Topology – Physicspupil

September 30, 2022October 24, 2022

202209300929 Theorem 4.1.3

Prove Theorem 4.1.3.

Let $(X,d)$ and $(Y,\rho )$ be metric spaces and $f:X\to Y$ be a function. Then, $f$ is continuous at a point $x_0$ if and only if $f(x_n)\to f(x_0)$ , for every sequence $\{ x_n\} \subset X$ with $x_n\to x_0$ .

_{Extracted from K. J. Pawan & A. Khalil. (2004). Metric Spaces.}

Background.

Define, by open spheres, continuity of a function $f$ at a point $x_0$ :

$\forall\, x,\exists\, x_0\in X, \forall\, \epsilon >0, \exists\,\delta >0,$ $d(x,x_0)<\delta\Rightarrow \rho (f(x),f(x_0))<\epsilon$

in other words,

$x\in S_\delta (x_0)\Rightarrow f(x)\in S_\epsilon (f(x_0))$ ,

or the same,

$f(S_\delta (x_0))\subset S_\epsilon (f(x_0))$ .

Proof.

$\begin{aligned} & f\textrm{ is continuous at }x_0 \\ \Leftrightarrow\enspace & f(S_\delta (x_0))\subset S_\epsilon (f(x_0))\textrm{ for some } \delta>0\textrm{ and any }\epsilon >0 \\ & \\ & \textrm{for every sequence }\{ x_n\}\subset X\textrm{ with }x_n\textrm{ tends to }x_0 \\ \Leftrightarrow\enspace & \textrm{there exists an }N>0\textrm{ such that }x_n\in S_{\delta}(x_0)\textrm{ for all }n>N \\ \end{aligned}$

$\begin{aligned} & x_n\in S_\delta (x_0) \Longrightarrow f(x_n)\in f(S_\delta (x_0))\subset S_\epsilon (f(x_0)) \\ \Leftrightarrow\enspace & f(x_n) \to f(x_0) \\ \end{aligned}$

October 23, 2020October 20, 2022

202010230206 Sidenote of Clopen

Observe that both $\emptyset$ and $X$ are open and closed in $X$ , i.e., clopen.

Proof. Pastime.

Remark. ( $S$ ‘s in several symbols )

$S$ is a set of points.
$S^{0}$ , the interior of set $S$ , contains all interior points.
$S'$ , the derived set of set $S$ , contains all accumulation/cluster/limit points.
$\bar{S}$ , the closure of set $S$ , contains all adherent points.
$\partial S$ (also denoted by $b(S)$ or $S^{b}$ ), the boundary of set $S$ , contains all boundary points.

October 23, 2020October 20, 2022

202010230028 Problem 2.4.11

Let $(X,d)$ be a metric space, $a\in X$ and $0<r<r'$ .

Prove that the set

$\{ x\in X:\enspace r<d(x,a)<r' \}$

is open in $(X,d)$ .

Setup.

Understand the definition given to each of the following:

First, what is meant by whether a set is open or not in some metric space?

Definition. (open set) Let $(X,d)$ be a metric space. A set $G\subset X$ is said to be an open set if it is a neighborhood of each of its points. (Equivalently, a set $G\subset X$ is said to be an open set
if for each $x\in G$ , there exists an $r>0$ such that $S_r(x)\subset G$ .)

_{Please refer to pg. 20, Jain and Ahmad’s Metric Spaces.}

Second, what is referred to as a neighborhood of some point(s)?

Definition. (neighborhood) Let $(X,d)$ be a metric space and $x\in X$ . A set $N\subset X$ is said to be a neighborhood (nbd) of $x$
if there exists an open sphere centred at $x$ and contained in $N$ ,
i.e., if $S_r(x)\subset N$ for some $r>0$ .

_{Please refer to pg. 19, Jain and Ahmad’s Metric Spaces.}

Third, what is an open sphere?

Definition. (open sphere) Let $(X,d)$ be a metric space. Let $x\in X$ and $r>0$ be a real number. The open sphere with centre $x$ and radius $r$ , denoted by $S_r(x)$ , the subset of $X$ given by $S_r(x)=\{ y\in X:\enspace d(x,y)<r \}$ N.b. An open sphere is always non-empty since it contains its centre at least.

