When I read about shift spaces in the book "An Introduction to Symbolic Dynamics and Coding" by Lind and Marcus, they introduce a kind of shift space called $S$-gap shifts.
Let $S$ be a subset of $\{0,1,2,\ldots\}$. There are two cases of $S$-gap shift: $S$ is finite and $S$ is infinite.
In the case $S$ is finite, let $X(S)$ be the collection of all bi-infinite sequence having the form $$x = \ldots 10^{n_{-1}}10^{n_{0}}10^{n_{1}}1\ldots,$$ where $n_i \in S$. Then $X=X(S)$ is the $S$-gap shift. In this case, everything is clear for me.
When $S$ is infinite, points in $X(S)$ having either this form $$x = \ldots 10^{n_{-1}}10^{n_{0}}10^{\infty}\ldots$$ or this form $$x = \ldots 0^{\infty}10^{n_{0}}10^{n_{1}}1\ldots.$$ However, I don't understand why we do not need to require $1$'s happen infinitely often in points in $X$. In fact, there is an exercise in the book (Exercise 1.2.8) that says if $X$ contains points as in the finite case, $X$ is not a shift space. Of course, I cannot prove this exercise because I have no intuition meaning why.
Does anyone have any ideas? I really appreciate your help.
After peering through the book you referenced in your question to make sure I understood the framework correctly, I believe I understand what your confusion resides in. Keeping the notation from the book, the authors say that a space $X$ is a shift space whenever $X = X_\mathcal{F},$ where $\mathcal{F}$ is a set of forbidden words. We also require the set $X$ to be shift-invariant and closed. These two things will help us figure out why this case is different. Thus let us show that $X(S)$ as defined in the case that $S$ is finite is not a shift space when $S$ is infinite.
Indeed, let $S = \{s_0 < s_2 < \dots \}$ be infinite and consider $x = \dots10^{n_{-1}}10^{n_{0}}10^{n_1}\ldots$ where $n_i = s_i$ whenever $i \geq 0$ and $n_i = s_0$ otherwise. Let $l_i := \sum_{1 \leq j \leq i}(n_j + 1).$ Since $X(S)$ must be closed and shift-invariant, it must contain $x_j = \sigma^{l_j}(x).$ This element starts with a trail of $s_j$ zeroes and since $s_j \to \infty,$ it follows that we cannot forbid any block of zeroes of length greater than or equal to $s_1.$ But then elements of the form $\dots 1 0^{n_{-1}} 1 0^\infty$ should be in $X(S).$ Mutatis mutandis, considering preimages of points in $X(S),$ it should also contain points of the form $0^\infty 1 0^{n_0} 1 \dots.$ Clearly this space is shift-invariant and closed, so it is the desired shift space. I hope this helps. :)