# Completeness

## Contents

# 3.7. Completeness#

The material in this section is primarily based on [2, 41].

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#

(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 \(x \in X\).

(Complete metric spaces)

\(\RR^n\) with the standard metric \(d(x,y) = | x-y|\) is complete.

The Euclidean space \(\RR^n\), with the standard Euclidean metric (\(\ell_2\) distance), is complete.

(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 \(D : B(X) \times B(X) \to \RR\) as:

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

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

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

By definition: \(D(f, g) = D(g, f)\).

For the triangle inequality, we proceed as follows:

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

\[ |f(x) - g(x)| \leq | f(x) - h(x)| + |h(x) - g(x)| \leq D(f,h) + D(h, g). \]Now, taking the supremum on the L.H.S. over all \(x \in X\), we get:

\[ D(f,g) \leq D(f,h) + D(h, g). \]

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

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

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

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

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

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

Since, \(\RR\) is complete, hence \(\{ f_n(x)\}\) converges at each \(x = X\).

We define a new function \(f: X \to \RR\) as \(f(x) = \lim f_n(x)\) for each \(x=X\).

Since \(|f_n(x) - f_m(x)| < \epsilon\) for all \(m,n>n_0\), hence \(|f_n(x) - f(x)| \leq \epsilon\) for all \(n > n_0\) and all \(x \in X\).

Thus, \(|f(x)| \leq \epsilon + |f_n(x)|\) for all \(x \in X\).

Thus, since \(f_n\) is bounded, hence \(f\) is also bounded.

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

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

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

Since \(D(f_n, f) \leq \epsilon\) for all \(n > n_0\), hence \(\lim f_n = f\).

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

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

## 3.7.2. Closed Subsets#

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.

Let \(\{ x_n \}\) be a Cauchy sequence of \(A\).

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

Thus, \(\{ x_n \}\) is convergent, since \(X\) is complete.

Let \(\lim x_n = x\).

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

Thus, every Cauchy sequence of \(A\) converges in \(A\).

Thus, \((A,d)\) is complete.

Let \(A\) be complete. We shall show that it contains all its closure points.

Let \(x \in X\) be a closure point of \(A\).

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

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

But then, \(\{ x_n \}\) is a Cauchy sequence of \(A\).

Since, \(A\) is complete, hence \(\{ x_n \}\) converges in \(A\).

Thus, \(x \in A\).

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.

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

Then the intersection \(\bigcap_{n=1}^{\infty}A_n\) 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:

We shall first prove that \(A\) cannot have more than one point.

Assume \(x, y \in A\).

Thus, \(x,y \in A_n \Forall n\).

Thus, \(0 \leq d(x,y) \leq \diam A_n \Forall n\).

But then, \(\lim_{n \to \infty} \diam A_n = 0\) implies that \(d(x, y) = 0\).

But, \(d(x,y) = 0 \implies x = y\) as \(d\) is a metric (Identity of indiscernibles).

Thus, if \(x,y \in \bigcap_{n=1}^{\infty}A_n\), then \(x=y\).

Thus, \(A\) contains at most one point.

We now show that \(A\) cannot be empty.

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

Since \(A_{n+p} \subseteq A_n\), hence \(d(x_{n+p}, x_n) \leq \diam A_n\) holds for every \(n, p\).

Thus, \(\{x_n\}\) is a Cauchy sequence of \(X\).

Since \(X\) is complete, \(\{x_n\}\) is convergent.

Hence, the limit \(x = \lim_{n\to\infty} x_n\) exists and \(x \in X\).

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

But since \(A_n\) is closed, hence \(x \in A_n\) for each \(n\).

Thus, \(x \in A\).

## 3.7.3. Nowhere Dense Sets#

(Nowhere dense/Rare)

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

It is also called a *rare* set.

\(A\) is nowhere dense if and only if \(X \setminus (\closure A)\) is dense in \(X\).

Proof. Let \(B = \closure A\).

Recall from Theorem 3.15 that:

Now,

The set of integers \(\ZZ\) is nowhere dense in \(\RR\).

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

Proof. Let \(A\) be open and dense.

\(B = X \setminus A\) is closed.

Thus, \(\closure B = B\).

