# Functional Analysis 5: Completion of Metric Spaces

This lecture will conclude our discussion on metric spaces with completion of metric spaces.

Definition. Let $X=(X,d)$ and $\tilde X=(\tilde X,d)$ be two metric space. A mapping $T: X\longrightarrow\tilde X$ is said to be an isometry of $T$ preserves distances, i.e.
$$\forall x,y\in X,\ \tilde d(Tx,Ty)=d(x,y).$$ The space $X$ is said to be isometric with $\tilde X$ if there exists a bijective isometry of $X$ onto $\tilde X$.

Theorem [Completion] For a metric space $(X,d)$ there exists a complete metric space $\hat X=(\hat X,\hat d)$ which has s subspace $W$ that is isometric with $X$ and is dense in $\hat X$. $\hat X$ is called the completion of $X$ and it is unique up to isometries.

Proof. This will be a lengthy proof and I have divided it into steps.

Step 1. Construction of $\hat X=(\hat X,\hat d)$.

Let $(x_n)$ and $(y_n)$ be Cauchy sequences in $X$. We say $(x_n)$ is equivalent to $(x_n’)$, and write $(x_n)\sim (x_n’)$, if
$$\lim_{n\to\infty}d(x_n,x_n’)=0.$$ $\sim$ is actually an equivalence relation on the set of all Cauchy sequences of $X$. Clearly $\sim$ is reflexive and symmetric. Let us show that $\sim$ is also transitive. Let $(x_n)\sim (x_n’)$ and $(x_n’)\sim (x_n^{\prime\prime})$. Then
$$\lim_{n\to\infty}d(x_nx_n’)=0\ \mbox{and}\ \lim_{n\to\infty}d(x_n’,x_n^{\prime\prime})=0.$$
\begin{align*}
\lim_{n\to\infty}d(x_n,x_n^{\prime\prime})&\leq\lim_{n\to\infty}d(x_n,x_n’)+\lim_{n\to\infty}d(x_n’,x_n^{\prime\prime})\\
&=0
\end{align*}
Let $\hat X$ be the set of all equivalence classes $\hat x,\hat y,\cdots$ of Cauchy sequences. Define
$$\hat d(\hat x,\hat y)=\lim_{n\to\infty}d(x_n,y_n),$$
where $x_n\in\hat x$ and $y_n\in\hat y$. We claim that $\hat d$ is a metric on $\hat X$. It suffices to show that $\hat d$ is well-defined. The conditions (M1)-(M3) hold due to the fact that $d$ is a metric. First we show that $\hat d(\hat x,\hat y)$ exists. It follows from (M3) that
$$d(x_n,y_n)\leq d(x_n,x_m)+d(x_m,y_m)+d(y_m,y_n)$$ and so we have
$$d(x_n,y_n)-d(x_m,y_m)\leq d(x_n,x_m)+d(y_m,y_n).$$ Similarly, we obtain
$$d(x_m,y_m)-d(x_n,y_n)\leq d(x_n,x_m)+d(y_m,y_n).$$ Hence,
$$|d(x_n,y_n)-d(x_m,y_m)|\leq d(x_n,x_m)+d(y_m,y_n)\rightarrow 0$$ as $n,m\rightarrow\infty$, i.e.
$$\lim_{n,m\to\infty}|d(x_n,y_n)-d(x_m,y_m)|=0.$$
Since $\mathbb{R}$ is complete, $\displaystyle\lim_{n\to\infty}d(x_n,y_n)$ exists. Now we show that the limit is independent of the choice of representatives $(x_n)$ and $(y_n)$. If $(x_n)\sim (x_n’)$ and $(y_n)\sim(y_n’)$, then
$$|d(x_n,y_n)-d(x_n’,y_n’)|\leq d(x_n,x_m)+d(y_m,y_n)\rightarrow 0$$ as $n\to\infty$.

Step 2. Construction of an isometry $T: X\longrightarrow W\subset \hat X$.

For each $b\in X$, let $\hat b$ be the equivalence class of the Cauchy sequence $(b,b,b,\cdots)$. Then $T(b):=\hat b\in\hat X$. Now,
$$\hat d(Tb,Tc)=\hat d(\hat b,\hat c)=d(b,c).$$ So, $T$ is an isometry. An isometry is automatically injective. $T$ is onto since since $T(X)=W$. Let us show that $W$ is dense in $\hat X$. Let $\hat x\in \hat X$ and let $(x_n)\in\hat x$. Since $(x_n)$ is Cauchy, given $\epsilon>0$ $\exists N$ such that $d(x_n,x_N)<\frac{\epsilon}{2}$ $\forall n\geq N$. Let $(x_N,x_N,\cdots)\in\hat x_N$. Then $\hat x_N\in W$.
\begin{align*}
\hat d(\hat x,\hat x_N)=\lim_{n\to\infty}d(x_n,x_N)\leq\frac{\epsilon}{2}<\epsilon&\Longrightarrow \hat x_N\in B(\hat x,\epsilon)\\
&\Longrightarrow B(\hat x,\epsilon)\cap W\ne\emptyset.
\end{align*}
Hence, $\bar W=\hat X$ i.e. $W$ is dense in $\hat X$.

Step 3. Completeness of $\hat X$.

