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Let $X$ be the set of all bounded sequences $x=(x_n)$ of real numbers and let $$d(x,y)=\sup{|x_n-y_n|}.$$ I need to show that $X$ is a complete metric space.

I need to show that all Cauchy sequences are convergent. I appreciate your help.

JohnD
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Klara
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2 Answers2

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HINT: Let $\langle x^n:n\in\Bbb N\rangle$ be a Cauchy sequence in $X$. The superscripts are just that, labels, not exponents: $x^n=\langle x^n_k:k\in\Bbb N\rangle\in X$. Fix $k\in\Bbb N$, and consider the sequence

$$\langle x^n_k:n\in\Bbb N\rangle=\langle x^0_k,x^1_k,x^2_k,\dots\rangle\tag{1}$$

of $k$-th coordinates of the sequences $x^n$. Show that for any $m,n\in\Bbb N$, $|x^m_k-x^n_k|\le d(x^m,x^n)$ and use this to conclude that the sequence $(1)$ is a Cauchy sequence in $\Bbb R$. $\Bbb R$ is complete, so $(1)$ converges to some $y_k\in\Bbb R$. Let $y=\langle y_k:k\in\Bbb N\rangle$; show that $y\in X$ and that $\langle x^n:n\in\Bbb N\rangle$ converges to $y$ in $X$.

Brian M. Scott
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  • Hi, I can prove that $y$ converges but I can't prove that $\langle x^n:n\in\Bbb N\rangle$ is converge, please help me.(I use standard definition of limits for metric space.) – Samvel Safaryan Sep 13 '23 at 14:50
  • @SamvelSafaryan: Fix $\epsilon>0$. $\langle x^n:n\in\Bbb N\rangle$ is Cauchy, so there is an $n_\epsilon\in\Bbb N$ such that $d(x^n,x^m)<\epsilon$ whenever $n,m\ge n_\epsilon$. In particular, $d(x^n,x^{n_0})<\epsilon$ for all $n\ge n_\epsilon$; use this to show that $d(y,x^{n_0})\le\epsilon$, and conclude that $d(y,x^n)<2\epsilon$ for all $n\ge n_\epsilon$. Since $\epsilon>0$ was arbitrary, it follows that $\langle x^n:n\in\Bbb N\rangle$ converges to $y$. – Brian M. Scott Sep 15 '23 at 05:54
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    Thank you, very much – Samvel Safaryan Sep 15 '23 at 15:06
  • @SamvelSafaryan: You’re very welcome. – Brian M. Scott Sep 15 '23 at 20:05
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It can be shown that $d$ is a metric on $X$. The metric space $(X,d)$ is generally denoted by $l^{\infty}$. We will show that every Cauchy sequence in $l^{\infty}$ converges. Let $(x_m)$ be any Cauchy sequence in $l^{\infty}$. For each $m\ge 1$, write $$x_m=(c_1^{(m)},c_2^{(m)},\cdots)\in l^{\infty}.$$

Its not hard to show that for each $j\in \mathbb{Z}^+$, the sequence $(c_j^{(m)})$ is a Cauchy sequence in $\mathbb{R}$and hence it converges (because $\mathbb{R}$ is complete) to $c_j\in \mathbb{R}$. Take $x=(c_j)$. With this, just show that $x$ is bounded, so that $x\in l^{\infty}$. Lastly, show that $x_m\to x$ as $m\to +\infty$.

Juniven Acapulco
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