!$ and $\{(n,f(n))\colon n\in\{1,\dots,n_5\}\}$ (red) for a "random" $f\in S$, showing how $g$ eventually "catches/meets" $f$ -- catches only by the last in the picture (negatively sloped) dotted blue line segment ($f$ was able to get through the previous blue barrier):