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3.7. Completeness

The material in this section is primarily based on [2, 23].

Recall from Sequences that in a Cauchy sequence, the points become increasingly closure to each other. We showed that every convergent sequence is a Cauchy sequence. However not every Cauchy sequence is a convergent sequence in a general metric space. This is an important issue to be examined in this section. There are metric spaces where every Cauchy sequence is also convergent. Such metric spaces are known as complete metric spaces.

In the remainder of this section, $(X,d)$ and $(Y,\rho )$ shall denote metric spaces.

3.7.1. Complete Metric Spaces

Definition 3.46 (Complete metric space)

A metric space $(X,d)$ is called complete if all of its Cauchy sequences converge in the space.

In other words, $X$ is complete if every Cauchy sequence $\{x_n\}$ of $X$ converges to a point $x\in X$.

Example 3.19 (Complete metric spaces)

1. ${\mathbb{R}}^n$ with the standard metric $d(x,y)=|x-y|$ is complete.

2. The Euclidean space ${\mathbb{R}}^n$, with the standard Euclidean metric (${\ell }_2$ distance), is complete.

Example 3.20 (The metric space of bounded functions)

Let $X$ be a non-empty set. Let $B(X)$ be the set of bounded (total) functions on $X$.

for any $f,g\in B(X)$, define $D:B(X)\times B(X)\to \mathbb{R}$ as:

$$D(f,g)\triangleq sup\{|f(x)-g(x)||x\in X\}.$$

Since both $f$ and $g$ are bounded, hence $D(f,g)$ is a real number. Thus $\mathrm{dom}D=B(X)\times B(X)$. We claim that $D$ is a metric:

1. $D(f,g)$ is nonnegative as it is the supremum of nonnegative numbers.

2. $D(f,f)=sup\{|f(x)-f(x)|\}=0$. Also, if $D(f,g)=0$ then $f(x)=g(x)\mathrm{\forall }x\in X$. Thus, $f=g$.

3. By definition: $D(f,g)=D(g,f)$.

4. For the triangle inequality, we proceed as follows:

   Let $f,g,h\in B(X)$. For each $x\in X$, we have:

   $$|f(x)-g(x)|\le |f(x)-h(x)|+|h(x)-g(x)|\le D(f,h)+D(h,g).$$

   Now, taking the supremum on the L.H.S. over all $x\in X$, we get:

   $$D(f,g)\le D(f,h)+D(h,g).$$

Next, we establish that $(B(X),D)$ is a complete metric space. Towards this, introduce a zero function $z:X\to \mathbb{R}$ as $z(x)=0$.

1. Let $\{f_n\}$ be a Cauchy sequence of $B(X)$.

2. Thus, for every $ϵ>0$, there exists $n_0$ such that $D(f_n,f_m)<ϵ$ for all $m,n>n_0$.

3. In particular, note that, by definition, $|f_n(x)-f_m(x)|\le D(f_n,f_m)$ for every $x\in X$.

4. Thus, for any $x\in X$, $|f_n(x)-f_m(x)|<ϵ$ for all $m,n>n_0$.

5. Thus, $\{f_n(x)\}$ is a Cauchy sequence of real numbers at each $x=X$.

6. Since, $\mathbb{R}$ is complete, hence $\{f_n(x)\}$ converges at each $x=X$.

7. We define a new function $f:X\to \mathbb{R}$ as $f(x)=lim{f}_n(x)$ for each $x=X$.

8. Since $|f_n(x)-f_m(x)|<ϵ$ for all $m,n>n_0$, hence $|f_n(x)-f(x)|\le ϵ$ for all $n>n_0$ and all $x\in X$.

9. Thus, $|f(x)|\le ϵ+|f_n(x)|$ for all $x\in X$.

10. Thus, since $f_n$ is bounded, hence $f$ is also bounded.

11. Thus, $f\in B(X)$.

12. Finally, $D(f_n,f)=sup\{|f_n(x)-f(x)|\}$.

13. Since $|f_n(x)-f(x)|\le ϵ$ for all $x\in X$, hence, taking the supremum on the L.H.S., $D(f_n,f)\le ϵ$.

14. Since $D(f_n,f)\le ϵ$ for all $n>n_0$, hence $lim{f}_n=f$.

15. Thus, every Cauchy sequence in $B(X)$ converges in $B(X)$.

16. Thus, $(B(X),D)$ is a complete metric space.

3.7.2. Closed Subsets

Theorem 3.58

Let $(X,d)$ be a complete metric space. Then a subset $A$ of $X$ is closed if and only if $(A,d)$ is a complete metric space in its own right.

Proof. Let $A$ be closed.

1. Let $\{x_n\}$ be a Cauchy sequence of $A$.

2. Then, $\{x_n\}$ is a Cauchy sequence of $X$ (since $A\subseteq X$).

3. Thus, $\{x_n\}$ is convergent, since $X$ is complete.

4. Let $lim{x}_n=x$.

5. Since $A$ is closed and $\{x_n\}$ of $A$ is convergent, hence $x\in A$ due to Theorem 3.33.

6. Thus, every Cauchy sequence of $A$ converges in $A$.

7. Thus, $(A,d)$ is complete.

Let $A$ be complete. We shall show that it contains all its closure points.

1. Let $x\in X$ be a closure point of $A$.

2. By Theorem 3.31, there is a sequence $\{x_n\}$ of $A$ that converges to $x$, i.e., $lim{x}_n=x$.

3. Since $\{x_n\}$ converges in $X$, hence $\{x_n\}$ is a Cauchy sequence of $X$ due to Proposition 3.19.

4. But then, $\{x_n\}$ is a Cauchy sequence of $A$.

5. Since, $A$ is complete, hence $\{x_n\}$ converges in $A$.

6. Thus, $x\in A$.

7. Thus, $A$ contains all its closure points. $A$ is closed.

Recall that the diameter of a set is defined to be the supremum of distances between all pairs of points of the set.

Theorem 3.59

Let $(X,d)$ be a complete metric space. Let $\{A_n\}$ be a sequence of closed, nonempty subsets of $X$ such that $A_{n+1}\subseteq A_n$ for each $n$ and

$$\underset{n\to \mathrm{\infty }}{lim}\mathrm{diam}{A}_n=0.$$

Then the intersection $\bigcap _{n=1}^{\mathrm{\infty }}{A}_n$ consists of precisely one point.

This result is due to G. Cantor. A similar result was seen in nested interval property for real line.

