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Let $X=[0,1]^{\mathbb{Z^+}}$, (i.e. sequences which take values from $X$). Let the metric be $d(x,y)=sup_{n\geq 1}2^{-n}|x_n-y_n|$.

Questions:

$(1)$ Show $sup$ can be replaced with $max$ since $\exists m\in \mathbb{Z^+}$ we have $d(x,y)=2^{-m}|x_n-y_n|$. (Stuck on this)

$(2)$ Show $d(x,y)$ is a metric.

Attempt: $(i)$ $d(x,x)=0$ clearly.

$d(x,y)>0$ since for any two sequences $x=(x_n)$, $y=(y_n)$ we have $|x_n-y_n|>0$ since these sequences are different for at least one $n$, so their distance is non-zero, in addition $sup_{n\geq 1}2^{-n}>0$ $\forall n$ so positivity is proven

$(ii)$ $d(x,y)=d(y,x)$ since $|x_n-y_n|=|y_n-x_n|$ $\forall n$, and so $sup_{n\geq 1}2^{-n}|y_n-x_n|$ holds. (A might need to elaborate on this a bit)

$(iii)$. $d(x,z)\leq d(x,y)+d(y,z)$

Thinking of doing $|x_n-y_n+y_n-z_n|\leq |x_n-y_n|+|y_n-z_n|$ and arguing the scaling factor $sup_{n\geq 1}2^{-n}$ just can be placed in front of all the quantities without affecting the algebra involved in the proof.

$(3)$ Show $H$ is a complete metric space

Feeling like this problem should be easy, but I am failing to see why every Cauchy sequence must converge in $H$.

$(4)$. Let $(x^{l}_n)=(\frac{n}{n+1})^\frac{1}{l}$. Show this converges in $H$, and compute the limit.

Assuming I had a proof of $(3)$, a sequence is convergent if and only if cauchy (in a complete metric space), then this sequence clearly converges, although I don't know how to compute the limit.

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  • $\begingroup$ I guess that $H=([0,1]^{\mathbb{Z}^+},d)$, right? But you didn't specify the limit of $l$ in (4). $\endgroup$ – Przemek Sep 15 '17 at 17:58
  • $\begingroup$ It actually turns out that sequence convergence in $H$ is precisely pointwise convergence of the sequences - i.e. $x^l \to x$ as $l \to \infty$ if and only if $\endgroup$ – Daniel Schepler Sep 15 '17 at 18:30
  • $\begingroup$ (sorry, got called away from keyboard) if and only if for all $n$, $x^l_n \to x_n$ as $l \to \infty$. $\endgroup$ – Daniel Schepler Sep 15 '17 at 18:31
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(1) Let $n$ be the first index such that $|x_{n}-y_{n}|>0.$ There exists some $m\geq 1$ such that $2^{-m}<|x_{n}-y_{n}|.$ Then for all $k\geq m+n,$ $2^{-k}|x_{k}-y_{k}|\leq 2^{-k}\leq 2^{-m-n}\leq 2^{-n}|x_{n}-y_{n}|.$ Thus, $$\sup_{k\geq 1}2^{-k}|x_{k}-y_{k}|=\max_{1\leq k\leq m+n}2^{-k}|x_{k}-y_{k}|.$$

(2) (i) You're mostly right: when $x\neq y$, there is some $n$ such that $|x_{n}-y_{n}|>0,$ so $d(x,y)\geq 2^{-n}|x_{n}-y_{n}|>0$, but what you were saying with "$\sup_{n\geq 1}2^{-n}>0\,\forall n$" didn't make sense.

(ii) I see no need to elaborate, but again be careful. When you say "so $\sup_{n\geq 1}2^{-n}|y_{n}-x_{n}|$ holds," the meaning is unclear.

(iii) As you suggest, $\sup_{n\geq 1}2^{-n}|x_{n}-y_{n}|\leq \sup_{n\geq 1}2^{-n}|x_{n}-z_{n}|+2^{-n}|z_{n}-y_{n}|.$ Now, taking the supremum over the first term and the second term separately cannot decrease this quantity, or put differently, $$\sup_{n\geq1}(2^{-n}|x_{n}-z_{n}|+2^{-n}|z_{n}-y_{n}|)\leq \sup_{n\geq 1}\left(2^{-n}|x_{n}-z_{n}|+\sup_{m\geq1}2^{-m}|z_{m}-y_{m}|\right)=\sup_{n\geq1}(2^{-n}|x_{n}-z_{n}|)+\sup_{m\geq 1}(2^{-m}|z_{m}-y_{m}|).$$

(3) Suppose $(x^{(k)})_{k=1}^{\infty}$ is a sequence of points in $H.$ Then for any $\varepsilon>0,$ there is some $N$ such that for all $j,k\geq N,$ $\sup_{n\geq 1}2^{-n}|x_{n}^{(j)}-x_{n}^{(k)}|<\varepsilon.$ In particular, for each $n,$ the sequence $(x_{n}^{(k)})_{k=1}^{\infty}$ is a Cauchy subsequence of $[0,1]$ ($|x_{n}^{(j)}-x_{n}^{(k)}|\leq 2^{n}d(x^{(j)},x^{(k)})<2^{n}\varepsilon$ when $j,k\geq N,$ and $\varepsilon>0$ was arbitrary), hence has a limit $x_{n}^{*}.$ The sequence $x^{*}=(x_{n}^{*})_{n\geq 1}$ clearly belongs to $H,$ and since $|x_{n}^{(k)}-x_{n}^{*}|<2^{n}\varepsilon$ for all $n$ when $k\geq N,$ we have $\sup_{n\geq 1}2^{-n}|x_{n}^{(k)}-x_{n}^{*}|\leq \varepsilon$ when $k\geq N,$ so $x^{(k)}\rightarrow x^{*}$ in the metric of $H.$

