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Complete Metric Space

A metric space \( (X, d) \) is called a complete metric space if every Cauchy sequence in \( X \) converges to a limit that is also in \( X \). That is, for every sequence \( (x_n) \) in \( X \), if

\forall \epsilon > 0, \exists N \in \mathbb{N} \text{ such that } m, n \geq N \Rightarrow d(x_m, x_n) < \varepsilon,

then there exists \( x \in X \) such that

\lim_{n \to \infty} x_n = x.

Space of Real Numbers is Complete

Prove that the set of real numbers \( \mathbb{R} \), equipped with the standard metric \( d(x, y) = |x - y| \), is a complete metric space.

We have

the metric space \( (\mathbb{R}, d) \), where the metric \( d \) is defined by:

d(x, y) = |x - y| \quad \text{for all } x, y \in \mathbb{R}.

Let \( \{x_n\} \) be a Cauchy sequence in \( \mathbb{R} \).

By definition of a Cauchy sequence, for every \( \varepsilon > 0 \), there exists a positive integer \( N \) such that for all \( m, n \geq N \):

d(x_n, x_m) = |x_n - x_m| < \varepsilon.

To show the limit exists.

By the Cauchy convergence criterion in \( \mathbb{R} \), there exists some \( x \in \mathbb{R} \) such that:

\lim_{n \to \infty} x_n = x.

To show that the sequence converges to \(x\) in the metric space.

Then for every \( \varepsilon > 0 \), there exists \( N_{0} \in \mathbb{N} \) such that

\begin{align*} & |x_n - x| \lt \varepsilon~\forall~n \geq N_{0} \\ \implies & d(x_n, x) \lt \varepsilon~\forall~n \geq N_{0} \end{align*}

Since the limit, \( x \), is a real number, we have \( x \in \mathbb{R} \). Therefore, the limit of the Cauchy sequence lies in the metric space.

Hence

\mathbb{R} \text{ is a complete metric space.}

Space \( \mathbb{R}^n \) is Complete

Prove that the Euclidean space \( \mathbb{R}^n \), with the metric \[ d(\mathbf{x}_m, \mathbf{x}_n) = \left[\sum_{i=1}^{n} \left(x_{m}^{(i)}-x_{n}^{(i)}\right)^{2}\right]^{\frac{1}{2}}, \] is a complete metric space.

We have

the metric space \( (\mathbb{R}^n, d) \), where the metric \( d \) is defined by:

d(\mathbf{x}_m, \mathbf{x}_n) = \left[ \sum_{i=1}^{n} \left( x_m^{(i)} - x_n^{(i)} \right)^2 \right]^{\frac{1}{2}}.

Let \( \{\mathbf{x}_k\} \) be a Cauchy sequence in \( \mathbb{R}^n \).

By definition of a Cauchy sequence, for every \( \varepsilon > 0 \), there exists a positive integer \( N \) such that for all \( m, n \geq N \):

d(\mathbf{x}_m, \mathbf{x}_n) = \left[ \sum_{i=1}^{n} \left( x_m^{(i)} - x_n^{(i)} \right)^2 \right]^{\frac{1}{2}} < \varepsilon.

Then for any \( \varepsilon > 0 \), there exists \( N_0 \in \mathbb{N} \) such that

\begin{align*} & \left[\sum_{i=1}^{n} \left(x_{m}^{(i)}-x_{n}^{(i)}\right)^{2}\right]^{\frac{1}{2}}< \varepsilon, \quad \forall~m, n \geq N_0 \\ \implies & \sum_{i=1}^{n} \left(x_{m}^{(i)}-x_{n}^{(i)}\right)^{2}< \varepsilon^{2}, \quad \forall~m, n \geq N_0 \\ \implies & \left(x_{m}^{(i)}-x_{n}^{(i)}\right)^{2}< \varepsilon^{2}, \quad \forall~m, n \geq N_0,~ i = 1, 2, \dots, n \\ \implies & \left|x_{m}^{(i)}-x_{n}^{(i)}\right|< \varepsilon, \quad \forall~m, n \geq N_0,~ i = 1, 2, \dots, n \end{align*}

To show the limit exists.

From the above inequality, for each coordinate \( i = 1, 2, \ldots, n \), the sequence \( \{x_k^{(i)}\} \subset \mathbb{R} \) is a Cauchy sequence in \( \mathbb{R} \).

Since \( \mathbb{R} \) is complete, each coordinate sequence converges. Let \( x^{(i)} = \lim_{k \to \infty} x_k^{(i)} \in \mathbb{R} \) for \( i = 1, \dots, n \).

Define the point \( \mathbf{x} = (x^{(1)}, x^{(2)}, \ldots, x^{(n)}) \in \mathbb{R}^n \).

