Sequential Criterion for Continuity in Metric Spaces

Metric Spaces

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Sequential Criterion for Continuity

Let \( (\mathbf{X}, d_{\mathbf{X}} )\) and \( (\mathbf{Y}, d_{\mathbf{Y}} )\) be two metric spaces, and let \( A \subseteq \mathbf{X} \).

A function \( f: A \mapsto \mathbf{Y} \) is said to be continuous at \( c \in A \) if and only if \( f(x_n) \) converges to \( f(c) \) in \( \mathbf{Y} \) for every sequence \( \{x_n\} \) in \( A \) converging to \( c \).

Given that \( (\mathbf{X}, d_{\mathbf{X}} )\) and \( (\mathbf{Y}, d_{\mathbf{Y}} )\) are two metric spaces and \( A \subseteq \mathbf{X} \). Let \( f: A \mapsto \mathbf{Y} \) be a mapping.

Assume \( f \) is continuous at \( c \in A \). Then for any \( \epsilon > 0 \), there exists \( \delta > 0 \) such that for all \( x \in A \),

d_{\mathbf{Y}}(f(x), f(c)) \lt \epsilon \quad \text{whenever} \quad d_{\mathbf{X}}(x, c) \lt \delta, \quad x \in A

To prove: \( f(x_n) \to f(c) \) in \( \mathbf{Y} \) for every sequence \( \{x_n\} \subset A \) with \( x_n \to c \).

Let \( \{x_n\} \) be a sequence in \( A \) such that

\lim_{n \to \infty} x_n = c \quad \text{in } \mathbf{X}.

Then for this \( \delta \gt 0 \), there exists \( N_0 \in \mathbb{N} \) such that

d_{\mathbf{X}}(x_n, c) \lt \delta \quad \forall n \geq N_0.

Therefore,

d_{\mathbf{Y}}(f(x_n), f(c)) \lt \epsilon \quad \forall n \geq N_0.

Hence,

\lim_{n \to \infty} f(x_n) = f(c) \quad \text{in } \mathbf{Y}.

Conversely, suppose that for every sequence \( \{x_n\} \subset A \) with \( x_n \to c \), we have \( f(x_n) \to f(c) \).

To prove: \( f \) is continuous at \( c \).

Assume the contrary: \( f \) is not continuous at \( c \). Then there exists \( \epsilon_0 > 0 \) such that for every \( \delta > 0 \), there exists \( x \in A \) with

d_{\mathbf{X}}(x, c) \lt \delta \quad \text{but} \quad d_{\mathbf{Y}}(f(x), f(c)) \geq \epsilon_0.

For each \( n \in \mathbb{N} \), let \( \delta = \frac{1}{n} \). Then there exists \( x_n \in A \) such that

d_{\mathbf{X}}(x_n, c) \lt \frac{1}{n}, \quad \text{but} \quad d_{\mathbf{Y}}(f(x_n), f(c)) \geq \epsilon_0.

So \( \{x_n\} \to c \) in \( \mathbf{X} \), but \( f(x_n) \not\to f(c) \) in \( \mathbf{Y} \):

d_{\mathbf{Y}}(f(x_n), f(c)) \geq \epsilon_0 \quad \forall n.

This contradicts our assumption. Therefore, \( f \) is continuous at \( c \).

Let \( ( \mathbf{X}, d_{\mathbf{X}} )\) and \( ( \mathbf{Y}, d_{\mathbf{Y}} )\) be metric spaces, and let \( A \subseteq \mathbf{X} \).

A function \( f: A \mapsto \mathbf{Y} \) is continuous on \( A \) if and only if for every sequence \( \{x_n\} \) in \( A \) and every point \( c \in A \), the condition

\lim_{n \to \infty} x_n = c \quad \text{implies} \quad \lim_{n \to \infty} f(x_n) = f(c).

Sequential Continuity in Real Numbers

Let \( \mathbb{R} \) be the set of real numbers with the standard metric

\[ d(x, y) = |x - y|. \]

Define the function \( f: \mathbb{R} \mapsto \mathbb{R} \) by \( f(x) = x^2 \). Show that \( f \) is continuous at \( c \in \mathbb{R} \) using the sequential criterion.

Solution

Let \( c \in \mathbb{R} \) be arbitrary. To use the sequential criterion, let \( \{x_n\} \subset \mathbb{R} \) be any sequence such that

\lim_{n \to \infty} x_n = c \quad \text{in the metric } d(x, y) = |x - y|.

We want to show:

\lim_{n \to \infty} f(x_n) = f(c) = c^2.

That is, for every \( \epsilon > 0 \), there exists \( N \in \mathbb{N} \) such that for all \( n \geq N \),

|x_n^2 - c^2| \lt \epsilon.

Now,

\[ |x_n^2 - c^2| = |x_n - c||x_n + c|. \]

Since \( x_n \to c \), the sequence \( \{x_n\} \) is convergent, hence bounded. So there exists \( M > 0 \) such that

|x_n| \leq M \quad \text{for all sufficiently large } n.

Therefore, there exists \( K > 0 \) such that

|x_n + c| \leq |x_n| + |c| \leq M + |c| = K.

Then,

|x_n^2 - c^2| = |x_n - c||x_n + c| \lt |x_n - c| \cdot K.

Let \( \epsilon > 0 \). Choose \( \delta = \frac{\epsilon}{K} \). Since \( x_n \to c \), there exists \( N \in \mathbb{N} \) such that for all \( n \geq N \),

|x_n - c| \lt \delta = \frac{\epsilon}{K}.

