Sets of binary sequences In my course on linear algebra we have recently introduced linear independent subsets of vector spaces. As an exercise I have been thinking about examples of infinite linearly independent sets and trying to prove that they are what I think they are. I am having trouble with the following:
$$V=\mathbb{R}^{\mathbb{N}},\quad \text{and}\quad S=\{ (\alpha_n)\in\{0,1\}^{\mathbb{N}}:\;(\alpha_n)\;\text{aperiodic}\}$$
I believe $S$ is linearly independent, and here are my thoughts:
Thoughts: by induction. Suppose any collection of $n$ sequences $p_k^{(1)},\dots, p_k^{(n)}\in S$ are linearly independent. Now consider a new sequence $q_k\in S$ is such that, for some $\lambda_i$ and for any $k:$
$$\lambda_{n+1}q_k+\sum_{i=1}^n \lambda_ip_k^{(i)}=0$$
Let $k_1,k_2\dots$ be the integers such that $q_{k_i}=0$. Since $q_k$ is aperiodic $k_i$ is as well. What I would like to conclude is that if $$\sum_{i=1}^n \lambda_ip_{k_j}^{(i)}=0$$
For some aperiodic sequence $k_j$ then $\lambda_i=0$ by the inductive hypothesis, but this simply does not follow (at least not immediately). Can someone help? Or is the statement I am trying to prove wrong?
 A: I think your conjecture is wrong, and $S$ is linearly dependent. Here's an example.
We define $x = (x_n)$ with $x_n = 1$ iff $n = 10^{2m}$ for any natural number $m$ and $x_n = 0$ otherwise.
$x$ is aperiodic because the lengths of the $0$-gaps between two $1$'s are strictly increasing. So we really have $x \in S$.
Similarly, we define $y = (y_n)$ with $y_n = 1$ iff $n = 10^{2m - 1}$ for any natural number $m$ and $y_n = 0$ otherwise. $y$ is aperiodic for the same reason as $x$ is. So we also have $y \in S$.
Now, by construction, there is no $n$ such that $x_n = 1 = y_n$.
This implies that $x + y = z = (z_n)$ with $z_n = 1$ iff $n = 10^m$ for any natural number $m$ and $z_n = 0$ otherwise.
Again, we see that $z$ is aperiodic. So $z \in S$ and $S$ is linearly dependent.
A: If $\mathbf a=(a_1,a_2,a_3,\dots)$ is any aperiodic sequence, then $\mathbf a':=(1-a_1,1-a_2,1-a_3,\dots)$ is also aperiodic. Choose any $\mathbf a\in S$ and any $\mathbf b\in S\setminus\{\mathbf a,\mathbf a'\}$. Then $\mathbf a,\mathbf a',\mathbf b,\mathbf b'$ are four distinct elements of S; the fact that $\mathbf a+\mathbf a'=\mathbf b+\mathbf b'=(1,1,1,\dots)$ shows that the set $S$ is not linearly independent.