Proof. Define:

$$A=\bigcap _{n=1}^{\mathrm{\infty }}{A}_n.$$

We shall first prove that $A$ cannot have more than one point.

1. Assume $x,y\in A$.

2. Thus, $x,y\in A_n\mathrm{\forall }n$.

3. Thus, $0\le d(x,y)\le \mathrm{diam}{A}_n\mathrm{\forall }n$.

4. But then, $\underset{n\to \mathrm{\infty }}{lim}\mathrm{diam}{A}_n=0$ implies that $d(x,y)=0$.

5. But, $d(x,y)=0⟹x=y$ as $d$ is a metric (Identity of indiscernibles).

6. Thus, if $x,y\in \bigcap _{n=1}^{\mathrm{\infty }}{A}_n$, then $x=y$.

7. Thus, $A$ contains at most one point.

We now show that $A$ cannot be empty.

1. For each $n$ choose $x_n\in A_n$. It is possible due to axiom of choice.

2. Since $A_{n+p}\subseteq A_n$, hence $d(x_{n+p},x_n)\le \mathrm{diam}{A}_n$ holds for every $n,p$.

3. Thus, $\{x_n\}$ is a Cauchy sequence of $X$.

4. Since $X$ is complete, $\{x_n\}$ is convergent.

5. Hence, the limit $x=\underset{n\to \mathrm{\infty }}{lim}{x}_n$ exists and $x\in X$.

6. Since $x_m\in A_n$ for all $m\ge n$, hence $x$ is a closure point of $A_n$ for each $n$ (Theorem 3.31).

7. But since $A_n$ is closed, hence $x\in A_n$ for each $n$.

8. Thus, $x\in A$.

3.7.3. Nowhere Dense Sets

Definition 3.47 (Nowhere dense/Rare)

A subset $A$ of $(X,d)$ is nowhere dense if its closure has an empty interior; i.e.,

$$\mathrm{int}\mathrm{cl}A=\varnothing .$$

It is also called a rare set.

Theorem 3.60

$A$ is nowhere dense if and only if $X\setminus (\mathrm{cl}A)$ is dense in $X$.

Proof. Let $B=\mathrm{cl}A$.

Recall from Theorem 3.15 that:

$$\begin{array}{r}\begin{array}{rl}& X\setminus (\mathrm{int}B)=\mathrm{cl}(X\setminus B)\\ ⟺& \mathrm{int}B=X\setminus (\mathrm{cl}(X\setminus B)).\end{array}\end{array}$$

Now,

$$\begin{array}{r}\begin{array}{rl}& \mathrm{int}\mathrm{cl}A=\varnothing \\ ⟺& \mathrm{int}B=\varnothing \\ ⟺& X\setminus (\mathrm{cl}(X\setminus B))=\varnothing \\ ⟺& \mathrm{cl}(X\setminus B)=X\\ ⟺& \mathrm{cl}(X\setminus (\mathrm{cl}A))=X\\ ⟺& X\setminus (\mathrm{cl}A)\text{ is dense in }X.\end{array}\end{array}$$

Example 3.21

The set of integers $\mathbb{Z}$ is nowhere dense in $\mathbb{R}$.

Corollary 3.1

If a set is open and dense, then its complement is rare (nowhere dense).

Proof. Let $A$ be open and dense.

1. $B=X\setminus A$ is closed.

2. Thus, $\mathrm{cl}B=B$.

3. Hence, $X\setminus (\mathrm{cl}B)=X\setminus B=A$.

4. But $A$ is dense, hence $B$ must be nowhere dense due to Theorem 3.60.

Theorem 3.61

The complement of a rare (nowhere dense) set is dense.

Proof. Let $A\subseteq X$ be rare (nowhere dense).

1. The interior of its closure is empty.

2. Thus, its closure $\mathrm{cl}A$ contains no open sets.

3. Thus, $A$ contains no open sets.

4. Thus, every open set in $X$ intersects with $X\setminus A$.

5. Thus, $X\setminus A$ is dense.

Proposition 3.20

The subset of a rare (nowhere dense) set is rare.

Proof. Let $A\subseteq B$ and $B$ be nowhere dense. Now

$$A\subseteq B⟹\mathrm{cl}A\subseteq \mathrm{cl}B⟹\mathrm{int}\mathrm{cl}A\subseteq \mathrm{int}\mathrm{cl}B.$$

Since $B$ is nowhere dense, hence $\mathrm{int}\mathrm{cl}B$ is empty. This in turn implies that $\mathrm{int}\mathrm{cl}A$ is empty. Thus, $A$ is nowhere dense.

Proposition 3.21

The boundary of a closed set is nowhere dense (rare).

Proof. Let $A$ be closed. Thus, $A=\mathrm{cl}A$. Since $\mathrm{bd}A=\mathrm{cl}A\setminus \mathrm{int}A$, hence $\mathrm{bd}A\subseteq A$ for a closed set.

1. Assume $\mathrm{bd}A$ has a nonempty interior.

2. Let $x$ be an interior point of $\mathrm{bd}A$.

3. Since $x\in \mathrm{bd}A\subseteq A$, hence $x$ is an interior point of $A$.

4. But boundary of $A$ doesn’t includes its interior points.

5. Hence, a contradiction.

6. Thus, $\mathrm{bd}A$ has an empty interior.

7. Finally $\mathrm{int}\mathrm{cl}\mathrm{bd}A$ = \interior \boundary A = \EmptySet$since$\boundary A$ is closed.

8. Thus, $\mathrm{bd}A$ is rare.

3.7.4. Cantor Set

Definition 3.48 (Cantor set)

Cantor set $C$ is a subset of $[0,1]$. It is constructed as follows:

1. Let $C_0=[0,1]$.

2. Trisect $C_0$ into $[0,\frac{1}{3}],(\frac{1}{3},\frac{2}{3}),[\frac{2}{3},1]$.

3. Remove the middle open interval $(\frac{1}{3},\frac{2}{3})$ and form $C_1=[0,\frac{1}{3}]\cup [\frac{2}{3},1]$.

4. Note that $C_1$ is a disjoint union of $2=2^1$ closed intervals.

5. Trisect each closed interval of $C_1$ and remove the middle open interval from each one of the trisections in an identical manner.