Let $(\hat x_n)$ be any Cauchy sequence in $\hat X$. Since $W$ is dense in $\hat X$, $\forall \hat x_n$, $\exists\hat z_n\in W$ such that $\hat d(\hat x_n,\hat z_n)<\frac{1}{n}$.
\begin{align*}
\hat d(\hat z_m,\hat z_n)&\leq \hat d(\hat z_n,\hat x_m)+\hat d(\hat x_m,\hat x_n)+\hat d(\hat x_n,\hat z_n)\\
&<\frac{1}{m}+\hat d(\hat x_m,\hat x_n)+\frac{1}{n}.
\end{align*}
Given $\epsilon>0$ by Archimedean property $\exists$ a positive integer $N_1$ such that $N>\frac{\epsilon}{3}$. Since $(\hat x_n)$ is a Cauchy sequence, $\exists$ a positive integer $N_2$ such that $\hat d(\hat x_m,\hat x_n)<\frac{\epsilon}{3}$ $\forall m,n\geq N$. Let $N=\max\{N_1,N_2\}$. Then $\forall m,n\geq N$, $\hat d(\hat z_m,\hat z_n)<\epsilon$ i.e. $(\hat z_m)$ is Cauchy. Since $T: X\longrightarrow W$ is an isometry and $\hat z_m\in W$, the sequence $(z_m)$, where $z_m=T^{-1}\hat z_m$, is Cauchy in $X$. Let $\hat x\in\hat X$ be the class to which $(z_m)$ belongs. Show that $\hat x$ is the limit of $(\hat x_n)$. For each $n=1,2,\cdots$,
\begin{align*}
\hat d(\hat x_n,\hat x)&\leq \hat d(\hat x_n,\hat z_n)+\hat d(\hat z_n,\hat x)\\
&<\frac{1}{n}+\hat d(\hat z_n,\hat x)\\
&=\frac{1}{n}+\lim_{m\to\infty}d(z_n,z_m)
\end{align*} since $(z_m)\in\hat x$ and $(z_n,z_n,\cdots)\in\hat z_n\in W$. This implies that $\displaystyle\lim_{n\to\infty}\hat d(\hat x_n,\hat x)=0$ i.e. the Cauchy sequence $(\hat x_n)$ in $\hat X$ has the limit $\hat x\in\hat X$. Therefore, $\hat X$ is complete.

Step 4. Uniqueness of $\hat X$ up to isometries.

Suppose that $(\tilde X,\tilde d)$ is another completion of $X$ i.e. it is a complete metric space with a subspace $\tilde W$ dense in $\tilde X$ and isometric with $X$. We show that $\hat X$ is isometric with $\tilde X$. Let $X$ is isometric with $W$ and $\tilde W$ via isometries $T$ and $\tilde T$, respectively. Then $W$ is isometric with $\tilde W$ via the isometry $\rho=\tilde T\circ T^{-1}$.

$$\begin{array}{ccc} & & W\\ & \nearrow &\downarrow\\ X & \longrightarrow & \tilde{W} \end{array}$$

Let $\hat x\in\hat X$. Then $\exists$ a sequence in $(\hat x_n)\in W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$. $(\hat x_n)$ is a Cauchy sequence and $\rho$ is an isometry, so $(\tilde x_n)$, where $\tilde x_n:=\rho\hat x_n$, is a Cauchy sequence in $\tilde W\subset \tilde X$. Since $\tilde X$ is complete, $\exists\tilde x\in\tilde X$ such that $\displaystyle\lim_{n\to\infty}\tilde x_n=\tilde x$. Define a mapping $\psi:\hat X\longrightarrow\tilde X$ by $\psi\hat x=\tilde x$. Then we claim that $\hat X$ is isometric with $\tilde X$ via $\psi$.

Step A. $\psi$ is well-defined.

It suffices to show that $T\hat x$ does not depend on the choice of $(\hat x_n)\in W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$. Let $(\hat x_n’)$ be another sequence in $W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n’=\hat x$. Then $(\tilde x_n’)$, where $\tilde x_n’=\rho\hat x_n’$, is a Cauchy sequence in $\tilde W$ and so $\exists\tilde x’\in\tilde X$ such that $\displaystyle\lim_{n\to\infty}\tilde x_n’=\tilde x’$. Now,
\begin{align*}
\tilde d(\tilde x,\tilde x’)&=\lim_{n\to\infty}\tilde d(\tilde x_n,\tilde x_n’)\\
&=\lim_{n\to\infty}\hat d(\hat x_n,\hat x_n’)\ (\rho\ \mbox{is an isometry})\\
&=\hat d(\hat x,\hat x)\\
&=0.
\end{align*}
Hence, $\tilde x=\tilde x’$.

Step B. $\psi$ is onto.

Let $\tilde x\in\tilde X$. Then $\exists$ a sequence $(\tilde x_n)$ in $\tilde W$ such that $\displaystyle\lim_{n\to\infty}\tilde x_n=\tilde x$. $(\tilde x_n)$ is Cauchy (since it is a convergent sequence) and $\rho^{-1}$ is an isometry, so the sequence $(\hat x_n)\subset \hat X$, where $\hat x_n=\rho^{-1}\tilde x_n$, is Cauchy. Since $\hat X$ is complete, $\exists\hat x\in\hat X$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$. Clearly $\psi\hat x=\tilde x$ and hence $\psi$ is onto.