_{Please refer to pg. 16, Jain and Ahmad’s Metric Spaces.}

October 2, 2020June 12, 2025

202010022249 Problem 2.1.10

In $C[0,1]$ , determine the values of $d_{\infty}(x,y)$ and $d_{1}(x,y)$ , when

(a) $x(t)=t^3+t+1$ and $y(t)=t^3+t^2+\frac{1}{2}t+1$ ;

(b) $x(t)=\sin t$ and $y(t)=t$ ;
(c) $x(t)=\sin t$ and $y(t)=t-\displaystyle{\frac{t^3}{6}}$ ;
(d) $x(t)=\mathrm{exp}(t)$ and $y(t)=\displaystyle{\sum_{m=0}^n}\frac{t^m}{m!}$ .

Recall.

Definition. (uniform metric) Let $C[a,b]$ be the set of all real-valued continuous functions defined on $[a,b]$ . For any $x,y\in C[a,b]$ , define the uniform metric $d_{\infty}$ :

$d_{\infty}(x,y)=\displaystyle{\max_{t\in [a,b]}}|x(t)-y(t)|$ .

N.b. If we let $B[a,b]$ be the set of all real-valued functions defined and bounded on $[a,b]$ , the uniform metric is then defined

$d_{\infty}(x,y)=\displaystyle{\sup_{t\in [a,b]}}|x(t)-y(t)|$ .

_{(cited from Examples 14 and 15, pg. 13, Pawan K. Jain and Khalil Ahmad’s Metric Spaces (2e) on Introductory Concepts)}

Definition. For any $x, y\in C[a,b]$ , define

$d_1(x,y)=\displaystyle{\int_a^b}|x(t)-y(t)|\,\mathrm{d}t$

N.b. $d_1(x,y)$ represents the absolute area between the functions $x$ and $y$ as a measure of the distance between these two functions.

_{(cited from Example 16, pg. 14, Pawan K. Jain and Khalil Ahmad’s Metric Spaces (2e) on Introductory Concepts)}

Solution.

(a)

$\begin{aligned} & \quad\, d_{\infty} (x,y) \\ &= \max_{t\in [0,1]} |x(t)-y(t)| \\ \end{aligned}$

Roughwork.

Approach.

To know the maximum value of $|x(t)-y(t)|$ , apply differentiation to

$f(t)=-t^2+\displaystyle{\frac{1}{2}}t$

and attain

$\begin{aligned} f(t) & = -t^2 + \frac{1}{2}t \\ f'(t) & = -2t+\frac{1}{2} \\ f'(t) & = 0 \Leftrightarrow t=\frac{1}{4} \\ \end{aligned}$

If the quadratic function $f(t)$ is plotted in a graph, a parabola admits of no inflexion points, needless to check on $f''(x)=0$ . So,

$\begin{array}{c|c|c|c|c|c} & t=0 & 0<t<\frac{1}{4} & t=\frac{1}{4} & \frac{1}{4}<t<1 & t=1 \\ &&&&&\\ \hline &&&&&\\ f(t) & 0 & \dots & \displaystyle{\frac{1}{16}} & \dots & -\displaystyle{\frac{1}{2}} \\ &&&&&\\ \hline &&&&&\\ f'(t) & \displaystyle{\frac{1}{2}}\enspace (>0) & \dots & 0\enspace (=0) & \dots & -\displaystyle{\frac{3}{2}}\enspace (<0) \\ &&&&&\\ \hline &&&&&\\ \textrm{plot} & \diagup & \dots & --- & \dots & \diagdown \\ &&&&&\\ \end{array}$

The continuous function $f(t)$ in the closed interval $[0,1]$ attains its maximum value $f(\frac{1}{4})=\frac{1}{16}$ when $t=\frac{1}{4}$ .

$\begin{aligned} & d_{\infty} (x,y) \\ = & \max_{t\in [0,1]} |x(t)-y(t)| \\ = & \max_{t\in [0,1]} |f(t)| \\ = & \frac{1}{16} \qquad\qquad\qquad \checkmark\\ \end{aligned}$

and should you think of what follows as quite right

$\begin{aligned} d_1(x,y) & = \int_0^1 |x(t)-y(t)| \,\mathrm{d}t \\ & = \int_0^1 f(t)\,\mathrm{d}t \\ & = \int_0^1 \bigg| -t^2 + \frac{1}{2}t \bigg| \, \mathrm{d}t \\ & = \bigg[ -\frac{t^3}{3} + \frac{t^2}{4} \bigg] \bigg|_0^1 \\ & = -\frac{1}{12}\qquad\qquad\qquad \times \\ \end{aligned}$

you might have rather mistaken calculus.

Correction.

Get back to the basics,

$\begin{aligned} f(t) & = 0 \\ \bigg| -t^2+\frac{1}{2}t \bigg| & = 0 \\ t^2-\frac{1}{2}t &= 0 \\ (t-\frac{1}{2})t & = 0 \\ t & = 0\quad \textrm{\scriptsize{OR}}\quad \frac{1}{2} \\ \end{aligned}$

From the previous graph of $C[0,1]$ , $f(t)$ is found to be positive when $t\in (0,0.5)$ , zero when $t\in \{ 0\} \cup\{ 0.5\}$ , and negative when $t\in (0.5,1]$ .

