Is the sequence space on two symbols Compact?? Consider the space of infinite sequences on two letters $0$ and $1$.
i.e. let  $\Omega = $\begin{cases} (s_0 s_1 s_2.....) : & \text{where each $s_i$ is $0$ or $1$} \end{cases}
Said differently, the elements of $\Omega$ are basically words of infinite letters.
Now define a metric on the given space $\Omega$ as 
$d(s,t)$ = $\sum_{n=0}^\infty \frac {|s_k - y_k |}{2^k}$
How could we prove that the metric space $(\Omega,$d$)$ is a compact metric  space?
Can we use the continuous shift map $\sigma : \Omega \to \Omega$ defined by 
$\sigma (s_0 s_1 s_2....)$ $=$ $(s_1 s_2 s_3.....)$.
I have proved that the above defined shift map is continuous by using Archimedean Property by choosing a natural number $n$ such that $\frac{1}{2^n}$ $\lt$ $\epsilon$ for any $\epsilon$.
Also the question is whether this map is uniformly continuous.
If this space is compact metric space then it would help me in proving that the dynamical system ($\Omega$,$\sigma$) will depend sensitively on initial conditions.
Any help would be appreciated.
 A: Yes, the space $(\Omega,d)$ is compact.

A metric space is compact if and only if it's complete and totally bounded.

First we show that $(\Omega ,d)$ is complete . . .

Suppose $x_1,x_2,x_3,...$ is a Cauchy sequence of elements of $\Omega$.

It follows that for each integer $k \ge 0$, the sequence $x_1[k],x_2[k],x_3[k],...$ is eventually constant, equal to $u_k$ say, for if not, we could find arbitrarily large positive integers $m,n$ which differ at position $k$, yielding 
$d(x_m,x_n) \ge {\large{\frac{1}{2^k}}}$.

Let $s$ be the element of $\Omega$ such that $s_k = u_k$, for all $k$.

Then it's easily seen that the sequence $x_1,x_2,x_3,...$ converges to $s$.

Thus,  $(\Omega ,d)$ is complete.

Next we show that $(\Omega ,d)$ is totally bounded . . .

Let $\epsilon > 0$, and let $m$ be a positive integer such that ${\large{\frac{1}{2^{m-1}}}} < \epsilon$.

Let $V$ be the set of $2^m$ points defined by
$$V = \{s \in \Omega\mid s_k = 0,\;\text{for all}\;k \ge m\}$$
Then it's easily seen that the set of open balls of radius $\epsilon$, centered at the points of $V$ cover $\Omega$.

Thus,  $(\Omega ,d)$ is totally bounded.

It follows that $(\Omega,d)$ is compact.

As to your second question, the answer is "yes", since every continuous function on a compact set is uniformly continuous. For a reference, see the following link:

$\qquad$https://en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem
A: It suffices to prove that any sequence $(x_n)_{n\in \Bbb N_0}$ of members of $\Omega$ has a sub-sequence converging to  a member of $\Omega.$ Let $x_n=(s_{n,j})_{j\in \Bbb N_0}$ be a sequence in $\Omega.$  
Take $y_0\in \{0,1\}$ such that $D(0)=\{n: x_{n,0}=y_0\}$ is infinite. Let $f(0)$ be the least (or any) $n\in D(0).$
For each $j\geq 0$ take $y_{1+j}\in \{0,1\}$ such that $$D(1+j)=\{n:\forall k\leq j\; (x_{n,k}=x_{f(j),k} \land x_{n,1+j}=y_{1+j} \text { is infinite })\}.$$ Let $f(1+j)$ be the least (or any) $n>f(j)$ such that $n\in D(1+j).$ 
Let $y=(y_j)_{j\in \Bbb N_0}.$ For all $j$ we have $f(1+j)>f(j)$  and  $$d(x_{f(j)},y)\leq \sum_{v=1+j}^{\infty}2^{-v}=2^{-j}.$$ So $(x_{f(j)})_{j\in \Bbb N_0} $ is a subsequence of $(x_n)_{n\in \Bbb N_0}$ that converges to $y$.