Hence, \(X \setminus (\closure B) = X \setminus B = A\).

But \(A\) is dense, hence \(B\) must be nowhere dense due to Theorem 3.60.

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

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

The interior of its closure is empty.

Thus, its closure \(\closure A\) contains no open sets.

Thus, \(A\) contains no open sets.

Thus, every open set in \(X\) intersects with \(X \setminus A\).

Thus, \(X \setminus A\) is dense.

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

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

Since \(B\) is nowhere dense, hence \(\interior \closure B\) is empty. This in turn implies that \(\interior \closure A\) is empty. Thus, \(A\) is nowhere dense.

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

Proof. Let \(A\) be closed. Thus, \(A = \closure A\). Since \(\boundary A = \closure A \setminus \interior A\), hence \(\boundary A \subseteq A\) for a closed set.

Assume \(\boundary A\) has a nonempty interior.

Let \(x\) be an interior point of \(\boundary A\).

Since \(x \in \boundary A \subseteq A\), hence \(x\) is an interior point of \(A\).

But boundary of \(A\) doesnâ€™t includes its interior points.

Hence, a contradiction.

Thus, \(\boundary A\) has an empty interior.

Finally \(\interior \closure \boundary A\) = \interior \boundary A = \EmptySet\( since \)\boundary A$ is closed.

Thus, \(\boundary A\) is rare.

## 3.7.4. Cantor Set#

(Cantor set)

*Cantor set* \(C\) is a subset of \([0,1]\). It is constructed as follows:

Let \(C_0 = [0,1]\).

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

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

Note that \(C_1\) is a disjoint union of \(2=2^1\) closed intervals.

Trisect each closed interval of \(C_1\) and remove the middle open interval from each one of the trisections in an identical manner.

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

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

Inductively build \(C_{n+1}\) from \(C_n\) using this procedure.

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

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

Clearly \(C_{n+1} \subseteq C_{n}\) by construction for all \(n\).

Define the Cantor set as:

\[ C \triangleq \bigcap_{n=1}^{\infty} C_n. \]

We next discuss different properties of the Cantor set.

(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:

Then, \(E=C\). In other words:

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

such that

For example

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,

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:

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

\(C \subseteq E\). 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:

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

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

\(E \subseteq C\)

Let \(x = .d_1 d_2 \dots\) with \(d_i \in \{0, 2\}\).

Start with \(A_0 = [0,1]\).

Given \(A_0\):

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

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

Given \(A_{n-1} = [a, a + \frac{1}{3^{n-1}}]\),

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

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

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

Thus, \(x \in C\).

\(C \subseteq E\)

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

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

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

Subsequently, whenever the first closed interval is chosen, then \(d_n = 0\) and whenever the second closed interval is chosen, then \(d_n = 2\).

Thus, \(x\) has a ternary expansion consisting entirely of 0s and 2s.

Thus, \(x \in E\).

Cantor set doesnâ€™t contain any open interval.

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

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

Then, the length of the interval is \(b-a > 0\).

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

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

We arrive at a contradiction.

Cantor set has an empty interior.

Proof. Assume \(C\) has a nonempty interior and \(x \in \interior C\).

Then there is a neighborhood \((x - \epsilon , x + \epsilon) \subset C\).

But, \(C\) doesnâ€™t contain any open intervals.

We arrive at a contradiction.

Thus, \(C\) has an empty interior.

Cantor set is a closed nowhere dense subset of \(\RR\).

Proof. \(C\) is an (infinite) intersection of closed sets. Hence \(C\) is closed. \(\closure C = C\). \(C\) has an empty interior. Thus, \(C\) is nowhere dense.

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:

Cantor set is uncountable. In particular:

Recall that \(\mathfrak{c}\) denotes the cardinality of the continuum (Definition 1.92). The notation \(2^{\Nat}\) was introduced in Theorem 1.34 to describe power sets. Theorem 1.33 established that

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:

Thus, each \(x \in C\) can be identified with a sequence \(d : \Nat \to \{ 0, 2\}\) given by \(d = \{ d_n \}\).

Thus, we have a bijective mapping between \(C\) and the set \(\{0, 2\}^{\Nat}\).