Doing it step-by-step,

$\begin{aligned} d_1(x,y) & = \int_0^1 \bigg| -t^2+\frac{1}{2}t \bigg| \,\mathrm{d}t \\ & = \int_{0}^{0.5}\mathrm{d}t\enspace \bigg| -t^2+\frac{1}{2}t \bigg| + \int_{0.5}^{1}\mathrm{d}t \enspace \bigg| -t^2+\frac{1}{2}t \bigg| \\ & = \int_{0}^{0.5}\mathrm{d}t\enspace \bigg( - t^2 + \frac{1}{2}t \bigg) + \int_{0.5}^{1}\mathrm{d}t \enspace \bigg( t^2-\frac{1}{2}t\bigg) \\ \end{aligned}$

Evaluating term-by-term, the first term being

$\begin{aligned} & \int_{0}^{0.5} \bigg( -t^2 + \frac{1}{2}t \bigg) \,\mathrm{d}t \\ = & \bigg[ -\frac{t^3}{3} + \frac{t^2}{4} \bigg] \bigg|_{0}^{0.5} \\ = & \bigg[ -\frac{(0.5)^3}{3} + \frac{(0.5)^2}{4} \bigg] - \bigg[ -\frac{(0)^3}{3} + \frac{(0)^2}{4} \bigg] \\ \dots & \enspace \textrm{by arithmetic}\enspace \dots \\ = & \frac{1}{48} \\ \end{aligned}$

and the second term being

$\begin{aligned} & \int_{0.5}^{1}\bigg( t^2-\frac{1}{2}t\bigg) \,\mathrm{d}t \\ = & \bigg[ \frac{t^3}{3} - \frac{t^2}{4} \bigg]\bigg|_{0.5}^{1} \\ = & \bigg[ \frac{(1)^3}{3} - \frac{(1)^2}{4} \bigg] - \bigg[ \frac{(0.5)^3}{3} - \frac{(0.5)^2}{4} \bigg] \\ \dots & \enspace \textrm{by arithmetic}\enspace \dots \\ = & \bigg( \frac{1}{12} \bigg) - \bigg( -\frac{1}{48} \bigg) \\ = & \frac{5}{48} \\ \end{aligned}$

In sum,

$d_{1}(x,y)=\displaystyle{\frac{1}{48}+\frac{5}{48}=\frac{6}{48}=\frac{1}{8}}=0.125$ .

Part (b), (c), and (d) are not chosen.

October 2, 2020October 24, 2022

202010020718 Problem 2.1.2

Let $(X,d)$ be a metric space and let $k$ be a fixed positive real number. For $x,\, y\in X$ , define

$d^{*}=kd(x,y)$ .

Prove that $d^{*}$ is a metric on $X$ .

Recall.

Definition. (metric) Let $X$ be a non-empty set. A metric on $X$ is a real-valued function $d:\enspace X\times X\rightarrow \mathbb{R}$ satisfying the following conditions i– iv:

i. $d(x,y)\ge 0$ ;
ii. $d(x,y)=0\Leftrightarrow x=y$ ;
iii. (Symmetry) $d(x,y)=d(y,x)$ ;
iv. (Triangle Inequality) $d(x,y)\le d(x,z)+d(z,y)$ for any $x,\, y,\, z\in X$ .

N.b. Given $x,\,y\in X$ , $d(x,y)$ is sometimes called the distance between $x$ and $y$ with respect to $d$ .

Proof.

i.

WTS (wish to show)

$d^{*}(x,y)\ge 0$

By definition $d^{*}(x,y)=kd(x,y)$ and in that the metric $d$ is let clear ( $\therefore d(x,y)\ge 0$ ) and $k$ a fixed positive real number ( $\therefore k>0$ ),

one can see

$\begin{aligned} d^{*}(x,y) & = kd(x,y) \\ \textrm{\dots because\enspace} & k>0 \enspace \textrm{and}\enspace d(x,y)\ge 0\textrm{\enspace \dots}\\ d^{*}(x,y) & \geqslant 0 \\ \end{aligned}$

$\therefore$ Condition i. is made.

ii.

