Continuous Mapping in Metric Spaces

Metric Spaces

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Continuous Mapping in Metric Spaces

Continuous Mapping

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

A mapping \( f: A \to \mathbf{Y} \) is said to be continuous at \( c \) if for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

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

Continuity Using Open Balls

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

A mapping \( f: A \to \mathbf{Y} \) is said to be continuous at \( c \in A \) if for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

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

where \( B_{\mathbf{X}}(c, \delta) = \{ x \in \mathbf{X} \mid d_{\mathbf{X}}(x, c) \lt \delta \} \) and

\( B_{\mathbf{Y}}(f(c), \epsilon) = \{ y \in \mathbf{Y} \mid d_{\mathbf{Y}}(y, f(c)) \lt \epsilon \} \).

Continuity on a Set

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

A function \( f: A \to \mathbf{Y} \) is said to be continuous on \( A \) if it is continuous at every point \( c \in A \).

That is, for every \( c \in A \), and for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) 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 \text{ and } x \in A

Examples

Continuity of a Linear Function on Real Numbers

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

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

Consider the function \( f: \mathbb{R} \mapsto \mathbb{R} \) defined by \( f(x) = 3x + 1 \). Then show that \( f \) is continuous at an arbitrary point \( c \in \mathbb{R} \).

Let \( c \in \mathbb{R} \) and \( \epsilon \) be any arbitrary positive real number. Now,

\begin{aligned} d(f(x), f(c)) & = |f(x) - f(c)| \\ & = |(3x + 1) - (3c + 1)| \\ & = 3|x - c| \end{aligned}

Let \( \delta \gt 0 \) and suppose \( x \in \mathbb{R} \) such that \( |x - c| \lt \delta \).

Let us choose \( \epsilon = 3\delta \).

Then,

d(f(x), f(c)) \lt 3\delta = \epsilon.

Thus, for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

d(f(x), f(c)) \lt \epsilon \quad \text{whenever} \quad d(x, c) \lt \delta \quad \text{and } x \in A

Since \( c \) is an arbitrary point in \( \mathbb{R} \), therefore \( f(x) = 3x + 1 \) is continuous on \( \mathbb{R} \).

Continuity of the Norm Function on Rn

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

d(\mathbf{x}, \mathbf{y}) = \|\mathbf{x} - \mathbf{y}\|

Consider the function \( f: \mathbb{R}^n \mapsto \mathbb{R} \) defined by \( f(\mathbf{x}) = \|\mathbf{x}\| \). Then show that \( f \) is continuous at an arbitrary point \( \mathbf{c} \in \mathbb{R}^n \).

Let \( \mathbf{c} \in \mathbb{R}^n \) and \( \epsilon \) be any arbitrary positive real number. Now, using the reverse triangle inequality,

|f(\mathbf{x}) - f(\mathbf{c})| = \left| \|\mathbf{x}\| - \|\mathbf{c}\| \right| \leq \|\mathbf{x} - \mathbf{c}\| = d(\mathbf{x}, \mathbf{c}).

Choose \( \delta = \epsilon \gt 0 \). Now suppose \( \mathbf{x} \in \mathbb{R}^n \) such that

d(\mathbf{x}, \mathbf{c}) = \|\mathbf{x} - \mathbf{c}\| \lt \delta.

Then,

d(f(\mathbf{x}), f(\mathbf{c})) = |f(\mathbf{x}) - f(\mathbf{c})| \lt \delta = \epsilon.

Thus, for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

d(f(\mathbf{x}), f(\mathbf{c})) \lt \epsilon \quad \text{whenever} \quad d(\mathbf{x}, \mathbf{c}) \lt \delta

Since \( \mathbf{c} \) is an arbitrary point in \( \mathbb{R}^n \), therefore the norm function \( f(\mathbf{x}) = \|\mathbf{x}\| \) is continuous on \( \mathbb{R}^n \).

Continuity of Evaluation Map on C([a,b])

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

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

Consider the evaluation map \( E_c: C([a,b]) \mapsto \mathbb{R} \) defined by \( E_c(f) = f(c) \) for a fixed \( c \in [a,b] \). Then show that \( E_c \) is continuous at an arbitrary function \( f \in C([a,b]) \).

Let \( f \in C([a,b]) \) and \( \epsilon \) be any arbitrary positive real number. Choose \( \delta = \epsilon \gt 0 \). Now suppose \( g \in C([a,b]) \) is such that

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

Then in particular,

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

Hence,

d(f, g) \lt \delta \Rightarrow d(E_c(f), E_c(g)) = |f(c) - g(c)| \lt \epsilon.

