Show that $(X,d)$ is compact

Question

Let $X=\{0,1\}^{\mathbb{N}}$. For $x=x_1x_2x_3\cdots$ and $y=y_1y_2y_3\cdots$ in $X$, define \begin{align*} d(x,y)=2^{-\min\{n\in\mathbb{N}:x_n\neq y_n\}} \end{align*}

Firstly, I showed $(X,d)$ is a metric space. Next, I knew that the topology generated by the metric d is same with the product topology of discrete space $\{0,1\}$.

Thus, I get the $(X,d)$ is compact by using Tychonoff's theorem.

But, I want to know how to prove this fact without using Tychonoff's theorem...

Any help is appreciated!!

Thank you!!

Answer 1

This is just Cantor's diagonal procedure. Start with a sequence in X and look at the first coordinates. You get a sequence of 0's and 1's. Obviously this has a convergent subsequence. Now look at the second coordinates along this subsequence and extract a convergent subsequence, etc. The 'diagonal' subsequence has the property that each coordinate converges along this subsequence. Convergence of each coordinate is equivalent to convergence in the given metric.