Think of a one-dimensional infinite *walk* as a map $$w\colon \mathbb{N}\to \{-1,1\}.$$ (If it is more convenient, you can think of a walk as a subset of $\mathbb{N}$, or as a binary word, or as any other convenient model.) Similarly, a *finite walk*, say of $n$-steps (for some integer $n\geq 1$), is a map $$v\colon [0,n-1]\cap \mathbb{N}\to\{-1,1\}.$$

Given a finite walk $v$, with exactly $n$ steps, let us say that is is *nondecreasing* if $\sum_{i=0}^{n-1}v(i)\geq 0$. Define *nonincreasing* finite walks dually, using $\leq$.

Given an infinite walk $w$, and a finite (nonempty) set $S$ of finite walks, let us say that $w$ *avoids using $S$ too often*, if there is some positive integer $N$ (depending on $w$ and $S$) so that if we concatenate any $N$ elements of $S$ (allowing repetitions), then such a concatenated finite walk does not appear as a (consecutive) subwalk in $w$.

For example, if $S=\{v_0\}$ where $v_0$ is the $1$ step walk where $v_0(0)=1$, and if $w$ is the walk where $w(i)=(-1)^{i}$, then we can take $N=2$, and we see that $w$ avoids using $S$ too often. Or in in terms of binary words, the infinite binary word $01010101\ldots$ does not have arbitrarily long strings of consecutive $1$s.

Of course, there exists an infinite walk that avoids using too often any finite set of nondecreasing finite walks. Just take $w$ to be the strictly decreasing walk where $w(i)=-1$.

My question is whether the following balancing act is possible.

Does there exist an infinite walk $w$, such that for

everyfinite set $S$ of nondecreasing finite walks, as well as foreveryfinite set $S$ of nonincreasing finite walks, $w$ avoids using $S$ too often?

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