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Let $Ξ£=\{0,1\}^β$ be the set of all sequences of 0 s and 1 s. If $Ο=\left(a_n\right)_{n=1}^β, Ο'=\left(b_n\right)_{n=1}^β β Ξ£$ define
$$
d\left(Ο, Ο'\right)=\frac{1}{\min \left\{n: a_n β b_n\right\}},
$$
where the right-hand side is to be interpreted as zero when $a_n=b_n$ for all $n$. Show that $(Ξ£, d)$ is a metric space.
Proof.
positivity: $d\left(Ο, Ο'\right)=0$ when $Ο=Ο'$; otherwise $ \min\left\{n: a_n β b_n\right\}β₯1βd\left(Ο, Ο'\right)>0$.
symmetry: $d\left(Ο, Ο'\right)=\frac{1}{\min \left\{n: a_n β b_n\right\}}=d\left(Ο', Ο\right)$
triangle inequality: Let $Ο''=\left(c_n\right)_{n=1}^β β Ξ£,m=\min\left\{\min \left\{n: a_n β b_n\right\},\min \left\{n: b_n β c_n\right\}\right\}$.
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.
$β\min \left\{n: a_n β c_n\right\} β₯m$
$β\frac{1}{\min \left\{n: a_n β c_n\right\}}β€\max\left\{\frac{1}{\min \left\{n: a_n β b_n\right\}},\frac{1}{\min \left\{n: b_n β c_n\right\}}\right\}β€\frac{1}{\min \left\{n: a_n β b_n\right\}}+\frac{1}{\min \left\{n: b_n β c_n\right\}}$
$βd(Ο,Ο'')β€d(Ο,Ο')+d(Ο',Ο'')$ -
Let $Ξ©=β^β$ be the space of all real sequences. If $π±=\left(x_n\right)_{n=1}^β,π²=\left(y_n\right)_{n=1}^β β Ξ©$, define
$$
d(π±,π²)=\sum_{n=1}^β \frac{1}{2^n} \frac{\left|x_n-y_n\right|}{1+\left|x_n-y_n\right|}
$$
Show that this defines a metric on $Ξ©$.
Proof.
$$ d(π±,π²)β€\sum_{n=1}^β \frac{1}{2^n}=1 $$ positivity: if $π±=π²$ then $βn,\left|x_n-y_n\right|=0βd(π±,π²)=0$; otherwise $βm,x_mβ y_mβ\frac{\left|x_m-y_m\right|}{1+\left|x_m-y_m\right|}>0 βd(π±,π²)>0$.
symmetry: $\left|x_n-y_n\right|=\left|y_n-x_n\right|βd(π±,π²)=d(π²,π±)$.
triangle inequality: The function $f(t)=\frac{t}{1+t},\ tβ₯0$ is increasing. Since ${|a+b|}β€{|a|}+{|b|}$, we have $f({|a+b|})β€f({|a|}+{|b|})$, so \begin{aligned} \frac{|a+b|}{1+{|a+b|}}&β€\frac{|a|}{1+{|a|}+{|b|}}+\frac{|b|}{1+{|a|}+{|b|}}\\ &β€\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 $d(π±, \mathbf{z})β€d(π±,π²)+d(π², \mathbf{z})$. -
Let $X=β^n$ and suppose that $d$ is one of $d_1, d_2, d_β$. 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 $(X, d)$?
$d_1, d_2, d_β$ all arise from norm, so satisfy linearity. The proof below applies to all of them.
Proof 1.
We will show $0βB\left(x_1, 0.9\right)$, then $0βB\left(x_2, 0.9\right)$, so they intersect.
Suppose $0βB\left(x_1, 0.9\right)$, then $d(x,0)β₯0.9$, multiplying by 1.5, $d\left(\frac32x_1,0\right)β₯1.35$.
$d\left(\frac32x_1,x_1\right)=d\left(\frac12x_1,0\right)<\frac12$, so $\frac32x_1βB(x_1,0.9)βB(0,1)$, so $d\left(\frac32x_1,0\right)< 1$, contradiction with previous line. Proof 2.