6. Let $C_2=[0,\frac{1}{9}]\cup [\frac{2}{9},\frac{1}{3}]\cup [\frac{2}{3},\frac{7}{9}]\cup [\frac{8}{9},1]$.

7. Note that $C_2$ is a union of $4=2^2$ disjoint closed intervals each of length $\frac{1}{9}=\frac{1}{3^2}$.

8. Inductively build $C_{n+1}$ from $C_n$ using this procedure.

9. Note that $C_n$ is a union of $2^n$ disjoint closed intervals of length $\frac{1}{3^n}$ each.

10. When $C_{n+1}$ is constructed from $C_n$, we get $2^{n+1}$ disjoint intervals of length $\frac{1}{3^{n+1}}$ each.

11. Clearly $C_{n+1}\subseteq C_n$ by construction for all $n$.

12. Define the Cantor set as:

    $$C\triangleq \bigcap _{n=1}^{\mathrm{\infty }}{C}_n.$$

We next discuss different properties of the Cantor set.

Property 3.1 (Characterization in ternary expansions)

Consider the set $E\subset [0,1]$ whose every element has a ternary (base 3) expansion of only 0s and 2s; i.e., for every $x\in E$, we have a representation:

$$x=.d_1{d}_2{d}_3\dots \text{ where }{d}_i\in \{0,2\}$$

Then, $E=C$. In other words:

$$C=E\triangleq \{\sum _{n=1}^{\mathrm{\infty }}\frac{d_n}{3^n},\text{ where }{d}_n\in \{0,2\}\}.$$

Proof. In the ternary (base 3), representation, each number $x\in [0,1]$ can be written as:

$$x=.d_1{d}_2{d}_3\dots \text{ where }{d}_i\in \{0,1,2\}$$

such that

$$x=\sum _{i=1}^{\mathrm{\infty }}\frac{d_i}{3^i}.$$

For example

$$1=.2222222...=2\sum _{i=1}^{\mathrm{\infty }}\frac{1}{3^i}=2\frac{1/3}{2/3}=2\frac{1}{2}=1.$$

It is possible that a number has two different ternary expansions. We can see that $\frac{1}{3}\in C_k$ for every $k$. Hence, $\frac{1}{3}\in C$.Interestingly,

$$\frac{1}{3}=.1=.022222....$$

has two different ternary representations. One is a finite representation, and the other is an infinite representation involving only 0s and 2s. We say that $\frac{1}{3}\in C$ as it has a ternary representation consisting of only 0s and 2s.

We have to prove two things:

1. $E\subseteq C$. Every number in $[0,1]$ which has a ternary expansion containing only 0s and 2s belongs to $C$.

2. $C\subseteq E$. Every number in $C$ has a ternary expansion containing only of 0s and 2s.

Note that if $A=[a,a+\frac{1}{3^n}]$ is a closed interval contained in $C_n$ then:

1. $[a,a+\frac{1}{3^{n+1}}]$ is the first closed interval drawn from $A$ in $C_{n+1}$.

2. $[a+\frac{2}{3^{n+1}},a+\frac{1}{3^n}]$ is the second closed interval drawn from $A$ in $C_{n+1}$.

$E\subseteq C$

1. Let $x=.d_1{d}_2\dots$ with $d_i\in \{0,2\}$.

2. Start with $A_0=[0,1]$.

3. Given $A_0$:

   1. If $d_1=0$, then let $A_1$ be the first closed interval drawn from $A_0$ : $[0,\frac{1}{3}]$ contained in $C_1$.

   2. Otherwise, if $d_1=2$, then let $A_1$ the second closed interval drawn from $A_0$ : $[\frac{2}{3},1]$ contained in $C_1$.

4. Given $A_{n-1}=[a,a+\frac{1}{3^{n-1}}]$,

   1. If $d_n=0$, then let $A_n$ be the first closed interval drawn from $A_{n-1}$ contained in $C_n$.

   2. Otherwise (if $d_n=2$), then let $A_n$ be the second closed interval drawn from $A_{n-1}$ contained in $C_n$.

5. Inductively, we can keep picking closed interval $A_n$ contained in $C_n$ for every digit in the ternary expansion of $x$.

6. Thus, $x\in C$.

$C\subseteq E$

1. For every $x\in C$, we can construct a ternary expansion as follows.

2. If $x\in [0,\frac{1}{3}]$, then $d_1=0$.

3. Otherwise if $x\in [\frac{2}{3},1]$, then $d_1=2$.

4. Subsequently, whenever the first closed interval is chosen, then $d_n=0$ and whenever the second closed interval is chosen, then $d_n=2$.

5. Thus, $x$ has a ternary expansion consisting entirely of 0s and 2s.

6. Thus, $x\in E$.

Property 3.2

Cantor set doesn’t contain any open interval.

Proof. Note that the total length of disjoint closed intervals in $C_n$ is $\frac{2^n}{3^n}={\left(\frac{2}{3}\right)}^n$.

1. Assume that there is an open interval $(a,b)\subseteq C$.

2. Then, the length of the interval is $b-a>0$.

3. But then there exists an $n$ such that ${\left(\frac{2}{3}\right)}^n<b-a$.

4. Then, for all $k\ge n$, $(a,b)$ cannot be contained in $C_k$.

5. We arrive at a contradiction.

Property 3.3

Cantor set has an empty interior.

Proof. Assume $C$ has a nonempty interior and $x\in \mathrm{int}C$.

1. Then there is a neighborhood $(x-ϵ,x+ϵ)\subset C$.

2. But, $C$ doesn’t contain any open intervals.

3. We arrive at a contradiction.

Thus, $C$ has an empty interior.

Property 3.4

Cantor set is a closed nowhere dense subset of $\mathbb{R}$.

Proof. $C$ is an (infinite) intersection of closed sets. Hence $C$ is closed. $\mathrm{cl}C=C$. $C$ has an empty interior. Thus, $C$ is nowhere dense.

Property 3.5

The total length of the removed intervals from $[0,1]$ to get $C$ equals 1.

Proof. At the n-th step, we remove $2^{n-1}$ open intervals of length $3^{-n}$ each.

Thus, total length removed in n-th step is $\frac{1}{2}{\left(\frac{2}{3}\right)}^n$.

Thus, total length removed is:

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2}{\left(\frac{2}{3}\right)}^n=\frac{1}{2}\frac{2/3}{1-2/3}=1.$$

Property 3.6

Cantor set is uncountable. In particular:

$$\mathrm{card}C=\mathrm{card}{2}^{\mathbb{N}}=\mathrm{card}\mathbb{R}=\mathfrak{c}.$$

Recall that $\mathfrak{c}$ denotes the cardinality of the continuum (Definition 1.84). The notation $2^{\mathbb{N}}$ was introduced in Theorem 1.17 to describe power sets. Theorem 1.16 established that

$$2^{\mathbb{N}}\sim \mathcal{P}(\mathbb{N})\sim \mathbb{R}.$$

Proof. Recall that two sets are called equivalent ($A\sim B$) if there is a bijective mapping between them.