Thus, \(C \sim \{0, 2\}^{\Nat}\).

This, in turn can be identified with a sequence \(c : \Nat \to \{0, 1\}\) where \(c = \{ c_n \}\) with \(c_n = 0\) if \(d_n = 0\) and \(c_n = 1\) if \(d_n = 2\).

This gives us a bijective mapping between \(\{0, 2\}^{\Nat}\) and \(\{0, 1\}^{\Nat}\).

Recall that \(2^{\Nat} = \{0, 1\}^{\Nat}\) with \(2 = \{0, 1\}\) (Theorem 1.34).

Thus \(C \sim 2^{\Nat}\) and

\[ \card C = \card 2^{\Nat} = \card \RR = \mathfrak{c}. \]

## 3.7.5. Meager Sets#

(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

(Co-meager set)

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

(Non-meager set)

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

Consider the metric space \(\RR\).

Singleton sets are rare (nowhere dense) in \(\RR\) as they are closed and their interior is empty.

The set of natural numbers \(\Nat\) is rare since it is closed and its interior is empty (no open intervals in \(\Nat\)).

The set \(\QQ\) is not rare since its closure is entire \(\RR\). It is meager since it is a countable union of rare singleton sets.

The set of irrational numbers \(\II\) is co-meager as \(\II = \RR \setminus \QQ\).

The set \(\RR\) is non-meager.

The subset of a meager set is meager.

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

where \(A_n\) are rare (nowhere dense).

But then

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

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

The union of countably many meager sets is meager.

Proof. Recall from Theorem 1.27 that countable union of countable sets is countable.

Let \(\{M_n\}\) be a countable collection of meager sets.

Then, each \(M_n\) is a countable union of rare sets.

Write \(M_n\) as \(\bigcup_i A_{n, i}\).

Thus,

\[ M = \bigcup_n M_n = \bigcup_n \bigcup_i A_{n, i} = \bigcup_{n, i} A_{n, i}. \]The family \(\{ A_{n, i} \}\) is countable.

Thus, \(M\) is a countable union of rare sets.

Thus, \(M\) is meager.

Since co-meager sets are complements of meager sets:

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

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

## 3.7.6. Baire Category Theorem#

(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:

For contradiction, we assume that \(X\) is meager with \(X = \bigcup_i A_i\).

We form a Cauchy sequence from points \(x_i \notin A_i\).

Since \(X\) is complete, we claim its convergence \(x = \lim x_i\).

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.

where \(A_k\) are rare (nowhere dense); i.e. \(\interior \closure A_k = \EmptySet\).

Recall from Theorem 3.60 that \(B_k = X \setminus \closure A_k\) 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.

\(B_1 = X \setminus \closure A_1\) is a nonempty open set.

Thus, there exists an interior point \(x_1\) in \(B_1\) with an open ball \(B(x_1, \epsilon_1) \subseteq B_1\) where we can make \(\epsilon_1 < \frac{1}{2}\).

\(B_2\) is nonempty, open and dense. Thus, \(C_2 = B_2 \cap B\left (x_1, \frac{\epsilon_1}{2}\right)\) is nonempty and open.

Thus, we can choose \(x_2 \in C_2\) such that \(x \in B(x_2, \epsilon_2) \subseteq C_2\) with \(\epsilon_2 \leq \frac{\epsilon_1}{2}\).

Proceeding inductively in this manner, we obtain a sequence \(\{ x_n \}\) such that

\[ x_n \in B(x_n, \epsilon_n) \subseteq B_n \cap B\left (x_{n-1}, \frac{\epsilon_{n-1}}{2} \right). \]We have \(A_n \cap B(x_n, \epsilon_n) = \EmptySet\). \(\epsilon_n \leq \frac{\epsilon_{n-1}}{2}\) with \(\epsilon_1 < \frac{1}{2}\).

By definition, \(x_m \in B\left (x_n, \frac{\epsilon_n}{2} \right)\) for every \(m > n\).

Thus, \(d(x_m, x_n) < \frac{\epsilon_n}{2}\) for every \(m > n\).

Since \(\epsilon_1 < \frac{1}{2}\), hence \(\epsilon_n < \frac{1}{2^{n+1}}\).

