Limit of a Function in Metric Spaces

Metric Spaces

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Definition of Limit of a Function at a Point in Metric Spaces

Understand the Limit of a Function in Metric Spaces through the Sequential Limit Theorem, including its formal proof and interpretation.

Limit of a Function at a Point

Let \( (\mathbf{X}, d_{\mathbf{X}}) \) and \( (\mathbf{Y}, d_{\mathbf{Y}}) \) be two metric spaces, and let \( A \subseteq \mathbf{X} \). Let \( f: A \mapsto \mathbf{Y} \), \( a \in \mathbf{X} \) be a limit point of \( A \), and \( b \in \mathbf{Y} \).

We say that

\lim_{x \to a} f(x) = b

if for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that for all \( x \in A \), we have

d_{\mathbf{Y}}(f(x), b) \lt \epsilon \quad \text{whenever} \quad 0 \lt d_{\mathbf{X}}(x, a) \lt \delta

Sequential Characterization of Limits in Metric Spaces

Let \( (\mathbf{X}, d_{\mathbf{X}}) \) and \( (\mathbf{Y}, d_{\mathbf{Y}}) \) be two metric spaces, and let \( A \subseteq \mathbf{X} \). Suppose \( a \in \mathbf{X} \) is a limit point of \( A \), and let \( f: A \to \mathbf{Y} \). Then

\lim_{x \to a} f(x) = b

if and only if for every sequence \( \{x_n\} \subset A \setminus \{a\} \) such that \( \lim_{n \to \infty} x_n = a \), we have

\lim_{n \to \infty} f(x_n) = b.

Given that \( (\mathbf{X}, d_{\mathbf{X}}) \) and \( (\mathbf{Y}, d_{\mathbf{Y}}) \) are two metric spaces, and \( A \subseteq \mathbf{X} \). Let \( a \in \mathbf{X} \) be a limit point of \( A \), and \( f: A \to \mathbf{Y} \).

Assume \( \lim\limits_{x \to a} f(x) = b \).

Let \( \{x_n\} \subset A \setminus \{a\} \) be any sequence such that

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

To prove: \( \lim\limits_{n \to \infty} f(x_n) = b \).

Let \( \epsilon \gt 0 \). Since \( \lim\limits_{x \to a} f(x) = b \), by definition of limit in a metric space, there exists \( \delta \gt 0 \) such that for all \( x \in A \),

d_{\mathbf{Y}}(f(x), b) \lt \epsilon \quad \text{whenever} \quad 0 \lt d_{\mathbf{X}}(x, a) \lt \delta

Since \( \lim\limits_{n \to \infty} x_n = a \), there exists \( N_{0} \in \mathbb{N} \) such that

d_{\mathbf{X}}(x_n, a) \lt \delta \quad \forall n \geq N_{0}

And by assumption, \( x_n \neq a \), so

0 \lt d_{\mathbf{X}}(x_n, a) \lt \delta \quad \forall n \geq N_{0}

Therefore,

d_{\mathbf{Y}}(f(x_n), b) \lt \epsilon \quad \text{for all } n \geq N_{0}.

So,

\lim_{n \to \infty} f(x_n) = b.

Conversely, suppose for every sequence \( \{x_n\} \subset A \setminus \{a\} \) such that \( \lim\limits_{n \to \infty} x_n = a \), we have

\lim_{n \to \infty} f(x_n) = b.

To prove: \( \lim\limits_{x \to a} f(x) = b \).

If not, suppose \( \lim\limits_{x \to a} f(x) \neq b \). Then there exists \( \epsilon_0 \gt 0 \) such that for every \( \delta \gt 0 \), there exists \( x \in A \) with

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

Construct a sequence \( \{x_n\} \subset A \setminus \{a\} \) as follows: for each \( n \in \mathbb{N} \), take \( \delta = \frac{1}{n} \), then there exists \( x_n \in A \setminus \{a\} \) such that

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

This gives a sequence \( \{x_n\} \subset A \setminus \{a\} \) such that

\lim_{n \to \infty} x_n = a \quad \text{but} \quad d_{\mathbf{Y}}(f(x_n), b) \geq \epsilon_0 \text{ for all } n.

Hence,

\lim_{n \to \infty} f(x_n) \neq b,

which contradicts our assumption. Therefore,

\lim_{x \to a} f(x) = b.

Image and Preimage Relation

Let \( (\mathbf{X}, d_{\mathbf{X}}) \) and \( (\mathbf{Y}, d_{\mathbf{Y}}) \) be metric spaces, and let \( f: \mathbf{X} \to \mathbf{Y} \) be a function.

Let \( A \subseteq \mathbf{X} \) and \( B \subseteq \mathbf{Y} \).

Then,

f(A) \subseteq B \quad \text{if and only if} \quad A \subseteq f^{-1}(B),

where \( f(A) = \{ f(x) \mid x \in A \} \) and \( f^{-1}(B) = \{ x \in \mathbf{X} \mid f(x) \in B \} \).

Assume \( f(A) \subseteq B \). To prove \( A \subseteq f^{-1}(B) \):

Let \( x \in A \). Then \( f(x) \in f(A) \subseteq B \), so \( x \in f^{-1}(B) \). Since \( x \in A \) was arbitrary, \( A \subseteq f^{-1}(B) \).

Conversely, assume \( A \subseteq f^{-1}(B) \). To prove \( f(A) \subseteq B \):

Let \( y \in f(A) \). Then \( y = f(x) \) for some \( x \in A \). Since \( x \in f^{-1}(B) \), \( f(x) \in B \). Hence \( y \in B \), so \( f(A) \subseteq B \).

Continuity and Limit Equality in Metric Spaces

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

Suppose \( a \in \mathbf{X} \) is a limit point of \( A \), and let \( f: A \to \mathbf{Y} \). Then the function \( f \) is continuous at \( a \in A \) if and only if

\lim_{x \to a} f(x) = f(a).

Assume \( f \) is continuous at \( a \in A \).

To prove \( \lim_{x \to a} f(x) = f(a) \):

By the definition of continuity, for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that for all \( x \in A \),

d_{\mathbf{Y}}(f(x), f(a)) \lt \epsilon \quad \text{whenever} \quad d_{\mathbf{X}}(x, a) \lt \delta.

This includes points where \( x \neq a \), hence satisfies the condition for limit.

Conversely, assume \( \lim_{x \to a} f(x) = f(a) \).

To prove \( f \) is continuous at \( a \):

By the definition of limit, for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

0 \lt d_{\mathbf{X}}(x, a) \lt \delta \Rightarrow d_{\mathbf{Y}}(f(x), f(a)) \lt \epsilon.

Also, when \( x = a \), the condition still holds since \( d_{\mathbf{Y}}(f(a), f(a)) = 0 \lt \epsilon \).

Hence, the function \( f \) is continuous at \( a \in A \).