(4) For any number $y\in(0,\infty),$ $y^{1/\ell}\rightarrow 1$ as $\ell\rightarrow\infty.$ In particular, $x_{n}^{\ell}\rightarrow 1$ as $\ell\rightarrow\infty$ for all $n\geq 1.$ Since $y\mapsto y^{1/\ell}$ is an increasing function for all $\ell\geq 1,$ and $\frac{n}{n+1}=1-\frac{1}{n+1}$ is increasing in $n,$ $|x_{n}^{\ell}-1|\leq |x_{1}^{\ell}-1|=|(1/2)^{1/\ell}-1|$ for all $n\geq 1.$ Thus $\sup_{n\geq 1}2^{-n}|x_{n}^{\ell}-1|\leq \sup_{n\geq 1}2^{-n}|(1/2)^{1/\ell}-1|=2^{-1}|(1/2)^{1/\ell}-1|\rightarrow 0$ as $\ell\rightarrow\infty,$ which proves that $x^{(\ell)}\rightarrow (1)_{n\geq 1}$ in the metric of $H$.

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  • $\begingroup$ how do you prove $|x_{n}^{(k)}-x_{n}^{*}|<2^{n}\varepsilon$ for all $n$ when $k\geq N$ ? $\endgroup$ – Gabriel Romon Sep 15 '17 at 18:07
  • $\begingroup$ Since $|x_{n}^{(j)}-x_{n}^{(k)}|< 2^{n}\varepsilon$ for all $j,k\geq N,$ when we let $j\rightarrow\infty,$ we get $|x_{n}^{(k)}-x_{n}^{*}|\leq 2^{n}\varepsilon.$ $\endgroup$ – RideTheWavelet Sep 15 '17 at 18:46
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  • For each $n$, $0\leq|x_n-y_n|\leq 2$, hence $2^{-n}|x_n-y_n|\to0$ as $n\to \infty$. There exists some $N$ such that $n\geq N\implies 2^{-n}|x_n-y_n|\leq \frac{d(x,y)}2$. Then $d(x,y)=\max_{1\leq n\leq N} 2^{-n}|x_n-y_n|$.

  • $d(x,x)=0 \implies x=0$ is easy to prove.

  • $d(x,y)=d(y,x)$ is easy to prove as well.

  • Given $x,y,z$, for all $n$, $2^{-n}|x_n-y_n|\leq 2^{-n}|x_n-z_n|+ 2^{-n}|z_n-y_n|\leq d(x,z)+d(z,y)$. Hence $\sup_{n\geq 1} 2^{-n}|x_n-y_n| \leq d(x,z)+d(z,y)$, that is $d(x,y)\leq d(x,z)+d(z,y)$

  • To prove the space is complete, consider a Cauchy sequence $x^{(n)}$. Let $k\geq 1$ be fixed. For integers $p,q$, we have $2^{-k}|x_k^{(p)}-x_k^{(q)}|\leq d(x^{(p)},x^{(q)})$. The sequence $x^{(n)}$ is therefore Cauchy in $\mathbb R$, hence convergent to some number that we note $x_k\in [0,1]$. Letting $k$ vary, this defines a sequence $x\in X$. It remains to prove that $x^{(n)}$ converges to $x$ with respect to the metric.

Let $\epsilon >0$. Since $x^{(n)}$ is Cauchy, there is some $N$ such that $p,q\geq N \implies d(x^{(p)},x^{(q)})\leq \epsilon/2$. Consider $p\geq N$.

Since $\lim_{n\to \infty} 2^{-n}|x_n^{(p)}-x_n|\to0$, there is some $N'$ such that $d(x^{(p)},x)=\max_{1\leq n \leq N'}2^{-n}|x_n^{(p)}-x_n|$. Since the sequences of real numbers $x_1^{(q)},\ldots,x_{N'}^{(q)}$ converge respectively to $x_1,\ldots, x_{N'}$ as $q\to \infty$, there exists $N''$ such that $q\geq N''\implies \forall i\in \{1,\ldots, N'\}, |x_i^{(q)}-x_i|\leq \epsilon/2$.

Consider $q\geq \max(N,N'')$. For $n\in \{1,\ldots, N'\}$, $$2^{-n}|x_n^{(p)}-x_n|\leq2^{-n}|x_n^{(p)}-x_n^{(q)}| + 2^{-n}|x_n^{(q)}-x_n| \leq d(x^{(p)},x^{(q)})+\epsilon/2\leq \epsilon$$

Since $d(x^{(p)},x)=\max_{1\leq n \leq N'}2^{-n}|x_n^{(p)}-x_n|$, we have $p\geq N \implies d(x^{(p)},x)\leq \epsilon$ and we're done.

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