Recall from Property 3.1 that:

$$C=\{\sum _{n=1}^{\mathrm{\infty }}\frac{d_n}{3^n},\text{ where }{d}_n\in \{0,2\}\}.$$

1. Thus, each $x\in C$ can be identified with a sequence $d:\mathbb{N}\to \{0,2\}$ given by $d=\{d_n\}$.

2. Thus, we have a bijective mapping between $C$ and the set $\{0,2{\}}^{\mathbb{N}}$.

3. Thus, $C\sim \{0,2{\}}^{\mathbb{N}}$.

4. This, in turn can be identified with a sequence $c:\mathbb{N}\to \{0,1\}$ where $c=\{c_n\}$ with $c_n=0$ if $d_n=0$ and $c_n=1$ if $d_n=2$.

5. This gives us a bijective mapping between $\{0,2{\}}^{\mathbb{N}}$ and $\{0,1{\}}^{\mathbb{N}}$.

6. Recall that $2^{\mathbb{N}}=\{0,1{\}}^{\mathbb{N}}$ with $2=\{0,1\}$ (Theorem 1.17).

7. Thus $C\sim 2^{\mathbb{N}}$ and

   $$\mathrm{card}C=\mathrm{card}{2}^{\mathbb{N}}=\mathrm{card}\mathbb{R}=\mathfrak{c}.$$

3.7.5. Meager Sets

Definition 3.49 (Meager set)

A union of countably many rare (nowhere dense) sets is said to be of first category or a meager set.

In other words, a subset $A$ is called meager (or of first category) if there exists a sequence $\{A_n\}$ of nowhere dense subsets such that

$$A=\bigcup _{n=1}^{\mathrm{\infty }}{A}_n.$$

Definition 3.50 (Co-meager set)

The complement of a meager set is called co-meager.

Definition 3.51 (Non-meager set)

A subset that is not meager is said to be non-meager or of second category.

Example 3.22

Consider the metric space $\mathbb{R}$.

1. Singleton sets are rare (nowhere dense) in $\mathbb{R}$ as they are closed and their interior is empty.

2. The set of natural numbers $\mathbb{N}$ is rare since it is closed and its interior is empty (no open intervals in $\mathbb{N}$).

3. The set $\mathbb{Q}$ is not rare since its closure is entire $\mathbb{R}$. It is meager since it is a countable union of rare singleton sets.

4. The set of irrational numbers $\mathbb{I}$ is co-meager as $\mathbb{I}=\mathbb{R}\setminus \mathbb{Q}$.

5. The set $\mathbb{R}$ is non-meager.

Theorem 3.62

The subset of a meager set is meager.

Proof. Let $A$ be a meager set and let $B\subseteq A$. Then,

$$A=\bigcup _{n=1}^{\mathrm{\infty }}{A}_n$$

where $A_n$ are rare (nowhere dense).

But then

$$B=B\cap A=B\cap \bigcup _{n=1}^{\mathrm{\infty }}{A}_n=\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\cap B.$$

Since subset of a nowhere dense set is nowhere dense, hence $B_n=A_n\cap B$ are nowhere dense. Hence

$$B=\bigcup _{n=1}^{\mathrm{\infty }}{B}_n$$

is meager (as it is a countable union of nowhere dense sets).

Theorem 3.63

The union of countably many meager sets is meager.

Proof. Recall from Theorem 1.10 that countable union of countable sets is countable.

1. Let $\{M_n\}$ be a countable collection of meager sets.

2. Then, each $M_n$ is a countable union of rare sets.

3. Write $M_n$ as $\bigcup _i{A}_{n,i}$.

4. Thus,

   $$M=\bigcup _n{M}_n=\bigcup _n\bigcup _i{A}_{n,i}=\bigcup _{n,i}{A}_{n,i}.$$

5. The family $\{A_{n,i}\}$ is countable.

6. Thus, $M$ is a countable union of rare sets.

7. Thus, $M$ is meager.

Corollary 3.2

Since co-meager sets are complements of meager sets:

1. Superset of a co-meager set is co-meager.

2. Countable intersection of co-meager sets is co-meager.

3.7.6. Baire Category Theorem

Theorem 3.64 (Baire category theorem)

A (nonempty) complete metric space is non-meager in itself.

In other words:

• A (nonempty) complete metric space is not a countable union of rare sets.

Proof. Our proof strategy is following:

1. For contradiction, we assume that $X$ is meager with $X=\bigcup _i{A}_i$.

2. We form a Cauchy sequence from points $x_i\notin A_i$.

3. Since $X$ is complete, we claim its convergence $x=lim{x}_i$.

4. We further show that the limit point $x$ cannot belong to any $A_i$.

Let us assume that $X$ is a complete metric space which is meager in itself; i.e., there exists a countable collections of rare sets (rare in $X$) such that $X$ is their union.

$$X=\bigcup _{k=1}^{\mathrm{\infty }}{A}_k$$

where $A_k$ are rare (nowhere dense); i.e. $\mathrm{int}\mathrm{cl}{A}_k=\varnothing$.

Recall from Theorem 3.60 that $B_k=X\setminus \mathrm{cl}{A}_k$ are dense in $X$. Also, $B_k$ are open since they are complement of closed sets. Also, recall from Theorem 3.23 that a dense set has a nonempty intersection with every nonempty open set.

1. $B_1=X\setminus \mathrm{cl}{A}_1$ is a nonempty open set.

2. Thus, there exists an interior point $x_1$ in $B_1$ with an open ball $B(x_1,{ϵ}_1)\subseteq B_1$ where we can make ${ϵ}_1<\frac{1}{2}$.

3. $B_2$ is nonempty, open and dense. Thus, $C_2=B_2\cap B(x_1,\frac{{ϵ}_1}{2})$ is nonempty and open.