Thus, \(d(x_m, x_n) \to 0\) as \(m,n \to \infty\).

Thus, \(\{x_n\}\) is a Cauchy sequence.

Since \(X\) is complete, hence every Cauchy sequence is convergent.

Thus there exists \(x \in X\) such that \(x = \lim x_n\). To which \(A_k\) does \(x\) belong then?

Fix some \(n \in \Nat\). For every \(m>n\) we have:

\[ d(x, x_n) \leq d(x, x_m) + d(x_n, x_m) < d(x, x_m) + \frac{\epsilon_n}{2}. \]Thus taking the limit \(m \to \infty\) leading to \(d(x, x_m) \to 0\), we get

\[ d(x, x_n) \leq \frac{\epsilon_n}{2} < \epsilon_n. \]Hence \(x \in B(x_n, \epsilon_n)\) for all \(n \in \Nat\).

Since \(A_n \cap B(x_n, \epsilon_n) = \EmptySet\), hence \(x \notin A_n\) for all \(n\).

But then \(x \notin X\) since \(X = \bigcup_n A_n\). A contradiction.

Hence, \(X\) must be non-meager.

If \((X, d)\) is complete and \(X = \bigcup_{n=1}^{\infty} A_n\), then at least one \(A_n\) is non-rare; i.e., \(\interior \closure A_n \neq \EmptySet\) 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.

The set of irrational numbers is non-meager.

Proof. Recall that \(\RR = \QQ \cup \II\) where \(\II\) is the set of irrational numbers.

\(\RR\) is complete. Hence \(\RR\) is non-meager.

\(\QQ\) is meager as it is a countable union of singletons which are rare sets.

Countable union of meager sets is meager.

Thus, if \(\II\) was meager, then \(\RR\) would be meager which is not true.

Hence, \(II\) must be non-meager.

(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

such that \(A_n\) are rare in \(X\).

Let \(B_n = X \setminus \closure A_n\). Then \(B_n\) are dense, nonempty and open in \(X\). Hence \(B_n \cap U\) is nonempty and open for every nonempty open set \(U \subseteq X\).

Let \(U\) be an arbitrary nonempty open set in \(X\).

There exists \(x_1 \in B_1 \cap U\) such that \(x_1 \in B(x_1, \epsilon_1) \subseteq B_1 \cap U\) where \(\epsilon_1 < \frac{1}{2}\).

\(B_2\) is dense and open so \(C_2 = B_2 \cap B\left(x_1, \frac{\epsilon_1}{2}\right)\) is nonempty and open.

We can choose a point \(x_2 \in B(x_2, \epsilon_2) \subseteq C_2\) where \(\epsilon_2 \leq \frac{\epsilon_1}{2}\).

Proceeding in this manner, we choose points

\[ x_n \in B(x_n, \epsilon_n) \subseteq B_n \cap B\left (x_{n-1}, \frac{\epsilon_{n-1}}{2}\right) \]to form a sequence \(\{ x_n \}\).

The sequence \(\{ x_n \}\) is Cauchy, \(X\) is complete, hence \(x = \lim x_n\) exists in \(X\).

Also, \(x \in B(x_n, \epsilon_n)\) for all \(n\).

In particular \(x \in B(x_1, \epsilon_1) \subseteq U\) and \(x \in B(x_n, \epsilon_n) \subseteq B_n\).

Thus, \(x \in B_n \cap U\) for all \(n\).

Thus,

\[\begin{split} \begin{aligned} & x \in U \cap \left ( \bigcap_{n} B_n \right )\\ & \implies x \in U \cap \left ( \bigcap_{n} (X \setminus \closure A_n) \right )\\ & \implies x \in U \cap \left ( X \setminus \left ( \bigcup_{n} (\closure A_n) \right ) \right )\\ & \implies x \in U \cap \left ( X \setminus \left ( \bigcup_{n} A_n \right ) \right )\\ & \implies x \in U \cap ( X \setminus M ). \end{aligned} \end{split}\]Thus, \(U\) has a nonempty intersection with \(X \setminus M\).

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

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

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

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

\(\QQ\) in \(\RR\)

The set \(\QQ\) is meager in \(\RR\).

\(\RR\) is complete.

