**Set-up**: By abuse, let $\sigma$ represent both the left shift operator on infinite bi-infinite strings and the cyclic left shift operator on finite strings. (Thus, for example, $\sigma(...01\bar{0}10...) = ...10\bar{1}01...$ (in the bi-infinite case, we place a bar $\bar{ }$ over the "central" character) and $\sigma(10010) = 00101$). Also let $\tilde{A}$ represent the orbit of the string $A$ under $\sigma$ in both the finite and the infinite case. For convenience let's call $A$ an "unfolding" of $\tilde{A}$ whenever $A$ is in the equivalence class $\tilde{A}$. Now let $\{\tilde{A}_i\}_{i=1}^{\infty}$ be a sequence of equivalence classes of finite binary strings.

Suppose that, for each $i$, some unfolding of $\tilde{A}_i$ is contained as a substring of some unfolding of $\tilde{A}_{i+1}$ with at least one character on either side of it; for convenience let's call this an "embedding" of $\tilde{A}_i$ into $\tilde{A}_{i+1}$. If a sequence of embeddings has been specified, then it is possible to take the "limit" of this sequence $\{\tilde{A}_i\}_{i=1}^{\infty}$ and arrive at a bi-infinite binary string $A$.

**Example 1**: The sequence of strings

```
0
101
01010
1010101
010101010
```

has as its limit: $...0101\bar{0}1010...$

**Example 2**: The sequence of strings

```
1
010
00100
0001000
000010000
```

has as its limit: $...0000\bar{1}0000...$

**Conjecture**: Given $\{\tilde{A}_i\}_{i=1}^{\infty}$, any sequences of embeddings lead to the same bi-infinite binary string, up to orbits of $\sigma$. That is, any two limits of this process $A$ and $B$ are related via $\sigma^k(A) = B$ for some integer $k$.

This result would be quite surprising to me if it were true, but I have not been able to construct a counterexample. It can be made more precise using the categorical language of projective limits, but I think the question is clear enough without going through this trouble.