4. Thus, we can choose $x_2\in C_2$ such that $x\in B(x_2,{ϵ}_2)\subseteq C_2$ with ${ϵ}_2\le \frac{{ϵ}_1}{2}$.

5. Proceeding inductively in this manner, we obtain a sequence $\{x_n\}$ such that

   $$x_n\in B(x_n,{ϵ}_n)\subseteq B_n\cap B(x_{n-1},\frac{{ϵ}_{n-1}}{2}).$$

6. We have $A_n\cap B(x_n,{ϵ}_n)=\varnothing$. ${ϵ}_n\le \frac{{ϵ}_{n-1}}{2}$ with ${ϵ}_1<\frac{1}{2}$.

7. By definition, $x_m\in B(x_n,\frac{{ϵ}_n}{2})$ for every $m>n$.

8. Thus, $d(x_m,x_n)<\frac{{ϵ}_n}{2}$ for every $m>n$.

9. Since ${ϵ}_1<\frac{1}{2}$, hence ${ϵ}_n<\frac{1}{2^{n+1}}$.

10. Thus, $d(x_m,x_n)\to 0$ as $m,n\to \mathrm{\infty }$.

11. Thus, $\{x_n\}$ is a Cauchy sequence.

12. Since $X$ is complete, hence every Cauchy sequence is convergent.

13. Thus there exists $x\in X$ such that $x=lim{x}_n$. To which $A_k$ does $x$ belong then?

14. Fix some $n\in \mathbb{N}$. For every $m>n$ we have:

    $$d(x,x_n)\le d(x,x_m)+d(x_n,x_m)<d(x,x_m)+\frac{{ϵ}_n}{2}.$$

15. Thus taking the limit $m\to \mathrm{\infty }$ leading to $d(x,x_m)\to 0$, we get

    $$d(x,x_n)\le \frac{{ϵ}_n}{2}<{ϵ}_n.$$

16. Hence $x\in B(x_n,{ϵ}_n)$ for all $n\in \mathbb{N}$.

17. Since $A_n\cap B(x_n,{ϵ}_n)=\varnothing$, hence $x\notin A_n$ for all $n$.

18. But then $x\notin X$ since $X=\bigcup _n{A}_n$. A contradiction.

19. Hence, $X$ must be non-meager.

Theorem 3.65

If $(X,d)$ is complete and $X=\bigcup _{n=1}^{\mathrm{\infty }}{A}_n$, then at least one $A_n$ is non-rare; i.e., $\mathrm{int}\mathrm{cl}{A}_n\ne \varnothing$ for some $n$.

Proof. If every $A_n$ were rare, then $X$ would be a meager set. But as per Baire category theorem, $X$ must be non-meager since it is complete.

Hence, at least one $A_n$ would be non-rare.

Proposition 3.22

The set of irrational numbers is non-meager.

Proof. Recall that $\mathbb{R}=\mathbb{Q}\cup \mathbb{I}$ where $\mathbb{I}$ is the set of irrational numbers.

1. $\mathbb{R}$ is complete. Hence $\mathbb{R}$ is non-meager.

2. $\mathbb{Q}$ is meager as it is a countable union of singletons which are rare sets.

3. Countable union of meager sets is meager.

4. Thus, if $\mathbb{I}$ was meager, then $\mathbb{R}$ would be meager which is not true.

5. Hence, $II$ must be non-meager.

Theorem 3.66 (Interior of a meager set)

A meager set has an empty interior in a complete metric space.

Proof. Let $M\subseteq X$ be meager. Then, we can write $M$ as

$$M=\bigcup _{n=1}^{\mathrm{\infty }}{A}_n$$

such that $A_n$ are rare in $X$.

Let $B_n=X\setminus \mathrm{cl}{A}_n$. Then $B_n$ are dense, nonempty and open in $X$. Hence $B_n\cap U$ is nonempty and open for every nonempty open set $U\subseteq X$.

1. Let $U$ be an arbitrary nonempty open set in $X$.

2. There exists $x_1\in B_1\cap U$ such that $x_1\in B(x_1,{ϵ}_1)\subseteq B_1\cap U$ where ${ϵ}_1<\frac{1}{2}$.

3. $B_2$ is dense and open so $C_2=B_2\cap B(x_1,\frac{{ϵ}_1}{2})$ is nonempty and open.

4. We can choose a point $x_2\in B(x_2,{ϵ}_2)\subseteq C_2$ where ${ϵ}_2\le \frac{{ϵ}_1}{2}$.

5. Proceeding in this manner, we choose points

   $$x_n\in B(x_n,{ϵ}_n)\subseteq B_n\cap B(x_{n-1},\frac{{ϵ}_{n-1}}{2})$$

   to form a sequence $\{x_n\}$.

6. The sequence $\{x_n\}$ is Cauchy, $X$ is complete, hence $x=lim{x}_n$ exists in $X$.

7. Also, $x\in B(x_n,{ϵ}_n)$ for all $n$.

8. In particular $x\in B(x_1,{ϵ}_1)\subseteq U$ and $x\in B(x_n,{ϵ}_n)\subseteq B_n$.

9. Thus, $x\in B_n\cap U$ for all $n$.

10. Thus,

    $$\begin{array}{r}\begin{array}{rl}& x\in U\cap \left(\bigcap _n{B}_n\right)\\ & ⟹x\in U\cap (\bigcap _n(X\setminus \mathrm{cl}{A}_n))\\ & ⟹x\in U\cap (X\setminus (\bigcup _n(\mathrm{cl}{A}_n)))\\ & ⟹x\in U\cap (X\setminus \left(\bigcup _n{A}_n\right))\\ & ⟹x\in U\cap (X\setminus M).\end{array}\end{array}$$

11. Thus, $U$ has a nonempty intersection with $X\setminus M$.

12. Since $U$ is arbitrary, hence $X\setminus M$ intersects with every nonempty open set in $X$.

13. Thus, $X\setminus M$ is dense in $X$ (Theorem 3.23).

14. Thus, $M$ has an empty interior (Theorem 3.24).

Corollary 3.3

If a set in a complete metric space has a non-empty interior, then it is not meager.

Example 3.23

$\mathbb{Q}$ in $\mathbb{R}$

1. The set $\mathbb{Q}$ is meager in $\mathbb{R}$.

2. $\mathbb{R}$ is complete.

3. The interior of $\mathbb{Q}$ in $\mathbb{R}$ is empty.

4. The closure of $\mathbb{Q}$ is $\mathbb{R}$.

5. Thus, while a meager set may have an empty interior, its closure need not have an empty interior. This is different from rare sets whose closure has an empty interior.

$\mathbb{Q}$ in $\mathbb{Q}$

1. $\mathbb{Q}$ by itself is not a complete metric space.

2. Singletons are rare sets in $\mathbb{Q}$.

3. Thus, $\mathbb{Q}$ is meager in $\mathbb{Q}$ as it is a countable union of rare sets.

4. However, the interior of $\mathbb{Q}$ is not empty. In fact it is whole of $QQ$.

5. Theorem 3.66 does not apply since $\mathbb{Q}$ is not a complete metric space.

Theorem 3.67

The set $[0,1]$ is uncountable.