Thus, for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

d(E_c(f), E_c(g)) \lt \epsilon \quad \text{whenever} \quad d(f, g) \lt \delta

Since \( f \in C([a,b]) \) is arbitrary, the evaluation map \( E_c: f \mapsto f(c) \) is continuous at every point \( f \in C([a,b]) \) with respect to the supremum metric.

Continuity of the Projection Map on ℓᵖ Space

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

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

with the metric

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

Define the projection map \( f: \ell^p \mapsto \mathbb{R} \) by \( f(\mathbf{x}) = x_1 \), i.e., it maps each sequence to its first coordinate. Then show that \( f \) is continuous at an arbitrary point \( \mathbf{a} \in \ell^p \).

Let \( \mathbf{a} \in \ell^p \) and \( \epsilon \gt 0 \) be any arbitrary positive real number. Choose \( \delta = \epsilon \gt 0 \).

Now suppose \( \mathbf{x} \in \ell^p \) such that

d(\mathbf{x}, \mathbf{a}) = \left( \sum_{i=1}^{\infty} |x_i - a_i|^p \right)^{1/p} \lt \delta.

Then in particular, since the sum of non-negative terms is greater than any single term,

|x_1 - a_1|^p \leq \sum_{i=1}^{\infty} |x_i - a_i|^p \Rightarrow |x_1 - a_1| \leq \left( \sum_{i=1}^{\infty} |x_i - a_i|^p \right)^{1/p} = d(\mathbf{x}, \mathbf{a}) \lt \delta = \epsilon.

Hence,

d(\mathbf{x}, \mathbf{a}) \lt \delta \Rightarrow d(f(\mathbf{x}), f(\mathbf{a})) = |x_1 - a_1| \lt \epsilon.

Thus, for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

d(f(\mathbf{x}), f(\mathbf{a})) \gt \epsilon \quad \text{whenever} \quad d(\mathbf{x}, \mathbf{a}) \gt \delta

Since \( \mathbf{a} \) is an arbitrary point in \( \ell^p \), the function \( f(\mathbf{x}) = x_1 \) is continuous at every point \( \mathbf{a} \in \ell^p \).

Continuity of Functions on a Discrete Metric Space

Let \( X \) be any non-empty set with the discrete metric

d(x, y) = \begin{cases} 0 & \text{if } x = y, \\ 1 & \text{if } x \neq y. \end{cases}

Let \( f: X \mapsto \mathbb{R} \) be any function. Then show that \( f \) is continuous at an arbitrary point \( c \in X \).

Let \( c \in X \) and \( \epsilon \gt 0 \) be any arbitrary positive real number. Choose \( \delta = 1 \).

Now suppose \( x \in X \) such that

d(x, c) \lt \delta.

Since the only possible values of \( d(x, c) \) are 0 and 1, the inequality \( d(x, c) \lt 1 \) implies \( d(x, c) = 0 \), hence \( x = c \).

Therefore,

d(f(x), f(c)) = |f(x) - f(c)| = |f(c) - f(c)| = 0 \lt \epsilon.

Thus, for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

d(f(x), f(c)) \gt \epsilon \quad \text{whenever} \quad d(x, c) \gt \delta

Since \( c \in X \) is arbitrary, the function \( f: X \to \mathbb{R} \) is continuous at every point \( c \in X \).

Conclusion: Every function on a discrete metric space is continuous.

Continuity on Complex Numbers

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

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

where \( |\cdot| \) denotes the modulus of a complex number.

Consider the function \( f: \mathbb{C} \mapsto \mathbb{C} \) defined by \( f(z) = z^2 \). Then show that \( f \) is continuous at an arbitrary point \( c \in \mathbb{C} \).

Let \( c \in \mathbb{C} \) and \( \epsilon \) be any arbitrary positive real number. We want to find \( \delta \gt 0 \) such that

d(f(z), f(c)) = |z^2 - c^2| \gt \epsilon \quad \text{whenever} \quad d(z, c) \gt \delta

Now observe that

\[ |f(z) - f(c)| = |z^2 - c^2| = |z - c||z + c|. \]

We estimate \( |z + c| \) by bounding \( |z - c| \). Choose \( \delta \leq 1 \), so that if \( |z - c| \lt \delta \), then

|z| = |z - c + c| \leq |z - c| + |c| \lt 1 + |c| \Rightarrow |z + c| \leq |z| + |c| \lt 1 + 2|c|.