Step C. $\psi$ is an isometry.

Let $\hat x,\hat y\in\hat X$. Then $\exists$ sequences $(\hat x_n)$, $(\hat y_n)$ in $W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$ and $\displaystyle\lim_{n\to\infty}\hat y_n=\hat y$, respectively.
\begin{align*}
\hat d(\hat x,\hat y)&=\lim_{n\to\infty}\hat d(\hat x_n,\hat y_n)\\
&=\lim_{n\to\infty}\tilde d(\tilde x_n,\tilde y_n)\ (\tilde x_n:=\rho\hat x_n,\ \tilde y_n:=\rho y_n)\\
&=\tilde d(\tilde x,\tilde y)\ (\lim_{n\to\infty}\tilde x_n=\tilde x,\ \lim_{n\to\infty}\tilde y_n=\tilde y)\\
&=\tilde d(\psi\hat x,\psi\hat y).
\end{align*}
Thus, $\psi$ is an isometry.

Remember that an isometry from a metric space into another metric space is automatically one-to-one. Therefore, $\hat X$ is isometric with $\tilde X$ via $\psi$.

Intuitively speaking, the completion of a metric space $X$ can be achieved by adding to $X$ all its limit points. Recall that if $x$ is a limit point of $X$, then there exists a sequence $(x_n)$ in $X$ such that $\displaystyle\lim_{n\to\infty}x_n=x$. This is a reminiscence of extending from rational numbers $\mathbb{Q}$ to real numbers $\mathbb{R}$ (which is complete) by adding to $\mathbb{Q}$ all its limit points (irrational numbers).

# Group Theory 6: Lagrange’s Theorem

In this lecture, we study Lagrange’s Theorem which is named after the Italian/French mathematician, physicists and astronomer Joseph-Louis Lagrange. It states that the order of a subgroup $H$ of a finite group $G$ divides the order of $G$. Lagrange’s Theorem has many important applications in group theory. Before we discuss the theorem we first need to study an important class of binary relations called equivalence relations.

Definition. A binary relation $R$ on a nonempty set $S$ is a subset $R\subset S\times S$. If $(a,b)\in R$, we say $a$ is $R$-related to $b$ and write $aRb$. A binary relation $R$ on a set $S$ is called an equivalence relation on $S$ if it satisfies:

1. $aRa$ $\forall a\in S$. ($R$ is reflexive.)

2. $\forall a,b,\in S$, $aRb\Longrightarrow bRa$. ($R$ is symmetric.)

3. $\forall a,b,c\in S$, $aRb$ and $bRc$ $\Longrightarrow$ $aRc$. ($R$ is transitive.)

Examples. 1. Define a binary relation $\equiv\mod n$ on $\mathbb{Z}$, the set of integers by
$$\forall a,b\in\mathbb{Z},\ a\equiv b\mod n\ \mbox{if}\ n|(a-b).$$ Then $\equiv\mod n$ is an equivalence relation in $\mathbb{Z}$, called the congruence relation modulo $n$. If $a\equiv b\mod n$, we say $a$ is congruent to $b$ modulo $n$. Note that $n|(a-b)$ if and only if $a-b=nk$ for some $k\in\mathbb{Z}$ if and only if $a$ and $b$ have the same remainder when they are divided by $n$ if and only if $a-b\in n\mathbb{Z}$ where
$$n\mathbb{Z}=\{nk: k\in\mathbb{Z}\},$$ the set of all integer multiples of $n$. $\forall n\in\mathbb{Z}$, $n\mathbb{Z}$ is a subgroup of the additive group $(\mathbb{Z},+)$.

2. Let $G$ be a group and $H\leq G$. Define a binary relation $R$ on $G$ by
$$\forall a,b\in G,\ aRb\ \mbox{if}\ ab^{-1}\in H.$$ Then $R$ is an equivalence relation on $G$.

Definition. If $R$ is an equivalence relation on a set $S$, then $[a]$, the equivalence class of $a\in S$, is defined by
$$[a]=\{b\in S: bRa\}.$$

Theorem. Let $R$ be an equivalence relation on a set $S$. Then

1. $S=\bigcup_{a\in S}[a]$.

2. $\forall a,b\in S$, either $[a]=[b]$ or $[a]\cap [b]=\emptyset$.

This means that an equivalence relation $R$ on a set $S$ partitions $S$ into equivalence classes. The set of all equivalence classes $\{[a]: a\in S\}$ is denoted by $S/R$ and is called the quotient set of $S$ modulo $R$.

The converse of the above theorem is also true, namely

Theorem. Let $\wp=\{A_i:i\in I\}$ be a partition of a set $S$ i.e. $S=\bigcup_{i\in I}A_i$ and $\forall i\ne j\in I$, $A_i\cap A_j=\emptyset$. Define a binary relation $R$ on $S$ by
$$\forall a,b\in S,\ aRb\ \mbox{if}\ a,b\in A_i\ \mbox{for some}\ i\in I.$$ Then $R$ is an equivalence relation on $S$, called the equivalence relation induced by the partition $\wp$.

Example. Let $f: X\longrightarrow Y$ be a surjective map. Then $\{f^{-1}(y): y\in Y\}$ forms a partition of $X$. The equivalence relation induced by this partition is
$$\ker f=\{(a,b)\in X\times X: f(a)=f(b)\}.$$ This equivalence relation is called the kernel of the function $f$.