Then for all \( n \geq N \),

|x_n^2 - c^2| \lt \frac{\epsilon}{K} \cdot K = \epsilon.

Hence,

\lim_{n \to \infty} f(x_n) = c^2 = f(c).

By the sequential criterion for continuity, \( f(x) = x^2 \) is continuous at every point \( c \in \mathbb{R} \).

Sequential Continuity in Complex Numbers

Let \( \mathbb{C} \) be the set of complex numbers with the standard metric

\[ d(z_1, z_2) = |z_1 - z_2|. \]

Define the function \( f: \mathbb{C} \mapsto \mathbb{C} \) by \( f(z) = \overline{z} \), the complex conjugate. Show that \( f \) is continuous at every point \( c \in \mathbb{C} \).

Let \( \{z_n\} \subset \mathbb{C} \) be a sequence such that

\lim_{n \to \infty} z_n = c.

We want to show

\lim_{n \to \infty} f(z_n) = \overline{c}.

Now,

|f(z_n) - f(c)| = |\overline{z_n} - \overline{c}| = |\overline{z_n - c}| = |z_n - c|.

Since \( |z_n - c| \to 0 \), it follows that \( |f(z_n) - f(c)| \to 0 \). Hence,

\lim_{n \to \infty} f(z_n) = \overline{c} = f(c).

So \( f \) is continuous at every \( c \in \mathbb{C} \).

Sequential Continuity in Rn

Let \( \mathbb{R}^n \) be the \( n \)-dimensional Euclidean space with the standard metric

d(\mathbf{x}, \mathbf{y}) = \sqrt{ \sum_{i=1}^n (x_i - y_i)^2 }.

Define the function \( f: \mathbb{R}^n \mapsto \mathbb{R} \) by \( f(\mathbf{x}) = x_1 \), the first coordinate. Show that \( f \) is continuous at any point \( \mathbf{c} \in \mathbb{R}^n \).

Let \( \{\mathbf{x}_n\} \subset \mathbb{R}^n \) be a sequence such that

\lim_{n \to \infty} \mathbf{x}_n = \mathbf{c}.

This means for every \( \epsilon > 0 \), there exists \( N \in \mathbb{N} \) such that for all \( n \geq N \),

d(\mathbf{x}_n, \mathbf{c}) \lt \epsilon.

By the definition of the Euclidean distance, we have

|x_n^{(1)} - c_1| \leq \sqrt{ \sum_{i=1}^n (x_n^{(i)} - c_i)^2 } = d(\mathbf{x}_n, \mathbf{c}) \to 0.

So,

\lim_{n \to \infty} f(\mathbf{x}_n) = \lim_{n \to \infty} x_n^{(1)} = c_1 = f(\mathbf{c}).

Thus, \( f \) is continuous at \( \mathbf{c} \in \mathbb{R}^n \).

Sequential Continuity in lp Space

Let \( \ell^p \) (for \( 1 \leq p \lt \infty \)) be the space of real sequences \( \mathbf{x} = (x_1, x_2, \dots) \) with

\sum_{i=1}^{\infty} |x_i|^p \lt \infty,

and metric

d(\mathbf{x}, \mathbf{y}) = \left( \sum_{i=1}^{\infty} |x_i - y_i|^p \right)^{1/p}.

Define \( f: \ell^p \mapsto \mathbb{R} \) by \( f(\mathbf{x}) = x_1 \). Show that \( f \) is continuous at every point \( \mathbf{a} \in \ell^p \).

Let \( \{\mathbf{x}_n\} \subset \ell^p \) be a sequence such that

\lim_{n \to \infty} \mathbf{x}_n = \mathbf{a} \text{ in the } \ell^p \text{ metric}.

Then,

\lim_{n \to \infty} d(\mathbf{x}_n, \mathbf{a}) = \lim_{n \to \infty} \left( \sum_{i=1}^{\infty} |x_n^{(i)} - a_i|^p \right)^{1/p} = 0.

In particular,

|x_n^{(1)} - a_1|^p \leq \sum_{i=1}^{\infty} |x_n^{(i)} - a_i|^p \to 0 \quad \Rightarrow \quad |x_n^{(1)} - a_1| \to 0.

Hence,

\lim_{n \to \infty} f(\mathbf{x}_n) = x_n^{(1)} = a_1 = f(\mathbf{a}).

Therefore, \( f \) is continuous at every \( \mathbf{a} \in \ell^p \).

Sequential Continuity in C([a,b])

Let \( C([a,b]) \) be the space of real-valued continuous functions on \( [a,b] \), equipped with the supremum metric

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

Fix \( c \in [a,b] \), and define the evaluation map \( E_c: C([a,b]) \mapsto \mathbb{R} \) by \( E_c(f) = f(c) \). Show that \( E_c \) is continuous at every function \( f \in C([a,b]) \).

Let \( \{f_n\} \subset C([a,b]) \) be a sequence such that

\lim_{n \to \infty} f_n = f \quad \text{in the supremum metric}.

Then,

\lim_{n \to \infty} \sup_{x \in [a,b]} |f_n(x) - f(x)| = 0.

In particular,

|f_n(c) - f(c)| \leq \sup_{x \in [a,b]} |f_n(x) - f(x)| \to 0.

Therefore,

\lim_{n \to \infty} E_c(f_n) = f(c) = E_c(f).

Hence, \( E_c \) is continuous at every function \( f \in C([a,b]) \).