$\begin{aligned} d^{*}(x,y) & = 0 \\ \Leftrightarrow kd(x,y) & = 0 \\ \dots\enspace \textrm{as}\enspace k>0 \enspace & \textrm{so}\enspace k\neq 0\enspace \dots \\ \Leftrightarrow d(x,y) & = 0 \\ \dots\enspace \textrm{as}\enspace d \enspace \textrm{was} &\enspace\textrm{foretold to be a metric}\enspace \dots \\ \Leftrightarrow x & = y \\ \end{aligned}$

$\therefore$ Condition ii. is made.

iii.

NTS (need to show)

$d^{*}(x,y)=d^{*}(y,x)$

$\begin{aligned} \textrm{LHS} & = d^{*}(x,y) \\ & = kd(x,y) \\ \dots \enspace & \textrm{by the symmetric property of }d\enspace \dots \\ & =kd(y,x) \\ & = d^{*}(y,x) \\ & = \textrm{RHS} \\ \end{aligned}$

$\therefore$ Condition iii. is made.

iv.

RTP (required to prove)

$d^{*}(x,y)\leqslant d^{*}(x,z)+d^{*}(z,y)$

One starts with the left hand side,

$\begin{aligned} \textrm{LHS} & = d^{*}(x,y) \\ & = kd(x,y) \\ \dots \textrm{as does}\enspace & d(x,y)\le d(x,z)+d(z,y)\enspace\textrm{the metric}\enspace d\enspace \textrm{do} \dots \\ & \le k\Big( d(x,z) + d(z,y) \Big) \\ & = kd(x,z) + kd(z,y) \\ & = d^{*}(x,z) + d^{*}(z,y) \\ & = \textrm{RHS} \\ \end{aligned}$

Condition iv. is made.

In conclusion, $(X,d^{*})$ is a metric space metered by a well-defined metric $d^{*}$ . This metric space shall simply be called $X$ hence.

September 24, 2020October 24, 2022

202009240421 Problem 2.1.1

For $x,\, y\in \mathbb{K}$ ( $\mathbb{R}$ or $\mathbb{C}$ ), define

$d(x,y)=\mathrm{min}\big\{ 1,\, |x-y| \big\}$ .

Prove that $d$ is a metric on $\mathbb{K}$ .

Motivation.

A ruler is marked by rules for the sake of measuring things. Not so much common to a ruler, but well worth the rule for the general, that a metric must measure its metric space, with the metric defined below:

Definition. (metric) Let $X$ be a non-empty set. A function $d: X\times X \rightarrow \mathbb{R}$ is said to be a metric on $X$ if it satisfies the following conditions:

i. $d(x,y)\ge 0\qquad \forall\, x,\, y\in X$ ;
ii. $d(x,y)=0 \Leftrightarrow x=y\qquad x,\, y\in X$ ;
iii. $d(x,y)=d(y,x) \qquad \forall\, x,\, y\in X$ ;
iv. $d(x,y)\le d(x,z)+d(z,y) \qquad \forall\, x,\, y,\, z\in X$ .

Remark.

i. As is known, distance should be either positive or zero (i.e., non-negative); ii. We are in one only by discrimination; iii. Fair and just from a symmetric point of view; and iv. The straighter the path, the shorter the distance (also known as the Triangle Inequality).

Proof.

Assume $x\, ,y\in\mathbb{C}$ for convenience. (Provided $\mathbb{R}\subset \mathbb{C}$ , the assumption is ready for reduction or extension.)

i.

If the minimum $\mathrm{min}\big\{ 1, |x-y|\big\} = 1$ , condition (i) $d(x,y)=1 \ge 0$ is seen. If the minimum $\mathrm{min}\big\{ 1,|x-y|\big\} = |x-y|$ , be it called the absolute value, the magnitude, the norm, the modulus, or whatsoever, a complex number is non-negative in norm.

Recall. (Norm of complex conjugate)

A complex number $z=x+\textrm{i}\, y$ contains two parts, the real part $x$ and the imaginary part $y$ . The norm $|z|$ (and the norm $|\bar{z}|$ of its complex conjugate $\bar{z}=x-\textrm{i}\, y$ )
is defined by the formula:

$|z|=|x+\textrm{i}\, y|=\sqrt{\big[\textrm{Re}(z)\big]^2+\big[\textrm{Im}(z)\big]^2}=\sqrt{x^2+y^2}$ .

$\dagger$ It turns out that $|z|=\sqrt{(x)^2+(y)^2}=\sqrt{(x)^2+(-y)^2}=|\bar{z}|$ (where $x,\, y\in\mathbb{R}$ ) is positive iff $x\enspace\textrm{\scriptsize OR}\enspace y\neq 0$ , and zero iff $x=y=0$ , but never negative.