Let $G$ be a group and $H\leq G$. Consider the equivalence relation $R$ on $G$ given by
$$\forall a,b\in G,\ aRb\ \mbox{if}\ ab^{-1}\in H.$$
Then for each $a\in G$, $[a]=Ha$ where $Ha=\{ha:h\in H\}$. $Ha$ is called a right coset of $H$ in $G$.

Example. Consider $\equiv\mod n$, the congruence relation modulo $n$ on $\mathbb{Z}$. For each $a\in \mathbb{Z}$, $[a]=n\mathbb{Z}+a$. The right coset $n\mathbb{Z}+a$ is called the congruence class of $a$ modulo $n$.

Now we are ready to discuss Lagrange’s Theorem.

Theorem [Lagrange's Theorem] If $G$ is a finite group and $H\leq G$, then $|H|||G|$.

Proof. $G=Ha_1\cup Ha_2\cup\cdots\cup Ha_k$ and $Ha_i\cup Ha_j=\emptyset$ if $i\ne j$, $i,j=1,2,\dots, k$. Due to the cancellation law, we see that $|H|=|Ha_i|$ for all $i=1,2,\cdots, k$. Therefore,
\begin{align*}
|G|&=\sum_{i=1}^k|Ha_i|\\
&=\sum_{i=1}^k|H|\\
&=k|H|.
\end{align*}
This completes the proof.

If $G$ is finite, the number of right cosets of $H$ in $G$, namely $\frac{|G|}{|H|}$ is called the index of $H$ in $G$ and is written as $|G:H|$ or $i_G(H)$. Here, we use the notation $|G:H|$.

It should be noted that the converse of Lagrange’s Theorem need not be true. For example, the alternating group of index 4
$$A_4=\{1,(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}$$ has no subgroup of order 6 though $|A_4|=12$ and $6|12$.

Theorem. A group $G$ of prime order is cyclic.

Proof. Let $|G|=p$, a prime. Let $H\leq G$. Then by Lagrange’s Theorem, $|H|||G|$ and so either $|H|=1$ or $|H|=p$. If $H\ne\{e\}$, then $H=G$. If $a\ne e\in G$, $\langle a\rangle=G$.

Let $G$ be a finite group and $a\ne e\in G$. Then $H=\langle a\rangle\leq G$. So, $a^m=e$ for some $m\in\mathbb{N}$.

Definition. If $G$ is finite, then the order of $a\ne e\in G$, written $|a|$ or $o(a)$, is the least positive integer $n$ such that $a^n=e$.

Corollary. If $G$ is finite and $a\ne e\in G$, then $|\langle a\rangle|||G|$.

Theorem. If $G$ is a finite group of order $n$, then $a^n=e$ for all $a\in G$.

Proof. Let $a\ne e\in G$ and $|\langle a\rangle|=k$. Then by Lagrange’s Theorem $k|n$ i.e. $n=kl$ for some $l\in\mathbb{N}$. So,
$$a^n=a^{kl}=(a^k)^l=e^l=e.$$

# Group Theory 5: Subgroups

In this lecture, we study subgroups. A nonempty subset $H$ of a group $G$ is called a subgroup of $G$ if $H$ itself forms a group relative to the operation $\cdot$ in $G$. If $H$ is a subgroup of $G$, we simply write $H\leq G$. Clearly $G\leq G$ and $\{e\}\leq G$. $G$ and $\{e\}$ are called trivial subgroups of $G$. Before we discuss subgroups further we need to go over some basic properties of a group as we need them.

Lemma. If $G$ is a group, then:

(a) Its identity element is unique.

(b) Each $a\in G$ has a unique inverse $a^{-1}\in G$.

(c) $\forall a\in G$, $(a^{-1})^{-1}=a$.

(d) $\forall a,b\in G$, $(ab)^{-1}=b^{-1}a^{-1}$.

The proof of this lemma is pretty much straightforward and is left to readers.

In order to show that a nonempty subset $H$ of a group $G$ is a subgroup we need to check:

1. $H$ is closed under $\cdot$, the operation in $G$. That is, $\forall a,b\in H$, $ab\in H$.

2. $e\in H$.

3. $\forall a\in H$, $a^{-1}\in H$.

Note that the associative law holds automatically. It turns out that there is a simpler criterion to check if a nonempty subset is a subgroup of a group.

Lemma. A nonempty subset $H\subset G$ is a subgroup of $G$ if and only if $\forall a,b\in H$, $ab^{-1}\in H$.

Proof. ($\Longrightarrow$) Clear.

($\Longleftarrow$) Let $a,b\in H$. Then by assumption, $aa^{-1}=e\in H$. Since $a,e\in H$, $ea^{-1}=a^{-1}\in H$. Since $b^{-1}\in H$, $a(b^{-1})^{-1}=ab\in H$. Therefore, $H$ is a subgroup of $G$.

Example. $\forall n\in\mathbb{N}\cup\{0\}$, $n\mathbb{Z}\leq(\mathbb{Z},+)$. Here, $n\mathbb{Z}=\{nx: x\in\mathbb{Z}\}$. In fact, they are all the subgroups of $(\mathbb{Z},+)$.