Proof. We prove this using Baire category theorem.

1. $[0,1]$ is a complete metric space with the standard metric.

2. Assume $[0,1]$ to be countable.

3. Then, $[0,1]=\{x_n{\}}_{n\in \mathbb{N}}$ is an enumeration of $[0,1]$.

4. The singleton set $\{x_n\}$ is rare in $[0,1]$.

5. Then $[0,1]$ being a countable union of rare sets would be meager.

6. But Baire category theorem says that a complete metric space is non-meager.

7. We have a contradiction.

8. Thus, $[0,1]$ must be uncountable.

3.7.7. Baire Spaces

Definition 3.52 (Baire space)

A metric space is called a Baire space if every nonempty open set is not a meager set.

Theorem 3.68

Every complete metric space is a Baire space.

Proof. Let $A$ be a nonempty open subset of a complete metric space $(X,d)$. If it was meager, then it would have an empty interior (Theorem 3.66). Thus, it must be non-meager.

Hence $X$ is a Baire space.

Theorem 3.69 (Characterization of Baire space)

For a metric space $X$, the following statements are equivalent:

1. $X$ is a Baire space.

2. Every countable intersection of open dense sets is also dense.

3. If $X=\bigcup _{n=1}^{\mathrm{\infty }}{F}_n$ and each $F_n$ is a closed set, then the open set $\bigcup _{n=1}^{\mathrm{\infty }}\mathrm{int}{F}_n$ is dense.

Proof. (1) $⟹$ (2)

1. Assume $X$ is a Baire space.

2. Let $\{A_n\}$ be a sequence of open dense sets in $X$.

3. Let $A=\bigcap _{n=1}^{\mathrm{\infty }}{A}_n$. We need to show that $A$ is dense.

4. We will show that $A$ has a nonempty intersection with every nonempty open set of $X$. Thus, claim that $A$ is dense due to Theorem 3.23.

5. Let $\mathcal{O}\subseteq X$ be an arbitrary nonempty open set.

6. Assume, for contradiction that $A\cap \mathcal{O}=\varnothing$.

7. Then, $X=X\setminus \varnothing =X\setminus (A\cap \mathcal{O})=(X\setminus A)\cup (X\setminus O)$.

8. Thus,

   $$\begin{array}{r}\begin{array}{rl}\mathcal{O}& =X\cap \mathcal{O}=((X\setminus A)\cup (X\setminus O))\cap \mathcal{O}=(X\setminus A)\cap \mathcal{O}\\ & =(X\setminus \left(\bigcap _{n=1}^{\mathrm{\infty }}{A}_n\right))\cap \mathcal{O}\\ & =\bigcup _{n=1}^{\mathrm{\infty }}((X\setminus A_n)\cap \mathcal{O}).\end{array}\end{array}$$

9. Due to Corollary 3.1, $X\setminus A_n$ are rare.

10. Due to Proposition 3.20, $(X\setminus A_n)\cap \mathcal{O}$ are rare.

11. Thus, $\mathcal{O}$ being a countable union of rare sets, is meager.

12. But, in a Baire space, every nonempty open set is not meager.

13. We have a contradiction.

14. Thus, $A\cap \mathcal{O}$ must be nonempty.

15. Thus, $A$ has a nonempty intersection with every nonempty open set in $A$.

16. Thus, $A$ is dense in $X$.

(2) $⟹$ (3)

1. We assume that every countable intersection of open dense sets is dense.

2. Let $\{F_n\}$ be a sequence of closed sets in $X$ satisfying $X=\bigcup _{n=1}^{\mathrm{\infty }}{F}_n$.

3. Let $A_n=\mathrm{int}{F}_n$ and let $A=\bigcup _{i=1}^{\mathrm{\infty }}{A}_n$. By definition, $A$ is open.

4. Since $F_n$ is closed, hence $E_n=F_n\setminus (\mathrm{int}{F}_n)$ is its boundary.

5. Due to Proposition 3.21, $E_n$ is rare.

6. Thus, the set $E=\bigcup _{n=1}^{\mathrm{\infty }}{E}_n$ is a meager set.

7. Since $E_n$ is closed and rare, hence $X\setminus E_n$ is open and dense (Theorem 3.61).

8. By our hypothesis (2), the set

   $$X\setminus E=X\setminus \left(\bigcup _{n=1}^{\mathrm{\infty }}{E}_n\right)=\bigcap _{n=1}^{\mathrm{\infty }}(X\setminus E_n)$$

   is also a dense set as it is a countable intersection of open dense sets $X\setminus E_n$.

9. Now, notice that:

   $$\begin{array}{r}\begin{array}{rl}X\setminus A& =X\setminus \left(\bigcup _{i=1}^{\mathrm{\infty }}{A}_n\right)\\ & =\bigcup _{n=1}^{\mathrm{\infty }}{F}_n\setminus (\bigcup _{i=1}^{\mathrm{\infty }}(\mathrm{int}{F}_n))\\ & \subseteq \bigcup _{n=1}^{\mathrm{\infty }}[F_n\setminus (\mathrm{int}{F}_n)]\\ & =\bigcup _{n=1}^{\mathrm{\infty }}{E}_n\\ & =E.\end{array}\end{array}$$

10. And $X\setminus A\subseteq E$ implies $X\setminus E\subseteq A$.

11. Since $X\setminus E$ is dense, hence $A$ is also dense.

(3) $⟹$ (1). We need to show that every nonempty open set is non-meager.

1. Let $V$ be a nonempty open set. Assume $V$ to be meager.

2. Then $V$ is a countable union of rare sets:

   $$V=\bigcup _{n=1}^{\mathrm{\infty }}{A}_n$$

   where $A_n$ are rare, thus $\mathrm{int}\mathrm{cl}{A}_n=\varnothing$.

3. We can write $X$ as:

   $$\begin{array}{r}\begin{array}{rl}X& =(X\setminus V)\cup V\\ & =(X\setminus V)\cup A_1\cup A_2\cup \dots \\ & =(X\setminus V)\cup (\mathrm{cl}{A}_1)\cup (\mathrm{cl}{A}_2)\cup \dots .\end{array}\end{array}$$

   The last expression is correct since $\mathrm{cl}{A}_n\subseteq X$.

