Let1, $k \geq 2$. Define $\Sigma_k=\{x=(x_k)_{-\infty}^{\infty} : x_n \in \{1,...,d\}\}$. Define the metric $d(x,y)= \sum^{\infty}_{n=-\infty} \frac{e(x_n,y_n)}{2^{|n|}}$, where $e(x_n,y_n)$ is one if $x_n \neq y_n$ and zero otherwise.
Let A be a $k\times k$ matrix with entries in $\{0,1\}$. Define $\Sigma_A=\{x \in \Sigma_k : A_{x_n,x_{n+1}} = 1\ \forall n \in \mathbb{N} \}$.
I have shown that $\Sigma_A$ is closed, and I now want to show it is also compact. Since I have no idea what an open set would be in this case, I'm guessing that I should be trying to show this set is sequentially compact, but I have no idea how to proceed. Any help would be appreciated.
We will actually show that $\Sigma_k$ is a compact metric space, which with the closedness of $\Sigma_A$ will imply compactness of $\Sigma_A$. First, we note that $\Sigma_k$ is complete, as for $(y_n)_{n=1}^{\infty}\subset \Sigma_k$, where $y_n = (x_{k,n})_{k=-\infty}^{\infty}$, if $(y_n)_{n=1}^{\infty}$ is Cauchy, then for any $r\geq 0$, we can take $N_r\in \mathbb{N}$ such that $d(y_m, y_n) < 2^{-r}$ for all $m, n\geq N_r$. Note that this implies $x_{k,m} = x_{k,n}$ for all $\lvert k\rvert\leq r$. Therefore, we can define $y = (x_k)_{k=-\infty}^{\infty}$ such that for $k = \pm r$ we let $x_k = x_{k,N_r}$, and we will have $d(y, y_n)\leq 2^{-r}$ for all $n\geq N_r$, so $y_n\to y$.
Now, we show that $\Sigma_k$ is totally bounded. For $R > 0$, we choose $r\in \mathbb{N}$ such that $2^{-r}\leq R/2$. Then, for each possible part $(x_k)_{k=-r}^r$ of a longer sequence in $\Sigma_k$, we take the closed ball of radius $2\cdot 2^{-r}$ around $(x_k^*)_{k=-\infty}^{\infty}$, where $x_k^* = x_k$ for $\lvert k\rvert\leq r$ and $x_k^* = 1$ otherwise. Note that the union of these closed balls of radius $2\cdot 2^{-r}$ will cover all of $\Sigma_k$, so $\Sigma_k$ is totally bounded and therefore compact. As a closed subset of $\Sigma_k$, $\Sigma_A$ will therefore also be compact.