Definition. The cyclic subgroup of $G$ generated by $a\in G$ is $\{a^k: k\in\mathbb{Z}\}$. It is denoted by $\langle a\rangle$.

Example. Let $G$ be a group and $a\in G$. Let $C(a)=\{g\in G: ga=ag\}$. Then $C(a)\leq G$. $C(a)$ is called the centralizer of $a$ in $G$.

Example. Let $G$ be a group and let $Z(G)=\{z\in G: zx=xz\ \forall x\in G\}$. Then $Z(G)$ is a subgroup of $G$, called the center of $G$.

Example. Let $G$ be a group and $H\leq G$. $\forall a\in G$, $a^{-1}Ha=\{a^{-1}ha: h\in H\}$ is also a subgroup of $G$.

Lemma. Suppose that $(G,\cdot)$ is a group and $H$ is a finite nonempty subset of $H$. If $H$ is closed under $\cdot$, then $H$ is a subgroup of $G$.

Proof. Let $H=\{a_1,a_2,\cdots,a_n\}$. Then $\forall a\in H$, $aH=\{aa_1,aa_2,\cdots,aa_n\}\subset H$. On the other hand, if $aa_i=aa_j$ then $a_i=a_j$. So, $|aH|=|H|$. This implies that $aH=H$ and hence
\begin{align*}
a\in aH &\Longrightarrow a=aa_i\ \mbox{for some}\ i\\
&\Longrightarrow a_i=e\in H.
\end{align*}
Since $e\in aH$, $e=aa_j$ for some $j$ $\Longrightarrow$ $a_j=a^{-1}$.

Example. The symmetric group $S_3=\{1,(123),(132),(23),(13),(12)\}$ has 5 subgroups $\{1\}$, $\{1,(12)\}$, $\{1,(13)\}$, $\{1,(23)\}$, and $A_3=\{1,(123),(132)\}$. $A_3$ is called alternating group of degree 3. Cycles of length 2 such as (12), (13), (23) are called transpositions. Any permutation can be written as a product of either an even number of transpositions or an odd number of transpositions. If a permutation is a product of an even number of transpositions, the permutation is called an even permutation. If a permutation is a product of an odd number of transpositions, the permutation is called an odd permutation. The identity permutation 1 is an even permutation as $\begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix}$ can be written as (12)(21)(23)(32)(31)(13). The permutation (123) can be written as (12)(13). So, (123) is an even permutation. Similarly (132) is an even permutation as well. So it turns out that the alternating group $A_3$ is the set of all even permutations. In general, the set $A_n$ of all even permutations of $\{1,2,3,\cdots,n\}$ form a subgroup of the symmetric group $S_n$. Remember that $S_n$ has order $n!$. Since there are exactly the same number of even permutations and odd permutations, the alternating group $A_n$ of degree $n$ has order $\frac{n!}{2}$.

Example. $\mathrm{SL}(n,\mathbb{R})=\{A\in\mathrm{GL}(n,\mathbb{R}): \det A=1\}$ is a subgroup of the general linear group $\mathrm{GL}(n,\mathbb{R})$ of degree $n$. $\mathrm{SL}(n,\mathbb{R})$ is called the special linear group of degree $n$. The general linear group $\mathrm{GL}(n,\mathbb{R})$ is the group of isometries of $\mathbb{R}^n$. An isometry of $\mathbb{R}^n$ is a linear bijective map from $\mathbb{R}^n$ onto itself which preserves the inner product. What this means is that if $T: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ is an isometry then for any two vectors $v,w\in\mathbb{R}^n$, $\langle Tv,Tw\rangle=\langle v,w\rangle$. The special linear group $\mathrm{SL}(n,\mathbb{R})$ is the group of all orientation preserving isometries of $\mathbb{R}^n$.

# Functional Analysis 4: Convergence, Cauchy Sequence, Completeness

The set $\mathbb{Q}$ of rational numbers is not complete (or not a continuum) since it has gaps or holes. For instance, $\sqrt{2}$ is not in $\mathbb{Q}$. On the other hand, the set $\mathbb{R}$ of real numbers has no gaps or holes, so it is complete (or is a continuum). Let $(x_n)$ be a sequence of real numbers. Suppose that $(x_n)$ converges to a real number $x$. Then by the triangle inequality, for any $m,n\in\mathbb{N}$, we have
$$|x_m-x_n|\leq |x_m-x|+|x-x_n|.$$
Hence, $\displaystyle\lim_{m,n\to\infty}|x_m-x_n|=0$, i.e. $(x_n)$ is a Cauchy sequence. Conversely, Georg Cantor introduced the completeness axiom that every Cauchy sequence of real numbers converges and defined a real number as the limit of a Cauchy sequence of rational numbers. For instance, consider the Cauchy sequence $(x_n)$ defined by
$$x_1=1,\ x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n},\ \forall n\geq 2.$$
If $(x_n)$ converges to a number $x$, it would satisfy $x^2=2$ i.e. $(x_n)$ converges to $\sqrt{2}$. There is another way to obtain the completeness of $\mathbb{R}$ by a Dedekind cut, though we are not going to delve into that here.

More generally, one can also consider a complete metric space and that is what we are going to study in this lecture.