The interior of \(\QQ\) in \(\RR\) is empty.

The closure of \(\QQ\) is \(\RR\).

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.

\(\QQ\) in \(\QQ\)

\(\QQ\) by itself is not a complete metric space.

Singletons are rare sets in \(\QQ\).

Thus, \(\QQ\) is meager in \(\QQ\) as it is a countable union of rare sets.

However, the interior of \(\QQ\) is not empty. In fact it is whole of \(QQ\).

Theorem 3.66 does not apply since \(\QQ\) is not a complete metric space.

The set \([0,1]\) is uncountable.

Proof. We prove this using Baire category theorem.

\([0,1]\) is a complete metric space with the standard metric.

Assume \([0,1]\) to be countable.

Then, \([0,1] = \{ x_n \}_{n \in \Nat}\) is an enumeration of \([0,1]\).

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

Then \([0,1]\) being a countable union of rare sets would be meager.

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

We have a contradiction.

Thus, \([0,1]\) must be uncountable.

## 3.7.7. Baire Spaces#

(Baire space)

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

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.

(Characterization of Baire space)

For a metric space \(X\), the following statements are equivalent:

\(X\) is a Baire space.

Every countable intersection of open dense sets is also dense.

If \(X = \bigcup_{n=1}^{\infty} F_n\) and each \(F_n\) is a closed set, then the open set \(\bigcup_{n=1}^{\infty} \interior F_n\) is dense.

Proof. (1) \(\implies\) (2)

Assume \(X\) is a Baire space.

Let \(\{ A_n\}\) be a sequence of open dense sets in \(X\).

Let \(A = \bigcap_{n=1}^{\infty}A_n\). We need to show that \(A\) is dense.

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.

Let \(\OOO \subseteq X\) be an arbitrary nonempty open set.

Assume, for contradiction that \(A \cap \OOO = \EmptySet\).

Then, \(X = X \setminus \EmptySet = X \setminus (A \cap \OOO) = (X \setminus A) \cup (X \setminus O)\).

Thus,

\[\begin{split} \begin{aligned} \OOO &= X \cap \OOO = ((X \setminus A) \cup (X \setminus O)) \cap \OOO = (X \setminus A) \cap \OOO\\ &= \left (X \setminus \left ( \bigcap_{n=1}^{\infty}A_n \right ) \right) \cap \OOO\\ &= \bigcup_{n=1}^{\infty} ((X \setminus A_n) \cap \OOO). \end{aligned} \end{split}\]Due to Corollary 3.1, \(X \setminus A_n\) are rare.

Due to Proposition 3.20, \((X \setminus A_n) \cap \OOO\) are rare.

Thus, \(\OOO\) being a countable union of rare sets, is meager.

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

We have a contradiction.

Thus, \(A \cap \OOO\) must be nonempty.

Thus, \(A\) has a nonempty intersection with every nonempty open set in \(A\).

Thus, \(A\) is dense in \(X\).

(2) \(\implies\) (3)

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

Let \(\{ F_n \}\) be a sequence of closed sets in \(X\) satisfying \(X = \bigcup_{n=1}^{\infty} F_n\).

Let \(A_n = \interior F_n\) and let \(A = \bigcup_{i=1}^{\infty} A_n\). By definition, \(A\) is open.

Since \(F_n\) is closed, hence \(E_n = F_n \setminus (\interior F_n)\) is its boundary.

Due to Proposition 3.21, \(E_n\) is rare.

Thus, the set \(E = \bigcup_{n=1}^{\infty}E_n\) is a meager set.

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

By our hypothesis (2), the set

\[ X \setminus E = X \setminus \left ( \bigcup_{n=1}^{\infty}E_n \right ) = \bigcap_{n=1}^{\infty} (X \setminus E_n) \]is also a dense set as it is a countable intersection of open dense sets \(X \setminus E_n\).

Now, notice that:

\[\begin{split} \begin{aligned} X \setminus A &= X \setminus \left ( \bigcup_{i=1}^{\infty} A_n \right)\\ &= \bigcup_{n=1}^{\infty} F_n \setminus \left ( \bigcup_{i=1}^{\infty} (\interior F_n) \right)\\ &\subseteq \bigcup_{n=1}^{\infty} [F_n \setminus (\interior F_n)]\\ &= \bigcup_{n=1}^{\infty} E_n\\ &= E. \end{aligned} \end{split}\]And \(X \setminus A \subseteq E\) implies \(X \setminus E \subseteq A\).

