Let $Σ=\{0,1\}^ℕ$ be the set of all sequences of 0 s and 1 s. If $σ=\left(a_n\right)_{n=1}^∞, σ'=\left(b_n\right)_{n=1}^∞ ∈ Σ$ define
$$
d\left(σ, σ'\right)=\frac{1}{\min \left\{n: a_n ≠ b_n\right\}},
$$
where the right-hand side is to be interpreted as zero when $a_n=b_n$ for all $n$. Show that $(Σ, d)$ is a metric space. Proof.
positivity: $d\left(σ, σ'\right)=0$ when $σ=σ'$; otherwise $ \min\left\{n: a_n ≠ b_n\right\}≥1⇒d\left(σ, σ'\right)>0$.
symmetry: $d\left(σ, σ'\right)=\frac{1}{\min \left\{n: a_n ≠ b_n\right\}}=d\left(σ', σ\right)$
triangle inequality: Let $σ''=\left(c_n\right)_{n=1}^∞ ∈ Σ,m=\min\left\{\min \left\{n: a_n ≠ b_n\right\},\min \left\{n: b_n ≠ c_n\right\}\right\}$.
The first $m-1$ terms of $(a_n),(b_n)$ are equal, and the first $m-1$ terms of $(b_n),(c_n)$ are equal, so the first $m-1$ terms of $(a_n),(c_n)$ are equal.
$⇒\min \left\{n: a_n ≠ c_n\right\} ≥m$
$⇒\frac{1}{\min \left\{n: a_n ≠ c_n\right\}}≤\max\left\{\frac{1}{\min \left\{n: a_n ≠ b_n\right\}},\frac{1}{\min \left\{n: b_n ≠ c_n\right\}}\right\}≤\frac{1}{\min \left\{n: a_n ≠ b_n\right\}}+\frac{1}{\min \left\{n: b_n ≠ c_n\right\}}$
$⇒d(σ,σ'')≤d(σ,σ')+d(σ',σ'')$
Let $Ω=ℝ^ℕ$ be the space of all real sequences. If $𝐱=\left(x_n\right)_{n=1}^∞,𝐲=\left(y_n\right)_{n=1}^∞ ∈ Ω$, define
$$
d(𝐱,𝐲)=\sum_{n=1}^∞ \frac{1}{2^n} \frac{\left|x_n-y_n\right|}{1+\left|x_n-y_n\right|}
$$
Show that this defines a metric on $Ω$. Proof.
$$
d(𝐱,𝐲)≤\sum_{n=1}^∞ \frac{1}{2^n}=1
$$
positivity: if $𝐱=𝐲$ then $∀n,\left|x_n-y_n\right|=0⇒d(𝐱,𝐲)=0$; otherwise $∃m,x_m≠y_m⇒\frac{\left|x_m-y_m\right|}{1+\left|x_m-y_m\right|}>0 ⇒d(𝐱,𝐲)>0$.
symmetry: $\left|x_n-y_n\right|=\left|y_n-x_n\right|⇒d(𝐱,𝐲)=d(𝐲,𝐱)$.
triangle inequality: The function $f(t)=\frac{t}{1+t},\ t≥0$ is increasing. Since ${|a+b|}≤{|a|}+{|b|}$, we have $f({|a+b|})≤f({|a|}+{|b|})$, so
\begin{aligned}
\frac{|a+b|}{1+{|a+b|}}&≤\frac{|a|}{1+{|a|}+{|b|}}+\frac{|b|}{1+{|a|}+{|b|}}\\
&≤\frac{|a|}{1+{|a|}}+ \frac{|b|}{1+{|b|}}
\end{aligned}
Let $a=x_n-y_n,b=y_n-z_n$, and summing over $n$, we get $d(𝐱, \mathbf{z})≤d(𝐱,𝐲)+d(𝐲, \mathbf{z})$.