Definition. A sequence $(x_n)$ is a metric space $(X,d)$ is said to converge or to be convergent to $x\in X$ if
$$\lim_{n\to\infty}d(x_n,x)=0.$$
$x$ is called the limit if $(x_n)$ and we write
$$\lim_{n\to\infty}x_n=x\ \mbox{or}\ x_n\rightarrow x.$$
If $(x_n)$ is not convergent, it is sad to be divergent. We can generalize the definiton of the convergence of a sequence we learned in calculus in terms of a metric as:

Definition. $\displaystyle\lim_{n\to\infty}d(x_n,x)=0$ if and only if given $\epsilon>0$ $\exists$ a positive integer $N$ s.t. $x_n\in B(x,\epsilon)$ $\forall n\geq N$.

A nonempty subset $M\subset X$ is said to be bounded if
$$\delta(M)=\sup_{x,y\in M}d(x,y)<\infty.$$

Lemma. Let $(X,d)$ be a metric space.

(a) A convergent sequence in $X$ is bounded and its limit is unique.

(b) If $x_n\rightarrow x$ and $y_n\rightarrow y$, then $d(x_n,y_n)\rightarrow d(x,y)$.

Proof. (a) Suppose that $x_n\rightarrow x$. Then one can find a positive integer $N$ such that $d(x_n,x)<1$ $\forall n\geq N$. Let $M=2\max\{d(x_1,x),\cdots,d(x_{N-1},x),1\}$. Then for all $m,n\in\mathbb{N}$,
\begin{align*}
d(x_m,x_n)&\leq d(x_m,x)+d(x,x_n)\ (\mbox{ (M3) triangle inequality)}\\
&\leq M.
\end{align*}
This means that $\delta((x_n))\leq M<\infty$ i.e. $(x_n)$ is bounded.

Suppose that $x_n\rightarrow x$ and $x_n\rightarrow y$. Then
\begin{align*}
0\leq d(x,y)&\leq d(x,x_n)+d(x_ny)\\
&\rightarrow 0
\end{align*}
as $n\to\infty$. So, $d(x,y)=0\Rightarrow x=y$ by (M1).

(b) By (M3),
$$d(x_n,y_n)\leq d(x_n,x)+d(x,y)+d(y,y_n)$$
and so we obtain
$$d(x_n,y_n)-d(x,y)\leq d(x_n,x)+d(y,y_n).$$
Similarly, we also obtain the inequality
$$d(x,y)-d(x_n,y_n)\leq d(x,x_n)+d(y_n,y).$$
Hence,
$$0\leq |d(x_n,y_n)-d(x,y)|\leq d(x_n,x)+d(y_n,y)\rightarrow 0$$
as $n\to\infty$.

Definition. A sequence $(x_n)\subset (X,d)$ is said to be Cauchy if given $\epsilon>0$ $\exists$ a positive integer $N$ such that
$$d(x_m,x_n)<\epsilon\ \forall m,n\geq N.$$
The space $X$ is said to be complete if every Cauchy sequence in $X$ converges.

Examples. The real line $\mathbb{R}$ and the complex plane $\mathbb{C}$ are complete.

Theorem. Every convergent sequence is Cauchy.

Proof. Suppose that $x_n\rightarrow x$. Then given $\epsilon>0$ $\exists$ a poksitive integer $N$ s.t. $d(x_n,x)<\frac{\epsilon}{2}$ for all $n\geq N$. Now, $\forall m,n\geq N$
$$d(x_m,x_n)\leq d(x_m,x)+d(x,x_n)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$
Therefore, $(x_n)$ is Cauchy.

Theorem. Let $M$ be a nonempty subset of a metric space $(X,d)$. Then

(a) $x\in\bar M\Longleftrightarrow \exists$ a seqence $(x_n)\subset M$ such that $x_n\rightarrow x$.

(b) $M$ is closed $\Longleftrightarrow$ given a sequence $(x_n)\subset M$, $x_n\rightarrow x$ implies $x\in M$.

Proof. (a) ($\Longrightarrow$) Since $x\in\bar M$, $\forall n\in\mathbb{N}$ $\exists x_n\in B\left(x,\frac{1}{n}\right)\cap M\ne\emptyset$. Let $\epsilon>0$ be given. Then by the Archimedean property, $\exists$ a positive integer $N$ s.t. $N\geq\frac{1}{\epsilon}$. Now,
$$n\geq N\Longrightarrow d(x_n,x)<\frac{1}{n}\leq\frac{1}{N}<\epsilon.$$

($\Longleftarrow$) Suppose that $\exists$ a sequence $(x_n)\subset M$ s.t. $x_n\rightarrow x$. Then given $\epsilon>0$ $\exists$ a positive integer $N$ s.t. $x_n\in B(x,\epsilon)$ $\forall n\geq N$. This means that $\forall\epsilon>0$, $B(x,\epsilon)\cap M\ne\emptyset$. So, $x\in\bar M$.

(b) ($\Longrightarrow$) Clear

($\Longleftarrow$) It suffices to show that $\bar M\subset M$. Let $x\in\bar M$. Then $\exists$ a sequence $(x_n)\subset M$ such that $x_n\rightarrow x$. By assumption, $x\in M$.
Theorem. A subspace $M$ of a complete metric space $X$ itself is complete if and only if $M$ is closed in $X$.