Since \(X \setminus E\) is dense, hence \(A\) is also dense.

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

Let \(V\) be a nonempty open set. Assume \(V\) to be meager.

Then \(V\) is a countable union of rare sets:

\[ V = \bigcup_{n=1}^{\infty} A_n \]where \(A_n\) are rare, thus \(\interior \closure A_n = \EmptySet\).

We can write \(X\) as:

\[\begin{split} \begin{aligned} X &= (X \setminus V) \cup V\\ &= (X \setminus V) \cup A_1 \cup A_2 \cup \dots\\ &= (X \setminus V) \cup (\closure A_1) \cup (\closure A_2) \cup \dots. \end{aligned} \end{split}\]The last expression is correct since \(\closure A_n \subseteq X\).

In this form, \(X\) is a countable union of closed sets.

By our hypothesis (3), the open set:

\[ (\interior (X \setminus V)) \cup (\interior (\closure A_1)) \cup (\interior (\closure A_2)) \cup \dots = \interior (X \setminus V) \]is dense in \(X\). Here, we used the fact that \(A_n\) are rare.

Since \(\interior (X \setminus V) \subseteq X \setminus V\), hence \(X \setminus V\) is also dense in \(X\).

In particular \(V \cap (X \setminus V) \neq \EmptySet\) since a dense set has a nonempty intersection with every nonempty open set (Theorem 3.23).

But this is impossible since \(V \cap (X \setminus V) = \EmptySet\).

Thus, \(V\) cannot be not a meager set.

We have established that any nonempty open \(V\) is not a meager set.

Hence, \(X\) is a Baire space.

## 3.7.8. Completion#

Let \((X, d)\) be a metric space and let \((Y, \rho)\) be a complete metric space. If \(f : X \to Y\) with \(A = \dom f\) 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 \(y = \lim f(x_n)\).

Let \(x \in \closure A\).

There exists a sequence \(\{ x_n \}\) of \(A\) such that \(\lim x_n = x\).

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

Choose \(\epsilon > 0\).

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

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

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

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

Since \(Y\) is complete, hence every sequence converges.

Thus, there is a limit \(y = \lim f(x_n)\).

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.

Assume that \(x = \lim x_n = \lim y_n\) where \(\{x_n \}\) and \(\{ y_n\}\) are two different sequences converging to \(x\).

Let \(u = \lim f(x_n)\) and \(v = \lim f(y_n)\). We claim that \(u = v\).

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

It is easy to show that \(\{z_n \}\) converges to \(x\).

Then, \(\lim f(z_n)\) exists in \(Y\).

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

Thus, \(\lim f(x_n) = \lim f(y_n) = \lim f(z_n)\).

Thus, \(u = v\).

Therefore \(\lim f(x_n)\) is independent of the choice of sequence \(\{ x_n \}\) as long as \(x = \lim x_n\).

We define a function \(f^* : X \to Y\) with \(\dom g = \closure A\) given the function \(f : X \to Y\) with \(\dom f = A\) as:

where \(\{x_n\}\) is any sequence converging to \(x \in \closure A\); i.e., \(x = \lim x_n\).

\(f^*\) is well defined since \(\lim f(x_n)\) is independent of the choice of the sequence \(\{ x_n \}\) converging to \(x \in \closure A\).

We next establish that \(f^*\) is uniformly continuous.

Let \(\epsilon > 0\).

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

Now let, \(x,y \in \closure A\) satisfying \(d(x,y) < \delta\).

Let \(\{x_n\}\) and \(\{y_n\}\) be convergent sequences of \(A\) with \(\lim x_n = x\) and \(\lim y_n = y\).

Then \(\lim d(x_n, y_n) = d(x, y)\) due to Theorem 3.34.

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

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

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.

Thus, \(\rho(f(x), f(y)) \leq \epsilon\).

Thus, \(f^*\) is uniformly continuous on \(\closure A\).

(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 \(\dom f = X\)
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 \(Y\).

