# Hilbert Cube is Complete

Let $X=[0,1]^\mathbb{N}$ consisting of sequences $x:=\{x_n\}_{n\in\mathbb{N}}$, $0\leq x_n \leq 1$. We define the metric $d(x,y)=\sum_{n\in\mathbb{N}}2^{-n}|x_n-y_n|$ Note that $(X,d)$ is a metric space. Show that $(X,d)$ is complete.

I am quite lost here. We say a set is complete all Cauchy sequences converge in $(X,d)$ are convergent. So let $(x_n)_{n\in\mathbb{N}}$ be a Cauchy-sequence, that is to say

$$\forall \varepsilon>0\;\exists N\in\mathbb{N}\;\forall n,m>N\;d(x_n,x_m)<\varepsilon$$

I am, however, totally lost here. Could anyone give me a hint on how to continue?

Let $(\mathbf{x}_{k})_{k \in \mathbb{N}} = \left( \left( x_{k}^{(j)} \right)_{j \in \mathbb{N}} \right)_{k \in \mathbb{N}}$ be Cauchy. We wanna find a sequence $\mathbf{y} = \left( y^{(j)} \right)_{j \in \mathbb{N}}$ such that $d( \mathbf{x}_k , \mathbf{y} ) \to 0$. The way we'll do this is as follows: (i) show that if the sequence is Cauchy, then the sequence $\left( x_{k}^{(j)} \right)_{k \in \mathbb{N}}$ is convergent for every $j_0 \in \mathbb{N}$, (ii) set $y^{(j)} = \lim_{k \to \infty} x_{k}^{(j)}$, and (iii) show $\lim_{k \to \infty} \mathbf{x}_{k} = \mathbf{y}$.

First, to demonstrate the claim in (i), fix $j_0 \in \mathbb{N}, \delta > 0$. Then by the Cauchy assumption, we know there exists $K = K(\delta / 2^{j_0}) \in \mathbb{N}$ such that if $k_1, k_2 \geq K$, then $$d( \mathbf{x}_{k_1} , \mathbf{x}_{k_2} ) = \sum_{j = 1}^{\infty} 2^{-j} \left| x_{k_1}^{(j)} - x_{k_2}^{(j)} \right| < \delta / 2^{j_0}.$$ But $2^{ - j_0} \left| x_{k_1}^{(j_0)} - x_{k_2}^{(j_0)} \right| \leq \sum_{j = 1}^{\infty} 2^{-j} \left| x_{k_1}^{(j)} - x_{k_2}^{(j)} \right|$, so we know in fact that $\left( x_{k}^{(j_0)} \right)_{k \in \mathbb{N}}$ is Cauchy, and thus has a limit, call it $y^{(j_0)}$. Therefore for every $j \in \mathbb{N} , \eta > 0$ exists $L(j, \eta)$ such that if $k \geq L(j, \eta)$, then $\left| x_{k}^{(j)} - y^{(j)} \right| < \eta$.

Now we show that $\mathbf{y} = \lim_{k \to \infty} \mathbf{x}_{k}$. Fix $\epsilon > 0$. Then there exists $J \in \mathbb{N}$ such that $\sum_{j = J + 1}^{\infty} 2^{-j} < \epsilon / 2$. Then let $L = \max \left\{ L \left( 1, \epsilon / 2 \right) , \ldots , L \left( J, \epsilon / 2 \right) \right\}$. Suppose $k \geq L$. Then \begin{align*} d( \mathbf{x}_k , \mathbf{y} ) & = \sum_{j = 1}^{\infty} 2^{-j} \left| x_{k}^{(j)} - y^{(j)} \right| \\ & = \left[ \sum_{j = 1}^{J} 2^{-j} \left| x_{k}^{(j)} - y^{(j)} \right| \right] + \sum_{j = J + 1}^{\infty} 2^{-j} \left| x_{k}^{(j)} - y^{(j)} \right| \\ & < \left[ \sum_{j = 1}^{J} 2^{-j} (\epsilon / 2) \right] + \sum_{j = J + 1}^{\infty} 2^{-j} \\ & \leq (\epsilon / 2) \left[ \sum_{j = 1}^{\infty} 2^{-j} \right] + \epsilon / 2 \\ & = \epsilon / 2 + \epsilon / 2 \\ & = \epsilon . \end{align*}

Suppose that $({\bf x}^{(n)})$ is a Cauchy sequence. Each element is of the form ${\bf x}^{(n)}=(x^n_1,x^n_2,\dots)$. Since for each $k$

$d(x^{(n)},x^{(m)}) \ge 2^{-k} |x^{(n)}_k-x^{(m)}_k|$, it follows that the sequence $(x^{n}_k:k\in \mathbb N)$ is Cauchy in ${\mathbb R}$. Since ${\mathbb R}$ is complete, it converges to an element $x^{\infty}_k$. Let $x^{(\infty)}=(x^{\infty}_1,x^{\infty}_2,\dots)$. It remains to show:

1. $x^{(\infty)}$ is in the Hilbert cube.

This is immediate because for each $k$, $(x^n_k:n\in{\mathbb N})$ is a sequence in $[0,1]$

2. $d (x^{(n)},x^{(\infty)}) \to 0$ as $n\to\infty$.

To show this, observe that for any $n,m,K\in {\mathbb N}$

$$d(x^{(n)},x^{(n+m)}) \ge \sum_{k=1}^K 2^{-k}|x^n_k- x^{n+m}_k|.$$

Letting $m\to\infty$ and using the fact that the RHS is a finite sum, we obtain

$$\liminf_{m\to\infty} d(x^{(n)},x^{(n+m)}) \ge \sum_{k=1}^K 2^{-k}|x^n_k - x^{\infty}_k|.$$

Since this is true for all $K$, we can replace the righthand side with its limit as $K\to\infty$:

$$\liminf_{m\to\infty} d(x^{(n)},x^{(n+m)}) \ge \sum_{k=1}^\infty 2^{-k} |x^n_k-x^{\infty}_k|.$$

The RHS is $d(x^{(n)},x^{(\infty)})$.