Proof. ($\Longrightarrow$) Let $M\subset X$ be complete. Let $(x_n)$ be a sequence in $M$ such that $x_n\rightarrow x$. Then $(x_n)$ is Cauchy. Since $M$ is complete, every Cauchy sequence must converge and hence $x\in M$. This means that $M$ is closed.

($\Longleftarrow$) Suppose that $M\subset X$ is closed. Let $(x_n)$ be a Cauchy sequence in $M\subset X$. Since $X$ is complete, $\exists x\in X$ such that $x_n\rightarrow x$. Since $M$ is closed, $x\in M$. Therefore, $M$ is complete.

Example. In $\mathbb{R}$ with Euclidean metric, the closed intervals $[a,b]$ are complete. $\mathbb{Z}$, the set of integers is also complete by the above theorem since it is closed in $\mathbb{R}$. One can directly see why $\mathbb{Z}$ is complete without quoting the theorem though. Let $(x_n)$ be a Cauchy sequence in $\mathbb{Z}$. Then we see that there exists a positive integer $N$ such that $x_N=x_{N+1}=x_{N+2}=\cdots$. Hence any Cauchy sequence in $\mathbb{Z}$ is a convergent sequence in $\mathbb{Z}$. Therefore, $\mathbb{Z}$ is complete.

Theorem. A mapping $T: X\longrightarrow Y$ is continuous at $x_0\in X$ if and only if $x_n\rightarrow x$ implies $Tx_n\rightarrow Tx_0$.

Proof. ($\Longrightarrow$) Suppose that $T$ is continuous and $x_n\rightarrow x$ in $X$. Let $\epsilon>0$ be given. Then $\exists\delta>0$ s.t. whenever $d(x,x_0)<\delta$, $d(Tx,Tx_0)<\epsilon$. Since $x_n\rightarrow x$, $\exists$ a positive integer $N$ s.t. $d(x_n,x_0)<\delta$ $\forall n\geq N$. So, $\forall n\geq N$, $d(Tx_n,Tx_0)<\epsilon$. Hence, $Tx_n\rightarrow Tx_0$.

($\Longleftarrow$) Suppose that $T$ is not continuous. Then $\exists\epsilon>0$ s.t. $\forall\delta>0$, $\exists x\ne x_0$ satisfying $d(x,x_0)<\delta$ but $d(Tx,tx_0)\geq\epsilon$. So, $\forall n=1,2,\cdots$, $\exists x_n\ne x_0$ satisfying $d(x_n,x_0)<\frac{1}{n}$ but $d(Tx_n,Tx_0)\geq\epsilon$. This means that $x_n\rightarrow x_0$ but $Tx_n\not\rightarrow Tx_0$.

# Group Theory 4: Examples of Groups

In the first lecture here, we defined a group. We have also seen an example of a group here, which is a symmetric group. In this lecture, we study some well-known examples of groups of finite and infinite orders. Recall that the order of a group os the number of elements in the group.

Examples of Groups

1. $(\mathbb{Z},+)$, $(\mathbb{Q},+)$, $(\mathbb{R},+)$, and $(\mathbb{C},+)$ are abelian groups of infinite order.
2. $(\mathbb{Q}\setminus\{0\},\cdot)$ is an abelian group of infinite order.
3. $(\mathbb{R}^+,\cdot)$ is an abelian group of infinite order. Here $\mathbb{R}^+$ denotes the set of all positive real numbers.
4. Let $E_n=\{e^{\frac{2k\pi i}{n}}: k=0,1,2,\cdots,n-1\}$. $e^{\frac{2k\pi i}{n}}$, $k=0,1,2,\cdots,n-1$ are the $n$-th roots of unity i.e. the zeros of $z^n=1$. $E_n$ forms an abelian group of order $n$. It is generated by a single element $e^{\frac{2\pi i}{n}}$. Such a group is called a cyclic group.

Definition. A group $G$ is said to be a finite group if it has a finite number of elements. The number of elements in $G$ is called the order of $G$ as mentioned previously and it is denoted by $|G|$ or $\mathrm{ord}(G)$. We use the notation $|G|$ for the order of the group $G$.

Example.Let $\mathbb{Z}_n=\{0,1,2,\cdots,n-1\}$. $\mathbb{Z}_n$ is the set of all possible remainders when an integer is divided by 2. The addition $+$ on $\mathbb{Z}_n$ is defined naturally as follows.
\begin{aligned} \mathbb{Z}_2\\ \begin{array}{|c||c|c|} \hline + & 0 & 1\\ \hline \hline 0 & 0 & 1\\ \hline 1 & 1 & 0\\ \hline \end{array} \end{aligned}
\begin{aligned} \mathbb{Z}_3\\ \begin{array}{|c||c|c|c|} \hline + & 0 & 1 & 2\\ \hline \hline 0 & 0 & 1 & 2\\ \hline 1 & 1 & 2 & 0\\ \hline 2 & 0 & 1 & 2\\ \hline \end{array} \end{aligned}
$(\mathbb{Z}_n,+)$ is an abelian group of order $n$. We will see later that the group $E_n$ and $\mathbb{Z}_n$ are indeed the same group.

Example. The set $M_{m\times n}(\mathbb{R})$ of all $m\times n$ matrices of real entries under matrix addition is an abelian group of infinite order.