Let $X=ℝ^n$ and suppose that $d$ is one of $d_1, d_2, d_∞$. Suppose we have two balls $B\left(x_1, 0.9\right), B\left(x_2, 0.9\right)$, both contained in $B(0,1)$. Show that they intersect. Is the same true in a general metric space $(X, d)$?
$d_1, d_2, d_∞$ all arise from norm, so satisfy linearity. The proof below applies to all of them. Proof 1.
We will show $0∈B\left(x_1, 0.9\right)$, then $0∈B\left(x_2, 0.9\right)$, so they intersect.
Suppose $0∉B\left(x_1, 0.9\right)$, then $d(x,0)≥0.9$, multiplying by 1.5, $d\left(\frac32x_1,0\right)≥1.35$.
$d\left(\frac32x_1,x_1\right)=d\left(\frac12x_1,0\right)<\frac12$, so $\frac32x_1∈B(x_1,0.9)⊂B(0,1)$, so $d\left(\frac32x_1,0\right)< 1$, contradiction with previous line.
Replace 0.9 with α and replace $\frac12$ with $\frac1C$. For the proof to remain valid, we need $\cases{Cα≥1\\d\left((C-1)x_1,0\right)< α}⇔\cases{Cα≥1\\C< 1+α}⇔α>\frac{\sqrt5-1}2≈0.618$.
Proof 2.
Let $v$ be the unit vector in the direction of $x_1$, then $x_1={‖x_1‖}v$. The fact that $B(x_1,0.9)⊂B(0,1)$ implies that $({‖x_1‖}+r)v∈B(0,1)$ for any $0< r< 0.9$. Thus ${\big‖({‖x_1‖}+r)v\big‖}={‖x_1‖}+r< 1$ for any $0< r< 0.9$. It follows that ${‖x_1‖}≤1-0.9=0.1$. Similarly ${‖x_2‖}≤0.1$. By triangle inequality ${‖x_1-x_2‖}≤0.2⇒x_1∈B(x_2,0.9)⇒x_1∈B(x_1,0.9)∩B(x_2,0.9)$.
This is false in a general metric space $(X, d)$, for example, when $d$ is the discrete metric, $B\left(x_1, 0.9\right)=\{x_1\}, B\left(x_2, 0.9\right)=\{x_2\}$, and $\{x_1\}∩\{x_2\}=∅$. Actually we can take $d$ be the discrete metric times a constant $α ∈(0.9,1]$, then we also have $B\left(x_1, 0.9\right)=\{x_1\}, B\left(x_2, 0.9\right)=\{x_2\}$, and $\{x_1\}∩\{x_2\}=∅$.
Let $(X, d)$ be a metric space, and let $‖·‖$ be a norm on $ℝ^2$. Define $\tilde{d}$ by
$$
\tilde{d}\left((x_1, x_2),(y_1, y_2)\right)=\left‖\left(d(x_1, y_1), d(x_2, y_2)\right)\right‖ .
$$
Show that $\tilde{d}$ is a metric on $X × X$ if $‖·‖$ is the $ℓ^1$-norm or the $ℓ^∞$-norm. Is this in fact true for an arbitrary norm $‖·‖$? Solution.
Positivity and symmetry are trivial. Triangle inequality:
For $ℓ^1$:
\begin{aligned}
&\tilde{d}\left((x_1,x_2),(y_1,y_2)\right)+\tilde{d}\left((y_1,y_2),(z_1,z_2)\right)\\
=&d(x_1, y_1)+d(x_2, y_2)+d(y_1, z_1)+d(y_2, z_2)\\
=&\left(d(x_1, y_1)+d(y_1, z_1)\right)+\left(d(x_2, y_2)+d(y_2, z_2)\right)\\
>&d(x_1, z_1)+d(x_2, z_2)\\
=&\tilde{d}\big((x_1,x_2), (z_1,z_2)\big).\end{aligned}
For $ℓ^2$, $\tilde{d}$ is the product metric.