To show that the sequence converges to \( \mathbf{x} \) in the metric space.

We claim that \( \lim_{k \to \infty} \mathbf{x}_k = \mathbf{x} \). Then for every \( \varepsilon > 0 \), since each \( x_k^{(i)} \to x^{(i)} \), there exists \( N_0 \in \mathbb{N} \) such that for all \( k \geq N_0 \):

\begin{align*} & \left|x_k^{(i)} - x^{(i)}\right| < \frac{\varepsilon}{\sqrt{n}} \quad \text{for each } i = 1, 2, \dots, n \\ \implies & \left(x_k^{(i)} - x^{(i)}\right)^2 < \frac{\varepsilon^2}{n} \quad \text{for each } i= 1, 2, \dots, n \\ \implies & \sum_{i=1}^{n} \left(x_k^{(i)} - x^{(i)}\right)^2 < n \cdot \frac{\varepsilon^2}{n} = \varepsilon^2 \\ \implies & \left[ \sum_{i=1}^{n} \left( x_k^{(i)} - x^{(i)} \right)^2 \right]^{\frac{1}{2}} < \varepsilon \quad \forall~k \geq N_0 \\ \implies & d(\mathbf{x}_k, \mathbf{x}) < \varepsilon \quad \forall~k \geq N_0 \end{align*}

This shows that \( \mathbf{x}_k \to \mathbf{x} \) in the metric defined on \( \mathbb{R}^n \).

Since the limit \( \mathbf{x} \in \mathbb{R}^n \), the sequence converges in the metric space.

Hence

\mathbb{R}^n \text{ is a complete metric space.}

Space \( C[a, b] \) is complete

Prove that the space \( C[a, b] \) of all real-valued continuous functions on a closed interval \([a, b]\), equipped with the metric

d(f, g) = \sup_{x \in [a, b]} |f(x) - g(x)|,

is a complete metric space.

We have

the metric space \( (C[a, b], d) \), where \( C[a, b] \) is the set of all real-valued continuous functions on the closed interval \([a, b]\), and the metric \( d \) is defined by:

d(f, g) = \sup_{x \in [a, b]} |f(x) - g(x)|, \quad \text{for all } f, g \in C[a, b].

Let \( \{f_n\} \) be a Cauchy sequence in \( C[a, b] \).

By definition of a Cauchy sequence, for every \( \varepsilon > 0 \), there exists a positive integer \( N \in \mathbb{N} \) such that for all \( m, n \geq N \):

\tag{1} d(f_n, f_m) = \sup_{x \in [a, b]} |f_n(x) - f_m(x)| < \varepsilon.

Then for any \( \varepsilon > 0 \), there exists \( N_0 \in \mathbb{N} \) such that

\begin{align} & \sup_{x \in [a, b]} |f_n(x) - f_m(x)| < \frac{\varepsilon}{2}, \quad \forall~n, m \geq N_0\\ \nonumber \implies & |f_n(x) - f_m(x)| < \varepsilon, \quad \forall~n, m \geq N_0,~x \in [a, b] \end{align}

To show the limit exists.

Fix any \( x \in [a, b] \). Then \( \{f_n(x)\} \subset \mathbb{R} \) is a Cauchy sequence of real numbers. Since \( \mathbb{R} \) is complete, for each \( x \in [a, b] \), there exists a real number \( f(x) \in \mathbb{R} \) such that:

\tag{2} \lim_{n \to \infty} f_n(x) = f(x).

Thus, we define a pointwise limit function \( f : [a, b] \to \mathbb{R} \) by \( f(x) = \lim_{n \to \infty} f_n(x) \).

To show that the limit function \( f \in C[a, b] \).

Let us prove that the pointwise limit function \( f(x) = \lim_{n \to \infty} f_n(x) \) is continuous on \([a, b]\), i.e., \( f \in C[a, b] \).

Recall that each \( f_n \in C[a, b] \), i.e., each \( f_n \) is continuous on \([a, b] \). We also know from equation (1) that the convergence \( f_n \to f \) is uniform, because for every \( \varepsilon > 0 \), there exists \( N \in \mathbb{N} \) such that:

\sup_{x \in [a, b]} |f_n(x) - f_m(x)| < \varepsilon, \quad \forall~n, m \geq N.

Hence, \( f_n \to f \) uniformly on \([a, b] \). That is, for every \( \varepsilon > 0 \), there exists \( N \in \mathbb{N} \) such that:

|f_n(x) - f(x)| < \varepsilon, \quad \forall~x \in [a, b],\; n \geq N.

Now, let us prove that \( f \) is continuous at every point \( x_0 \in [a, b] \). Fix such an \( x_0 \).

