Prove a metric space is totally bounded

Question

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?

Answer 1

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.

Answer 2

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 $>N$.