4. In this form, $X$ is a countable union of closed sets.

5. By our hypothesis (3), the open set:

   $$(\mathrm{int}(X\setminus V))\cup (\mathrm{int}(\mathrm{cl}{A}_1))\cup (\mathrm{int}(\mathrm{cl}{A}_2))\cup \cdots =\mathrm{int}(X\setminus V)$$

   is dense in $X$. Here, we used the fact that $A_n$ are rare.

6. Since $\mathrm{int}(X\setminus V)\subseteq X\setminus V$, hence $X\setminus V$ is also dense in $X$.

7. In particular $V\cap (X\setminus V)\ne \varnothing$ since a dense set has a nonempty intersection with every nonempty open set (Theorem 3.23).

8. But this is impossible since $V\cap (X\setminus V)=\varnothing$.

9. Thus, $V$ cannot be not a meager set.

10. We have established that any nonempty open $V$ is not a meager set.

11. Hence, $X$ is a Baire space.

3.7.8. Completion

Theorem 3.70

Let $(X,d)$ be a metric space and let $(Y,\rho )$ be a complete metric space. If $f:X\to Y$ with $A=\mathrm{dom}f$ is a uniformly continuous function on $A$ then $f$ has a unique uniformly continuous extension to the closure of $A$.

Proof. If a sequence $\{x_n\}$ of $A$ converges to a closure point $x=lim{x}_n$, then the sequence $\{f(x_n)\}$ of $Y$ converges to some limit $y=limf(x_n)$.

1. Let $x\in \mathrm{cl}A$.

2. There exists a sequence $\{x_n\}$ of $A$ such that $lim{x}_n=x$.

3. Consider the sequence $\{f(x_n)\}$ of $Y$.

4. Choose $ϵ>0$.

5. Since $f$ is uniformly continuous on $A$, hence there exists $\delta >0$ such that $\rho (f(x),f(y))<ϵ$ whenever $d(x,y)<\delta$.

6. Since $\{x_n\}$ is convergent and hence Cauchy, we can pick $n_0$ such that $d(x_m,x_n)<\delta$ for all $m,n>n_0$.

7. Thus, $\rho (f(x_m),f(x_n))<ϵ$ for all $m,n>n_0$.

8. Thus, $\{f(x_n)\}$ is a Cauchy sequence of $Y$.

9. Since $Y$ is complete, hence every sequence converges.

10. Thus, there is a limit $y=limf(x_n)$.

For any sequence $\{x_n\}$ of $A$ converging to $x=lim{x}_n$, the corresponding sequence $\{f(x_n)\}$ of $Y$ has the same limit.

1. Assume that $x=lim{x}_n=lim{y}_n$ where $\{x_n\}$ and $\{y_n\}$ are two different sequences converging to $x$.

2. Let $u=limf(x_n)$ and $v=limf(y_n)$. We claim that $u=v$.

3. Consider the sequence $\{z_n\}$ defined as $z_{2n}=x_n$ and $z_{2n-1}=y_n$.

4. It is easy to show that $\{z_n\}$ converges to $x$.

5. Then, $limf(z_n)$ exists in $Y$.

6. If a sequence converges, then all its subsequences converge to the same limit.

7. Thus, $limf(x_n)=limf(y_n)=limf(z_n)$.

8. Thus, $u=v$.

9. Therefore $limf(x_n)$ is independent of the choice of sequence $\{x_n\}$ as long as $x=lim{x}_n$.

We define a function $f^{\ast }:X\to Y$ with $\mathrm{dom}g=\mathrm{cl}A$ given the function $f:X\to Y$ with $\mathrm{dom}f=A$ as:

$$f^{\ast }(x)=limf(x_n)$$

where $\{x_n\}$ is any sequence converging to $x\in \mathrm{cl}A$; i.e., $x=lim{x}_n$.

$f^{\ast }$ is well defined since $limf(x_n)$ is independent of the choice of the sequence $\{x_n\}$ converging to $x\in \mathrm{cl}A$.

We next establish that $f^{\ast }$ is uniformly continuous.

1. Let $ϵ>0$.

2. We can choose $\delta >0$ such that $\rho (f(x),f(y))<ϵ$ whenever $d(x,y)<\delta$ for any $x,y\in A$.

3. Now let, $x,y\in \mathrm{cl}A$ satisfying $d(x,y)<\delta$.

4. Let $\{x_n\}$ and $\{y_n\}$ be convergent sequences of $A$ with $lim{x}_n=x$ and $lim{y}_n=y$.

5. Then $limd(x_n,y_n)=d(x,y)$ due to Theorem 3.34.

6. Since $d(x,y)<\delta$, we can pick $n_0$ such that $d(x_n,y_n)<\delta$ for all $n>n_0$.

7. But since $x_n,y_n\in A$ and $f$ is uniformly continuous, we have, $\rho (f(x_n),f(y_n))<ϵ$ for all $n>n_0$.

8. Since $\{f(x_n)\}$ and $\{f(y_n)\}$ are convergent sequences of $Y$, hence $lim\rho (f(x_n),f(y_n))=\rho (f(x),f(y))$ again due to Theorem 3.34.

9. Thus, $\rho (f(x),f(y))\le ϵ$.

10. Thus, $f^{\ast }$ is uniformly continuous on $\mathrm{cl}A$.

Definition 3.53 (Completion of a metric space)

A complete metric space $(Y,\rho )$ is called a completion of a metric space $(X,d)$ if there exists an isometry $f:(X,d)\to (Y,\rho )$ with $\mathrm{dom}f=X$ such that $f(X)$ is dense in $Y$.

If we think of $X$ and $f(X)$ as identical (up to an isometry), then we can think of $X$ as a subset of $Y$.

Theorem 3.71

Any two completions of a metric space are isometric.

Proof. Let $(Y_1,{\rho }_1)$ and $(Y_2,{\rho }_2)$ be two different completions of $(X,d)$.

1. Then there are isometries $f:X\to Y_1$ and $g:X\to Y_2$ with $\mathrm{dom}f=\mathrm{dom}g=X$.

2. $g$ is an isometry with $\mathrm{dom}g=X$ and $\mathrm{range}g=g(X)$.

3. $f^{-1}$ is an isometry with $\mathrm{dom}{f}^{-1}=f(X)$ and $\mathrm{range}{f}^{-1}=X$.

4. Then, $h=g\circ f^{-1}$ is an isometry from $Y_1$ to $Y_2$ with $\mathrm{dom}h=f(X)$ and $\mathrm{range}h=g(X)$.

5. $f(X)$ is dense in $Y_1$. Hence $\mathrm{cl}f(X)=Y_1$.