Let $v$ be the unit vector in the direction of $x_1$, then $x_1={βx_1β}v$. The fact that $B(x_1,0.9)βB(0,1)$ implies that $({βx_1β}+r)vβB(0,1)$ for any $0< r< 0.9$. Thus ${\big\|({βx_1β}+r)v\big\|}={βx_1β}+r< 1$ for any $0< r< 0.9$. It follows that ${βx_1β}β€1-0.9=0.1$. Similarly ${βx_2β}β€0.1$. By triangle inequality ${βx_1-x_2β}β€0.2βx_1βB(x_2,0.9)βx_1βB(x_1,0.9)β©B(x_2,0.9)$.
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\}β©\{x_2\}=β $.
Actually we can take $d$ be the discrete metric times a constant $Ξ± β(0.9,1]$, 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\}β©\{x_2\}=β $. -
Let $(X, d)$ be a metric space, and let $βΒ·β$ be a norm on $β^2$. 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 $X Γ X$ if $βΒ·β$ is the $β^1$-norm or the $β^β$-norm. Is this in fact true for an arbitrary norm $βΒ·β$?
Solution.
Positivity and symmetry are trivial. Triangle inequality:
For $β^1$: \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 $β^2$, $\tilde{d}$ is the product metric.
For $β^β$: \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)\\ β₯&\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 $ββ β$.
Take $X=\Bbb R$ with $d(a,b)={|a-b|}$ and consider the norm ${β(x,y)β}={|x|}+{|x-y|}$ on $\Bbb R^2$. Then$$\tilde{d}\big((2,0),(0,0)\big)={\big\|\big(d(2,0),d(0,0)\big)\big\|}={β(2,0)β}=4$$is greater than$$\tilde{d}\big((2,0),(1,1)\big)+\tilde{d}\big((1,1),(0,0)\big)={\big\|\big(d(2,1),d(0,1)\big)\big\|}+{\big\|\big(d(1,0),d(1,0)\big)\big\|}={β(1,1)β}+{β(1,1)β}=2$$ -
Consider the 2-adic metric on the integers defined in the lecture notes. Show that the sequence $9,99,999,β¦$ converges.
Proof.
$d(10^n-1,-1)=2^{-n}$ converges to 0 as $nββ$, so $10^n-1$ converges to $-1$. -
Let $X$ be a metric space and suppose $f: X β β$ is continuous. Show that if $f(a) β 0$, then there is an $Ξ΅>0$ such that $1/f$ is defined and is continuous on $B(a, Ξ΅)$.
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 $f$, $βΞ΅>0,βxβB(a,Ξ΅):{|f(x)-f(a)|}<\frac12{|f(a)|}β{|f(x)|}>\frac12{|f(a)|}>0β1/f$ is defined on $B(a,Ξ΅)$.
For all $Ξ΅_0>0$, by continuity of $f$, $βΞ΄>0,βxβB(a,Ξ΄):{|f(x)-f(a)|}< Ξ΅_0\left(\frac12f(a)\right)^2β\left|\frac1{f(x)}-\frac1{f(a)}\right|=\frac{|f(x)-f(a)|}{|f(a)|β {|f(x)|}}<\frac{|f(x)-f(a)|}{\left(\frac12f(a)\right)^2}< Ξ΅_0β1/f$ is continuous. -
Suppose that $(X, d)$ is a metric space and that $X Γ X$ is endowed with the product metric defined in lectures. Show that the metric $d$, viewed as a map from $X Γ X$ to $β$, is continuous.
Proof.
Let $x, y, z, tβX$. By triangle inequality,$${|d(x, y)-d(z, t)|}β€{|d(x, y)-d(z, y)|}+{|d(z,y)-d(z,t)|}β€d(x, z)+d(y, t)=d_1((x, y),(z, t))β€\sqrt2d_2((x, y),(z, t))$$So given $Ξ΅>0$ we may take $Ξ΄=Ξ΅/\sqrt2$, and if $d_2((x, y),(z, t))<\delta$ then ${|d(x, y)-d(z, t)|}< Ξ΅$, so $d:XΓXββ$ is continuous. -
Show that the map $f:β€ββ€$ defined by $f(n)=n^2$ is continuous in the 2-adic metric.
