While looking at a solution to a math problem, I came across this, let $f:\mathbb N\to \{0,1\}$ be a function such that $$f(x)=\cases{1\quad\text{if $x$ is balanced}\\ 0 \quad \text{otherwise.}}$$
I won't explain what is a balanced number because it's irrelevant to the purpose of my question. Now consider $$S=\{(f(n+1),f(n+2),...,f(n+m))\mid n\in \mathbb N\}$$
Here the solution reads : $S\subseteq \{0,1\}^m$ and $\mid \{0,1\}^m\mid =2^m<\mid\mathbb N \mid $ hence by the pigeonhole principal $\exists a\ne b$ such that $f(a+n)=f(b+n), \forall n\le m$.
But I don't know how one can conclude $S\subseteq \{0,1\}^m$ I mean $S$ should be infinite because $n$ runs over all the natural numbers.