Example. The set $\mathrm{GL}(n,\mathbb{R})$ of alll $n\times n$ non-singular matrices of real entries under matrix multiplication is a non-abelian group of infinite order. $\mathrm{GL}(n,\mathbb{R})$ is called the general linear group of degree $n$. It can be viewed as the set of all invertible linear transformations $T: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ that preserve the standard Euclidean inner product. A map $T: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ is said to be an isometry if it is a linear isomorphism (i.e linear, one-to-one and onto, but remember from linear algebra that a linear map $T: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ is one-to-one, it is also onto) and it preserves inner product i.e. for any vectors $v$ and $w$ in $\mathbb{R}$, $\langle Tv,Tw\rangle=\langle v,w\rangle$.

Example. [Klein four-group (Vierergruppe in German)]

Klein four-group

Let $V=\{e,a,b,c\}$ where $a$ is counterclockwise rotation of the rectangle in the picture about the $x$-axis by $180^\circ$, $b$ is counterclockwise rotation of the rectangle about the $y$-axis by $180^\circ$ and $c$ is counterclockwise rotation of the rectangle about the $z$-axis (the axis coming out of the origin toward you) by $180^\circ$. Then $a$, $b$ and $c$ satisfy the following relationship
\begin{align*}
a^2=b^2=c^2=e,\ ab&=ba=c,\ bc=cb=a,\\
ca&=ac=b.
\end{align*}
Here, $\cdot$ is the function composition i.e. successive application of rotations. By labling the vertices of the rectangle as 1, 2, 3, 4 as seen in the picture, we can define each rotation as a permutation of $\{1,2,3,4\}$.
\begin{align*}
e&=\begin{pmatrix}
1 & 2 & 3 & 4\\
1 & 2 & 3 & 4
\end{pmatrix},\ a=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 1 & 4 &3
\end{pmatrix},\\
b&=\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix},\ c=\begin{pmatrix}
1 & 2 & 3 & 4\\
3 & 4 & 1 &2
\end{pmatrix}.
\end{align*}
For instance, we calculate $ab$:
\begin{align*}
ab&=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 1 & 4 &3
\end{pmatrix}\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}\\
&=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 1 & 4 &3
\end{pmatrix}\begin{pmatrix}
4 & 3 & 2 & 1\\
3 & 4 & 1 & 2
\end{pmatrix}\\
&=\begin{pmatrix}
1 & 2 & 3 & 4\\
3 & 4 & 1 &2
\end{pmatrix}\\
&=c.
\end{align*}

Example. [The $n$-th dihedral group $D_n$, $n\geq 3$]

Consider an equilateral triangle shown in the following picture.

Dihedral group D3

$\rho$ is counterclockwise rotation about the axis coming out of the origin by $\frac{360^\circ}{3}=120^\circ$ and $\mu_i$, $i=1,2,3$ is counterclockwise rotation about the axis of rotation through each vertex $i$ by $180^\circ$. As permutations of the vertices $\{1,2,3\}$, $\rho$ and $\mu_i$ , $i=1,2,3$ are given by
\begin{align*}
\rho&=\begin{pmatrix}
1 & 2 & 3\\
2 & 3 & 1
\end{pmatrix},\ \mu_1=\begin{pmatrix}
1 & 2 & 3\\
1 & 3 & 2
\end{pmatrix}\\
\mu_2&=\begin{pmatrix}
1 & 2 & 3\\
3 & 2 & 1
\end{pmatrix},\ \mu_3=\begin{pmatrix}
1 & 2 & 3\\
2 & 1 &3
\end{pmatrix}.
\end{align*}
$D_3=\{e,\rho,\rho^2,\mu_1,\mu_2,\mu_3\}$ is a group called the 3rd dihedral group. $$\mu_1\mu_2=\rho^2\ne\rho=\mu_2\mu_1.$$
So, we see that $D_3$ is not an abelian group. $D_3$ is the group of symmetries of an equilateral triangle.

Now this time let us consider a square as shown in the following picture.

Dihedral group D4

$\rho$ is counterclockwise rotation about the $z$-axis coming out of the origin by $\frac{360^\circ}{4}=90^\circ$. $\mu_i$, $i=1,2$ are counterclockwise rotations about $y$-axis and $x$-axis, respectively by $180^\circ$. $\delta_i$, $i=1,2$ are counterclockwise rotations about the axis through the vertices 2 and 4 and the vertices through 1 and 3, respectively by $180^\circ$. As permutations of the vertices $\{1,2,3,4\}$, $\rho$, $\mu_i$ and $\delta_i$, $i=1,2$ are given by
\begin{align*}
\rho&=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 3 & 4 & 1
\end{pmatrix},\\
\mu_1&=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 1 & 4 & 3
\end{pmatrix},\ \mu_2=\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix},\\
\delta_1&=\begin{pmatrix}
1 & 2 & 3 & 4\\
3 & 2 & 1 & 4
\end{pmatrix},\ \delta_2=\begin{pmatrix}
1 & 2 & 3 & 4\\
1 & 4 & 3 & 2
\end{pmatrix}.
\end{align*}
$D_4=\{e,\rho,\rho^2,\rho^3,\mu_1,\mu_2,\delta_1,\delta_2\}$ is the group of symmetries of a square, called the 4th dihedral group. It is also called the octic group. Note that $|D_n|=2n$, $n\geq 3$.