Proof.
Let $2^mβ£x-a$, then $2^mβ£x^2-a^2=(x-a)(x+a)$, so $d\big(f(x),f(a)\big)< d(x,a)$.
$βaββ€,Ξ΅>0$, then $βxβB(a,Ξ΅):d\big(f(x),f(a)\big)< d(x,a)< Ξ΅$. So $f$ is continuous at $a$. -
Let $X, Y$ and $Z$ be metric spaces and equip $Y Γ Z$ with the product metric defined in lectures. If $F: X β Y Γ Z$ 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 $Ξ΅>0$, we can find $Ξ΄>0$, such that $βaβB(x,Ξ΄):f_1(a)βB\left(f_1(x),\fracΞ΅{\sqrt2}\right)$ and $f_2(a)βB\left(f_2(x),\fracΞ΅{\sqrt2}\right)$.
So $βaβB(x,Ξ΄):d_{YΓZ}\Big(\big(f_1(a),f_2(a)\big),\big(f_1(x),f_2(x)\big)\Big)=\sqrt{d_Y\big(f_1(a),f_1(x)\big)^2+d_Z\big(f_2(a),f_2(x)\big)^2}<\sqrt2\fracΞ΅{\sqrt2}=Ξ΅$. Therefore $F$ is continuous.
If $F$ is continuous, for any $Ξ΅>0$, we can find $Ξ΄>0$, such that $βaβB(x,Ξ΄):Ξ΅>d_{YΓZ}\Big(\big(f_1(a),f_2(a)\big),\big(f_1(x),f_2(x)\big)\Big)=\sqrt{d_Y\big(f_1(a),f_1(x)\big)^2+d_Z\big(f_2(a),f_2(x)\big)^2}β₯d_Y\big(f_1(a),f_1(x)\big)$.
So $f_1$ is continuous. Similarly $f_2$ is continuous. -
Let $X$ denote the vector space of sequences $π±=\left(x_n\right)_{n=1}^β$ with $x_n β β$ and $\sum_{n=1}^β x_n^2<β$. Explain why ${βπ±β}_β:=\sup_n\left|x_n\right|$ is a well-defined norm on $X$. Is the metric induced by this norm equivalent to the metric induced by the $β^2$-norm?
Proof.
${βπ±β}_β=\sup_n\left|x_n\right|β€\sqrt{\sum_{n=1}^β x_n^2}={βπ±β}_2$, so it is finite. Clearly ${βπ±β}_β=0$ iff $π±=0$; ${βΞ»π±β}_β = {|Ξ»|}{βπ±β}_β$ for all $Ξ»ββ,π±βX$; ${βπ±+π²β}_β=\sup_n\left|x_n+y_n\right|β€\sup_n\left|x_n\right|+\sup_n\left|y_n\right|={βπ±β}_β+{βπ²β}_β$ for all $π±,π²β X$. So ${βπ±β}_β$ is a well-defined norm on $X$.
The metric induced by $β^β$-norm and $β^2$-norm are not equivalent. Let ${\bf 0}=(0,0,β¦)$. Suppose $B_2({\bf0},1)βB_β({\bf0},r)$ for some $r>0$.
Let $0< a< r$ and ${\bf a}_k=(\underbrace{a,a,β¦,a}_k,0,0,β¦)$. Then ${β{\bf a}_kβ}_β=aβ{\bf a}_kβB_β({\bf0},r)$ but ${β{\bf a}_kβ}_2=a\sqrt kβ{\bf a}_kβB_2({\bf0},1)$ for $k>a^{-2}$, contradicting to the assumption $B_2({\bf0},1)βB_β({\bf0},r)$.