$\ddagger$ The sum, the difference, the product, and the quotient of two complex numbers is one another complex number for the complex number field is algebraically closed.

ii.

(only-if) Giving a try straightforth:

$\begin{aligned} d(x,y) & = 0 \\ \textrm{min}\big\{ 1, |x-y| \big\} & = 0 \\ |x-y| & = 0\qquad \textrm{\scriptsize OR\qquad\quad } 1 = 0 \qquad \textrm{(rejected for\enspace }1\neq 0) \\ |x-y| & = 0 \\ x-y & = 0 \\ x & = y \end{aligned}$

(if) In reverse from backward:

$\begin{aligned} x & = y \\ x-y & = 0 \\ |x-y| & = 0 \\ \textrm{min}\big\{ 1, |x-y| \big\} & = 0 \\ d(x,y) & = 0 \end{aligned}$

If provided with appropriate explanation, the proof can be shortened by use of the two-way if-and-only-if.

iii.

Suffice it to check whether $d(x,y)=d(y,x)$ is true or not.

$\begin{aligned} \textrm{LHS} & = d(x,y) \\ & = \textrm{min}\big\{ 1,\, |x-y| \big\} \\ & = \textrm{min}\big\{ 1,\, |y-x| \big\} \\ & = d(y,x) \\ & = \textrm{RHS} \end{aligned}$

Roughwork.

$\forall\, x,\,y \in\mathbb{C}$ , let $x=a+\textrm{i}\, b$ and $y=c+\textrm{i}\, d$ where $a$ , $b$ , $c$ , and $d$ are real numbers. Then,
$\begin{aligned} x-y & = (a+\textrm{i}\, b)-(c+\textrm{i}\, d) \\ x-y & = (a-c) + \textrm{i}\, (b-d) \\ |x-y| & = \sqrt{(a-c)^2+(b-d)^2} \\ |x-y| & = \sqrt{(c-a)^2+(d-b)^2} \\ |x-y| & = |(c-a) + \textrm{i}\, (d-b)| \\ |x-y| & = |(c+\textrm{i}\, d) - (a+\textrm{i}\, b)| \\ |x-y| & = |y-x| \\ \end{aligned}$

iv.

Given here are some equations, I write out all them lest I might forget any:

$d(x,y)=\textrm{min}\big\{ 1, |x-y| \big\}$ ,
$d(x,z)=\textrm{min}\big\{ 1, |x-z| \big\}$ ,
$d(z,y)=\textrm{min}\big\{ 1, |z-y| \big\}$ .

RTP (i.e. required to prove):

$d(x,y)\le d(x,z) + d(z,y)$

$\begin{aligned} \textrm{RHS} & = d(x,z) + d(z,y) \\ & = \textrm{min}\big\{ 1,|x-z|\big\} + \textrm{min}\big\{ 1,|z-y|\big\} \\ & = \textrm{min}\big\{ 2,\, 1+|z-y|,\, 1+|x-z|,\, |x-z|+|z-y|\big\} \\ \end{aligned}$

If $d(x,y)=\textrm{min}\big\{ 1, |x-y|\big\} \stackrel{?}{=}1$ , so what have I done with?

WTS (i.e. wish to show):

$1\le \textrm{min}\big\{ 2, 1+|z-y|, 1+|x-z|, |x-z|+|z-y|\big\}$

The following inequalities hold evidently:

$\begin{aligned} 1 & \le 2 \\ 1 & \le 1+|z-y| \\ 1 & \le 1+|x-z| \\ \end{aligned}$

But $1\le |x-z|+|z-y|$ has not yet been ascertained.

The Argand diagram above replaces the usual $x$ – and $y$ -axes of the Cartesian plane with the real and the imaginary axes of Argand plane.

Owing to my giving too raw and rude maybe a proof, the problem should have otherwise been treated case-by-case. I.e., they are in either case:

$1\le |x-y| \qquad \qquad \textrm{\scriptsize OR}\qquad\qquad 1> |x-y|$

Just do it by rote:

$\begin{aligned} & \quad\, d(x,y) \\ & =\textrm{min}\big\{ 1, |x-y| \big\} \\ & =\textrm{min}\big\{ 1, |x-z+z-y| \big\} \\ \dots \, & \textrm{by the Triangle Inequality}\, \dots \\ & \leqslant \textrm{min} \big\{ 1, |x-z| + |z-y| \big\} \\ & \leqslant \textrm{min} \big\{ 1, |x-z| \big\} + \textrm{min} \big\{ 1, |z-y| \big\} \\ & \leqslant d(x,z) + d(z,y) \\ \end{aligned}$

$d(x,y) \le d(x,z) + d(z,y)$

QED