Since \( f_N \) is continuous at \( x_0 \), given \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that for all \( x \in [a, b] \) with \( |x - x_0| < \delta \):

|f_N(x) - f_N(x_0)| < \frac{\varepsilon}{3}. \tag{3}

Also, from uniform convergence, we have:

|f(x) - f_N(x)| < \frac{\varepsilon}{3}, \quad \text{and} \quad |f(x_0) - f_N(x_0)| < \frac{\varepsilon}{3}, \tag{4}

Now estimate \( |f(x) - f(x_0)| \) as follows:

\begin{align*} |f(x) - f(x_0)| &\leq |f(x) - f_N(x)| + |f_N(x) - f_N(x_0)| + |f_N(x_0) - f(x_0)| \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. \end{align*}

Hence, for all \( x \in [a, b] \) such that \( |x - x_0| < \delta \), we have \( |f(x) - f(x_0)| < \varepsilon \). This shows that \( f \) is continuous at \( x_0 \).

Since \( x_0 \in [a, b] \) was arbitrary, \( f \) is continuous on the entire interval \([a, b] \), i.e., \( f \in C[a, b] \).

To show that the sequence converges to \( f \) in the metric space.

Given \( \varepsilon > 0 \), from (1) and the definition of \( f \) in (2), there exists \( N_0 \in \mathbb{N} \) such that for all \( n \geq N_0 \):

\begin{align*} & |f_n(x) - f(x)| < \varepsilon \quad \text{for all } x \in [a, b] \\ \implies & \sup_{x \in [a, b]} |f_n(x) - f(x)| < \varepsilon \\ \implies & d(f_n, f) < \varepsilon \quad \forall~n \geq N_0 \end{align*}

This shows that \( f_n \to f \) in the metric defined on \( C[a, b] \).

Since the limit function \( f \in C[a, b] \), the sequence converges in the metric space.

Hence

C[a, b] \text{ is a complete metric space.}

Space of all bounded sequences of real numbers is complete

d(f, g) = \sup_{n\in \mathbb{N} } |x_{n} - y_{n}|,

is a complete metric space.

Let \( \ell^\infty = \{ x = (x_n)_{n \in \mathbb{N}} \subset \mathbb{R} \mid \sup_n |x_n| < \infty \} \).

Metric:

d(x, y) = \sup_{n \in \mathbb{N}} |x_n - y_n|.

To prove: \( (\ell^\infty, d) \) is a complete metric space.

Let \( \{x^{(k)}\} \subset \ell^\infty \) be a Cauchy sequence. That is, for every \( \varepsilon > 0 \), there exists \( N \in \mathbb{N} \) such that for all \( k, m \geq N \),

d(x^{(k)}, x^{(m)}) = \sup_n |x^{(k)}_n - x^{(m)}_n| < \varepsilon.

Fix \( n \in \mathbb{N} \). Then \( \{x^{(k)}_n\}_{k \in \mathbb{N}} \subset \mathbb{R} \) is a Cauchy sequence in \( \mathbb{R} \). Since \( \mathbb{R} \) is complete, define

x_n = \lim_{k \to \infty} x^{(k)}_n.

Define \( x = (x_n) \). We show \( x \in \ell^\infty \) and \( x^{(k)} \to x \) in \( d \).

Step 1: Show \( x \in \ell^\infty \).

Since each \( x^{(k)} \in \ell^\infty \), we have for each \( k \in \mathbb{N} \):

\sup_{n \in \mathbb{N}} |x^{(k)}_n| \leq M_k.

As \( \{x^{(k)}\} \) is Cauchy in \( \ell^\infty \), it is eventually bounded uniformly. That is, there exists \( M > 0 \) such that for all large \( k \):

\sup_{n \in \mathbb{N}} |x^{(k)}_n| \leq M.

Then for each fixed \( n \in \mathbb{N} \),

|x_n| = \lim_{k \to \infty} |x^{(k)}_n| \leq \limsup_{k \to \infty} \sup_n |x^{(k)}_n| \leq M.

Hence, \( \sup_n |x_n| \leq M \), and so \( x \in \ell^\infty \).

Step 2: Show \( d(x^{(k)}, x) \to 0 \).

Given \( \varepsilon > 0 \), since \( \{x^{(k)}\} \) is Cauchy, choose \( N \in \mathbb{N} \) such that for all \( k, m \geq N \):

\sup_n |x^{(k)}_n - x^{(m)}_n| < \varepsilon.

Let \( m \to \infty \). Then \( x^{(m)}_n \to x_n \) for each \( n \), and we get:

\sup_n |x^{(k)}_n - x_n| \leq \varepsilon.

Hence, \( d(x^{(k)}, x) \to 0 \) as \( k \to \infty \).

Conclusion: \( x^{(k)} \to x \in \ell^\infty \). Thus, \( \ell^\infty \) is complete.