6. $h$ is uniformly continuous (since it is an isometry).

7. $Y_2$ is complete.

8. Then, due to Theorem 3.70, there exists a uniformly continuous extension $h^{\ast }$ of $h$ to all of $Y_1$.

We have established that $h^{\ast }$ is uniformly continuous.

We next show that $h^{\ast }$ is an isometry.

1. Let $u,v\in Y_1$ and $z=h^{\ast }(u)$ and $w=h^{\ast }(v)$.

2. There is a sequence $\{x_n\}$ of $X$ such that $limf(x_n)=u$.

3. There is a sequence $\{y_n\}$ of $X$ such that $limf(y_n)=v$.

4. Let $u_n=f(x_n)$ and $v_n=g(y_n)$.

5. Let $z_n=g(u_n)$ and $w_n=g(y_n)$.

6. We have $lim{z}_n=z$ and $lim{w}_n=w$.

7. Since $f$ is an isometry, hence $d(x_n,y_n)={\rho }_1(u_n,v_n)$.

8. Since $g$ is an isometry, hence $d(x_n,y_n)={\rho }_2(z_n,w_n)$.

9. Thus, ${\rho }_1(u_n,v_n)={\rho }_2(z_n,w_n)$.

10. Taking limits, we get ${\rho }_1(u,v)={\rho }_2(z,w)$.

11. Thus, $h^{\ast }$ is an isometry.

12. Since it is an isometry, hence it is injective.

We next show that $h^{\ast }$ is surjective (onto).

1. Let $v\in Y_2$.

2. There exists a sequence $\{x_n\}$ such that $limg(x_n)=v$.

3. There exists $u\in Y_1$ such that $limf(x_n)=u$.

4. Let $f(x_n)=u_n$ and $g(x_n)=v_n$.

5. Then, $h(u_n)=(g\circ f^{-1})(u_n)=g(x_n)=v_n$.

6. Thus, $h^{\ast }(u_n)=v_n$.

7. Then, since $h^{\ast }$ is uniformly continuous, $lim{u}_n=u$ \implies $lim{h}^{\ast }(u_n)=lim{v}_n=v=h^{\ast }(u)$.

8. Thus, for every $v\in Y_2$, there exists $u\in Y_1$ such that $h^{\ast }(u)=v$.

9. Thus, $h^{\ast }(u)$ is surjective.

Together since $h^{\ast }$ is an isometric which is onto, hence $Y_1$ and $Y_2$ are isometric.

Remark 3.10

If $f:(X,d)\to (Y,\rho )$ is an isometry with $\mathrm{dom}f=X$ and $Y$ is a complete metric space, then $\mathrm{cl}f(X)$ is a completion of $X$.

Proof. Let Z = $\mathrm{cl}f(X)$. Then $Z$ is a closed subset of $Y$. By Theorem 3.58, $(Z,\rho )$ is a complete subspace.

Now define $g:(X,d)\to (Z,\rho )$ as:

$$g(x)=f(x).$$

Then, $g$ is an isometry and $g(X)$ is dense in $Z$. Thus, $Z$ is a completion of $X$.

Theorem 3.72

Every metric space has a unique (up to an isometry) completion.

Proof. We prove this theorem by constructing a metric space which is a completion of a given metric space.

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

2. Fix an element $a\in X$.

3. Now, for every $x\in X$, we introduce a function $f_x:X\to \mathbb{R}$ defined as:

   $$f_x(y)=d(x,y)-d(y,a)\mathrm{\forall }y\in X.$$

   Note that $\mathrm{dom}{f}_x=X$.

4. From the triangular inequality we have:

   $$|f_x(y)|=|d(x,y)-d(y,a)|\le d(x,a)\mathrm{\forall }y\in X.$$

5. Thus, for a given $x$ and fixed $a$, the parametrized function $f_x$ is bounded by $d(x,a)$.

6. Thus, $f_x\in B(X)$, the space of bounded functions from $X$ to $\mathbb{R}$.

7. We established in Example 3.20 that the metric space $B(X)$ is a complete metric space.

We next show that the mapping $f:X\to B(X)$ given by $x\mapsto f_x$ is an isometry. Recall that the distance between two functions $g,h\in B(X)$ is given by:

$$D(g,h)=sup\{|g(x)-h(x)||x\in X\}.$$

Now for some $x,z\in X$ with corresponding function $f_x,f_z\in B(X)$, for any $y\in X$, we have:

$$\begin{array}{r}\begin{array}{rl}|f_x(y)-f_z(y)|& =|d(x,y)-d(y,a)-[d(z,y)-d(y,a)]|\\ & =|d(x,y)-d(z,y)|\le d(x,z).\end{array}\end{array}$$

At the same time:

$$|f_x(z)-f_z(z)|=|d(x,z)-d(z,z)|=d(x,z).$$

Thus,

$$D(f_x,f_z)=sup\{|f_x(y)-f_z(y)||y\in X\}=d(x,z).$$

Thus, $f$ is an isometry. Since $(B(X),D)$ is a complete metric space, hence as per Remark 3.10, $(\mathrm{cl}f(X),D)$ is a completion of $X$.

We have already established in Theorem 3.71 that any two completions of $X$ are isometric to each other. Thus, the completion is unique up to an isometry.

3.7.9. Isometries

Theorem 3.73 (Isometries preserve closed sets)

Let $(X,d)$ and $(Y,\rho )$ be complete metric spaces. Let $f:(X,d)\to (Y,\rho )$ be an isometry. Let $C\subset X$ be a closed set in $X$. Then, $f(C)$ is closed.

Proof. 1. Let $\{y_n\}$ be a convergent sequence of $f(C)$.

1. Let $y=lim{y}_n$.

2. Recall that an isometric is injective.

3. Thus, $y_n=f(x_n)$ for some $x_n\in C$ for every $n$.

4. Since $\{y_n\}$ is convergent, hence $\{y_n\}$ is Cauchy.

5. Now, for any $m,n\in \mathbb{N}$

   $$d(x_m,x_n)=\rho (y_m,y_n)$$

   as $f$ is isometric.

6. Thus, $\{x_n\}$ is a Cauchy sequence of $C$.

7. Since $X$ is complete, hence every Cauchy sequence converges.

8. Let $x=lim{x}_n$.

9. Since $C$ is closed, hence $x\in C$.

10. By continuity of $f$, we have $f(x)=y$.

11. Thus, $y=f(x)\in f(C)$.

12. Thus, every convergent sequence of $C$ converges in $C$.

13. Thus, $f(C)$ is closed.

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