# Prove a metric space is totally bounded

Let $$X = 2^{\mathbb{Z+}}$$ be the space of binary sequences $$(x_k)_{k\ge1}$$ with each $$x_k \in \{0,1\}.$$ Define a metric on $$X$$ by $$d(x,y) = \sum^\infty_{k=1} |x_k−y_k|/2^k.$$ I am trying to use the definition of a totally bounded space but I haven't found the $$\varepsilon$$-net of $$X.$$ My question is how to find the $$\varepsilon$$-net so that we can prove that $$X$$ is totally bounded?

• Your space is homeomorphic to $\prod_{n\in\mathbb{Z}_+}\{0,1\}$ with the product topology. The product t of compact spaces is compact. As for he specifics of your question, notice that $\sum_{n\geq m}\frac{1}{2^n}\xrightarrow{m\rightarrow\infty}0$. – Oliver Diaz Jul 22 '20 at 20:39

If I'm understanding the question correctly, what you're looking for is, for each $$\varepsilon>0,$$ a finite set of points in your space such that every point in the space is within a distance $$\varepsilon$$ of some point in that finite set.
Find the smallest positive integer $$k$$ such that $$2^{-k}<\varepsilon.$$ Then consider the set of all sequences of the following form: $$x_1, x_2, x_3, \ldots,x_k,\,\underbrace{0, 0, 0, 0, 0, \ldots\ldots}$$ There are only $$2^k$$ of these, a finite number. And every point is within $$\varepsilon$$ of one of these.
It is enough to consider the special case of an $$\epsilon$$ of the shape: $$\epsilon = \frac 1{2^{N-1}}\ ,\qquad N\in\Bbb Z_{>0}\ .$$ So let us fix such an $$N$$ and a corresponding $$\epsilon$$. Consider all (finitely many) elements $$x(a)=(a_1,a_2,\dots,a_N,0,0,\dots)$$ with $$a=(a_1,a_2,\dots,a_N)\in\{0,1\}^{\times N}$$ that have all possible starts on the first $$N$$ places, followed by zeros.
Consider now an $$x$$ in the metric space, select the $$x(a)$$ with the match with $$x$$ on the first $$N$$ places and estimate the distance from $$x$$ to $$x(a)$$ by being generous on the places on positions $$>N$$.