Let $\Sigma = \{a_i: i \geq 1\}$ be a countable alphabet. Define $s_n = \prod_{j=1}^{\infty} a_1 \cdots a_n$. Then juxtapose sequences $s_2, s_3, \dots$ :
$$ a_1a_2a_1a_2a_1a_2 \cdots \\ a_1a_2a_3a_1a_2a_3 \cdots \\ a_1a_2a_3a_4a_1a_2 \cdots \\ \vdots $$
Now apply a shift to any number of the sequences: $T_k(s_n) = a_k a_{k+1} \cdots a_n s_n, \ k = 2, \dots n$. Then is it true that the sequences, shifted and juxtaposed, will always contain column of all $a_1$'s?
On
I was wondering if we could take a different approach to this. In the original juxtaposition we can see that the $k^{th}$ apperance of $a_{1}$ make a line. For example the very first column of $a_1$ makes a vertical line. The next one makes a line of slope -1, and the next -1/2. It we index starting from $k = 0$ we can see that the $k^{th}$ appearance makes a line of slope $-1/k.$ Clearly, without any shifts we cannot draw a vertical line that intersects an infinite number of lines (with the exception of $k = 0$). And suppose there is a non-trivial shift applied to at least 1 row that lets us draw such a vertical line that intersects an infinite number of lines. It is clear that if we draw such a line in the position on the diagram without shift that it would require an infinite number of shifts as there are infinite number of points that do not intersect the line.
Just a thought. I don't know if this would work.
No, for two reasons:
First, if $n$ and $m$ have a common divisor, then it's possible to shift $s_n$ so that it never has an $a_1$ in the same position as $s_m$ does. For example, if we align $s_2$ and $s_4$ to form the grid \begin{array}{ccccc} a_1 & a_2 & a_1 & a_2 & \dots \\ a_2 & a_3 & a_4 & a_1 & \dots \end{array} then an $a_1$ in the $s_4$ row will always be below an $a_2$ in the first row.
If we fix this problem, and (for example) restrict ourselves to rows with a prime period, then the Chinese Remainder Theorem guarantees that any finite set of rows eventually all have an $a_1$ in the same position. So we have columns that begin with arbitrarily many $a_1$'s.
However, we still are not guaranteed a column that is entirely $a_1$, by the following construction: choose shifts so that the $k^{\text{th}}$ row does not have an $a_1$ in its $k^{\text{th}}$ position. Then the $k^{\text{th}}$ column also does not have an $a_1$ in its $k^{\text{th}}$ position, so in particular it does not consist of all $a_1$'s.