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 \((X,d)\).

Then there are isometries \(f : X \to Y_1\) and \(g : X \to Y_2\) with \(\dom f = \dom g = X\).

\(g\) is an isometry with \(\dom g = X\) and \(\range g = g(X)\).

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

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

\(f(X)\) is dense in \(Y_1\). Hence \(\closure f(X) = Y_1\).

\(h\) is uniformly continuous (since it is an isometry).

\(Y_2\) is complete.

Then, due to Theorem 3.70, there exists a uniformly continuous extension \(h^*\) of \(h\) to all of \(Y_1\).

We have established that \(h^*\) is uniformly continuous.

We next show that \(h^*\) is an isometry.

Let \(u,v \in Y_1\) and \(z = h^*(u)\) and \(w = h^*(v)\).

There is a sequence \(\{ x_n \}\) of \(X\) such that \(\lim f(x_n) = u\).

There is a sequence \(\{ y_n \}\) of \(X\) such that \(\lim f(y_n) = v\).

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

Let \(z_n = g(u_n)\) and \(w_n = g(y_n)\).

We have \(\lim z_n = z\) and \(\lim w_n = w\).

Since \(f\) is an isometry, hence \(d(x_n, y_n) = \rho_1(u_n, v_n)\).

Since \(g\) is an isometry, hence \(d(x_n, y_n) = \rho_2(z_n, w_n)\).

Thus, \(\rho_1(u_n, v_n) = \rho_2(z_n, w_n)\).

Taking limits, we get \(\rho_1(u, v) = \rho_2(z, w)\).

Thus, \(h^*\) is an isometry.

Since it is an isometry, hence it is injective.

We next show that \(h^*\) is surjective (onto).

Let \(v \in Y_2\).

There exists a sequence \(\{ x_n \}\) such that \(\lim g(x_n) = v\).

There exists \(u \in Y_1\) such that \(\lim f(x_n) = u\).

Let \(f(x_n) = u_n\) and \(g(x_n) = v_n\).

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

Thus, \(h^*(u_n) = v_n\).

Then, since \(h^*\) is uniformly continuous, \(\lim u_n = u\) \implies \(\lim h^*(u_n) = \lim v_n = v = h^*(u)\).

Thus, for every \(v \in Y_2\), there exists \(u \in Y_1\) such that \(h^*(u) = v\).

Thus, \(h^*(u)\) is surjective.

Together since \(h^*\) is an isometric which is onto, hence \(Y_1\) and \(Y_2\) are isometric.

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

Proof. Let Z = \(\closure f(X)\). 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:

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

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.

Let \((X,d)\) be a metric space.

Fix an element \(a \in X\).

Now, for every \(x \in X\), we introduce a function \(f_x : X \to \RR\) defined as:

\[ f_x(y) = d(x,y) - d(y,a) \Forall y \in X. \]Note that \(\dom f_x = X\).

From the triangular inequality we have:

\[ |f_x(y)| = |d(x,y) - d(y,a) | \leq d(x,a) \Forall y \in X. \]Thus, for a given \(x\) and fixed \(a\), the parametrized function \(f_x\) is bounded by \(d(x,a)\).

Thus, \(f_x \in B(X)\), the space of bounded functions from \(X\) to \(\RR\).

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:

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

At the same time:

Thus,

Thus, \(f\) is an isometry. Since \((B(X), D)\) is a complete metric space, hence as per Remark 3.10, \((\closure f(X), D)\) 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#

(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 \(f(C)\).

Let \(y = \lim y_n\).

Recall that an isometric is injective.

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

Since \(\{ y_n \}\) is convergent, hence \(\{ y_n \}\) is Cauchy.

Now, for any \(m,n \in \Nat\)

\[ d(x_m, x_n) = \rho(y_m, y_n) \]as \(f\) is isometric.

Thus, \(\{ x_n \}\) is a Cauchy sequence of \(C\).

Since \(X\) is complete, hence every Cauchy sequence converges.

Let \(x = \lim x_n\).

Since \(C\) is closed, hence \(x \in C\).

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

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

Thus, every convergent sequence of \(C\) converges in \(C\).

Thus, \(f(C)\) is closed.