https://en.wikipedia.org/wiki/Shift_space for definition of a shift space. My first attempt was to take an infinite binary string, in which every word occurs in it. Then close it under the shift operator, however this space is not closed, and contains every periodic point in it's closure, so that doesn't work. In fact the set of periodic points are dense in the full shift, so the closure of the set of period points is the full shift.
My second attempt is to create a forbidden word set, as follows. Enumerate all words say $\{w_n\}_{n\in\mathbb{N}}$, then include $w_n^{2^n+2}$ the $(2^n+2)$-fold concatenation of $w_n$. Then this space is defined to be a shift space, and won't include any period points by design. My only issue now is to show it is not empty, which I hope that I can demonstrate an element inductively, because I spaced the $w_n$ far enough apart.
My third attempt is to fix my first attempt, by finding an irrational number, whose binary expansion has a bound for the number of consecutive 0's and 1's.
My last attempt is to consider the binary string $x$ where $x_{i}=0$ if $\sin(i)>0$, 1 otherwise. Then I should be able to use some irrationality of integer values of $\sin(z)$ to conclude that for any binary block $w$, there exist some $n$ where $w^n$ never appears in $x$. For instance $1111$ can't occur because $4>pi$, similarly $111000^5$ can't occur by experimentation. How can I make this precise, maybe with Taylor series remainder estimation? Taking the closure with respect to the shift map, and then the closure with respect to the product topology will yield the desired shift space without periodic points.
