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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.