For $ℓ^∞$:
\begin{aligned}
&\tilde{d}\left((x_1,x_2),(y_1,y_2)\right)+\tilde{d}\left((y_1,y_2),(z_1,z_2)\right)\\
=&\max\left(d(x_1, y_1), d(x_2, y_2)\right)+\max\left(d(y_1, z_1), d(y_2, z_2)\right)\\
≥&\max\left(d(x_1, y_1)+d(y_1, z_1), d(x_2, y_2)+d(y_2, z_2)\right)\\
>&\max\big(d(x_1, z_1), d(x_2, z_2)\big)\\=&\tilde{d}\big((x_1,x_2), (z_1,z_2)\big).
\end{aligned}
This is not true for an arbitrary norm $‖⋅‖$.
Take $X=\Bbb R$ with $d(a,b)={|a-b|}$ and consider the norm ${‖(x,y)‖}={|x|}+{|x-y|}$ on $\Bbb R^2$. Then$$\tilde{d}\big((2,0),(0,0)\big)={\big‖\big(d(2,0),d(0,0)\big)\big‖}={‖(2,0)‖}=4$$is greater than$$\tilde{d}\big((2,0),(1,1)\big)+\tilde{d}\big((1,1),(0,0)\big)={\big‖\big(d(2,1),d(0,1)\big)\big‖}+{\big‖\big(d(1,0),d(1,0)\big)\big‖}={‖(1,1)‖}+{‖(1,1)‖}=2$$
Consider the 2-adic metric on the integers defined in the lecture notes. Show that the sequence $9,99,999,…$ converges. Proof.
$d(10^n-1,-1)=2^{-n}$ converges to 0 as $n→∞$, so $10^n-1$ converges to $-1$.
Let $X$ be a metric space and suppose $f: X → ℝ$ is continuous. Show that if $f(a) ≠ 0$, then there is an $ε>0$ such that $1/f$ is defined and is continuous on $B(a, ε)$. Proof.
Just compose $f$ with $1/x$, since $1/x$ is continuous on the image of $f$. Or we prove it explicitly:
By continuity of $f$, $∃ε>0,∀x∈B(a,ε):{|f(x)-f(a)|}<\frac12{|f(a)|}⇒{|f(x)|}>\frac12{|f(a)|}>0⇒1/f$ is defined on $B(a,ε)$.
For all $ε_0>0$, by continuity of $f$, $∃δ>0,∀x∈B(a,δ):{|f(x)-f(a)|}< ε_0\left(\frac12f(a)\right)^2⇒\left|\frac1{f(x)}-\frac1{f(a)}\right|=\frac{|f(x)-f(a)|}{{|f(a)|}⋅{|f(x)|}}<\frac{|f(x)-f(a)|}{\left(\frac12f(a)\right)^2}< ε_0⇒1/f$ is continuous.
Suppose that $(X, d)$ is a metric space and that $X × X$ is endowed with the product metric defined in lectures. Show that the metric $d$, viewed as a map from $X × X$ to $ℝ$, is continuous. Proof.
Let $x, y, z, t∈X$. By triangle inequality,$${|d(x, y)-d(z, t)|}≤{|d(x, y)-d(z, y)|}+{|d(z,y)-d(z,t)|}≤d(x, z)+d(y, t)=d_1((x, y),(z, t))≤\sqrt2d_2((x, y),(z, t))$$So given $ε>0$ we may take $δ=ε/\sqrt2$, and if $d_2((x, y),(z, t))<\delta$ then ${|d(x, y)-d(z, t)|}< ε$, so $d:X×X→ℝ$ is continuous.
Introduction to Metric and Topological Spaces - Wilson A. Sutherland - Oxford University Press (2009) Solutions to Chapter 5 exercises (Need to sign up and log in)
Exercise 5.17 is about $d_1$, but here "product metric" refers to $d_2$.
Show that the map $f:ℤ→ℤ$ defined by $f(n)=n^2$ is continuous in the 2-adic metric. Proof.