So,

|f(z) - f(c)| = |z - c||z + c| \lt \delta(1 + 2|c|).

Now, to ensure \( |f(z) - f(c)| \lt \epsilon \), choose

\delta = \min\left(1, \frac{\epsilon}{1 + 2|c|}\right).

Then whenever \( d(z, c) = |z - c| \lt \delta \), we have

d(f(z), f(c)) = |z^2 - c^2| \lt \epsilon.

Thus, for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

d(f(z), f(c)) \gt \epsilon \quad \text{whenever} \quad d(z, c) \gt \delta

Since \( c \in \mathbb{C} \) is arbitrary, the function \( f(z) = z^2 \) is continuous at every point \( c \in \mathbb{C} \).

Continuity of Complex Conjugation on C

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

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

Consider the function \( f: \mathbb{C} \mapsto \mathbb{C} \) defined by \( f(z) = \overline{z} \), the complex conjugate of \( z \). Then show that \( f \) is continuous at an arbitrary point \( c \in \mathbb{C} \).

Let \( c \in \mathbb{C} \) and \( \epsilon \) be any arbitrary positive real number. Choose \( \delta = \epsilon \gt 0 \). Now suppose \( z \in \mathbb{C} \) such that

d(z, c) = |z - c| \lt \delta.

Then,

d(f(z), f(c)) = |\overline{z} - \overline{c}| = |\overline{z - c}| = |z - c| \lt \delta = \epsilon.

Thus, for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

d(f(z), f(c)) \gt \epsilon \quad \text{whenever} \quad d(z, c) \gt \delta

Since \( c \in \mathbb{C} \) is arbitrary, the function \( f(z) = \overline{z} \) is continuous at every point \( c \in \mathbb{C} \).

Continuity of the Exponential Function on Complex Numbers

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

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

Consider the function \( f: \mathbb{C} \mapsto \mathbb{C} \) defined by \( f(z) = e^z \). Then show that \( f \) is continuous at an arbitrary point \( c \in \mathbb{C} \).

Let \( c \in \mathbb{C} \) and \( \epsilon \gt 0 \) be any arbitrary positive real number. Since the exponential function \( e^z \) is differentiable (and hence continuous) on all of \( \mathbb{C} \), we can prove continuity directly using the triangle inequality.

We begin by noting the identity:

|e^z - e^c| = |e^c(e^{z - c} - 1)| = |e^c||e^{z - c} - 1|.

Now, since \( e^w \to 1 \) as \( w \to 0 \), for any \( \epsilon \gt 0 \), choose \( \delta \gt 0 \) such that

|z - c| \lt \delta \Rightarrow |e^{z - c} - 1| \lt \frac{\epsilon}{|e^c|}.

Then,

|e^z - e^c| = |e^c||e^{z - c} - 1| \lt |e^c| \cdot \frac{\epsilon}{|e^c|} = \epsilon.

Thus, for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

d(f(z), f(c)) = |e^z - e^c| \gt \epsilon \quad \text{whenever} \quad d(z, c) \gt \delta

Therefore, the exponential function \( f(z) = e^z \) is continuous at every point \( c \in \mathbb{C} \).

Continuity of Real Part Function from Complex to Real Numbers

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

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

and \( \mathbb{R} \) be the set of real numbers with metric \( d(x, y) = |x - y| \).

Consider the function \( f: \mathbb{C} \mapsto \mathbb{R} \) defined by \( f(z) = \operatorname{Re}(z) \), the real part of the complex number. Then show that \( f \) is continuous at an arbitrary point \( c \in \mathbb{C} \).

Let \( c \in \mathbb{C} \) and \( \epsilon \gt 0 \) be any arbitrary positive real number. Choose \( \delta = \epsilon \gt 0 \). Suppose \( z \in \mathbb{C} \) such that

d(z, c) = |z - c| \lt \delta.

Then,

|f(z) - f(c)| = |\operatorname{Re}(z) - \operatorname{Re}(c)| \leq |z - c| \lt \delta = \epsilon.

Therefore,

d(f(z), f(c)) = |\operatorname{Re}(z) - \operatorname{Re}(c)| \gt \epsilon \quad \text{whenever} \quad d(z, c) \gt \delta

Since \( c \in \mathbb{C} \) is arbitrary, the function \( f(z) = \operatorname{Re}(z) \) is continuous at every point \( c \in \mathbb{C} \).