Let $2^m∣x-a$, then $2^m∣x^2-a^2=(x-a)(x+a)$, so $d\big(f(x),f(a)\big)< d(x,a)$.
$∀a∈ℤ,ε>0$, then $∀x∈B(a,ε):d\big(f(x),f(a)\big)< d(x,a)< ε$. So $f$ is continuous at $a$.
Let $X, Y$ and $Z$ be metric spaces and equip $Y × Z$ with the product metric defined in lectures. If $F: X → Y × Z$ is a function and we write $F(x)=\left(f_1(x), f_2(x)\right)$, show that $F$ is continuous if and only if $f_1$ and $f_2$ are continuous. Proof.
If $f_1$ and $f_2$ are continuous, for any $ε>0$, we can find $δ>0$, such that $∀a∈B(x,δ):f_1(a)∈B\left(f_1(x),\fracε{\sqrt2}\right)$ and $f_2(a)∈B\left(f_2(x),\fracε{\sqrt2}\right)$.
So $∀a∈B(x,δ):d_{Y×Z}\Big(\big(f_1(a),f_2(a)\big),\big(f_1(x),f_2(x)\big)\Big)=\sqrt{d_Y\big(f_1(a),f_1(x)\big)^2+d_Z\big(f_2(a),f_2(x)\big)^2}<\sqrt2\fracε{\sqrt2}=ε$. Therefore $F$ is continuous.
If $F$ is continuous, for any $ε>0$, we can find $δ>0$, such that $∀a∈B(x,δ):ε>d_{Y×Z}\Big(\big(f_1(a),f_2(a)\big),\big(f_1(x),f_2(x)\big)\Big)=\sqrt{d_Y\big(f_1(a),f_1(x)\big)^2+d_Z\big(f_2(a),f_2(x)\big)^2}≥d_Y\big(f_1(a),f_1(x)\big)$.
So $f_1$ is continuous. Similarly $f_2$ is continuous.
Let $X$ denote the vector space of sequences $𝐱=\left(x_n\right)_{n=1}^∞$ with $x_n ∈ ℝ$ and $\sum_{n=1}^∞ x_n^2<∞$. Explain why ${‖𝐱‖}_∞:=\sup_n\left|x_n\right|$ is a well-defined norm on $X$. Is the metric induced by this norm equivalent to the metric induced by the $ℓ^2$-norm? Proof.
${‖𝐱‖}_∞=\sup_n\left|x_n\right|≤\sqrt{\sum_{n=1}^∞ x_n^2}={‖𝐱‖}_2$, so it is finite. Clearly ${‖𝐱‖}_∞=0$ iff $𝐱=0$; ${‖λ𝐱‖}_∞ = {|λ|}{‖𝐱‖}_∞$ for all $λ∈ℝ,𝐱∈X$; ${‖𝐱+𝐲‖}_∞=\sup_n\left|x_n+y_n\right|≤\sup_n\left|x_n\right|+\sup_n\left|y_n\right|={‖𝐱‖}_∞+{‖𝐲‖}_∞$ for all $𝐱,𝐲∈ X$. So ${‖𝐱‖}_∞$ is a well-defined norm on $X$.
The metric induced by $ℓ^∞$-norm and $ℓ^2$-norm are not equivalent. Let ${\bf 0}=(0,0,…)$. Suppose $B_2({\bf0},1)⊃B_∞({\bf0},r)$ for some $r>0$. Let $0< a< r$ and ${\bf a}_k=(\underbrace{a,a,…,a}_k,0,0,…)$. Then ${‖{\bf a}_k‖}_∞=a⇒{\bf a}_k∈B_∞({\bf0},r)$ but ${‖{\bf a}_k‖}_2=a\sqrt k⇒{\bf a}_k∉B_2({\bf0},1)$ for $k>a^{-2}$, contradicting to the assumption $B_2({\bf0},1)⊃B_∞({\bf0},r)$.