Continuity Characterizations in Metric Spaces

Metric Spaces

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Continuity Characterization

Open Ball Characterization of Continuity

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

Then the function \( f \) is continuous at \( a \in \mathbf{X} \) if and only if for every \( \epsilon \gt 0 \), there exists \( \delta \gt 0 \) such that

B(a,\delta) \subseteq f^{-1}\left(B(f(a), \epsilon)\right),

where \( B(a, \delta) = \{ x \in \mathbf{X} \mid d_{\mathbf{X}}(x, a) \lt \delta \} \) is the open ball centered at \( a \), and \( B(f(a), \epsilon) = \{ y \in \mathbf{Y} \mid d_{\mathbf{Y}}(y, f(a)) \lt \epsilon \} \) is the open ball centered at \( f(a) \) in \( \mathbf{Y} \).

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

Let \( f \) be continuous at \( a \in \mathbf{X} \).

Let \( \epsilon \gt 0 \), then there exists \( \delta \gt 0 \) such that for all \( x \in \mathbf{X} \),

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

To prove: \( B(a,\delta) \subseteq f^{-1}\left(B(f(a), \epsilon)\right) \)

\begin{align*} & x \in B(a, \delta) \\ \Rightarrow\ & d_{\mathbf{X}}(x, a) \lt \delta \\ \Rightarrow\ & d_{\mathbf{Y}}(f(x), f(a)) \lt \epsilon \\ \Rightarrow\ & f(x) \in B(f(a), \epsilon), \\ \Rightarrow\ & x \in f^{-1}(B(f(a), \epsilon)). \end{align*}

Therefore,

B(a, \delta) \subseteq f^{-1}(B(f(a), \epsilon)).

Conversely

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

B(a, \delta) \subseteq f^{-1}(B(f(a), \epsilon)).

To prove: \( f \) is continuous at \( a \in \mathbf{X} \).

Let \( x \in \mathbf{X} \) with \( d_{\mathbf{X}}(x, a) \lt \delta \).

Then \( x \in B(a, \delta) \), so by assumption:

f(x) \in B(f(a), \epsilon) \Rightarrow d_{\mathbf{Y}}(f(x), f(a)) \lt \epsilon.

Thus, \( f \) is continuous at \( a \in \mathbf{X} \).

Topological Characterization of Continuity

Let \( (\mathbf{X}, d_{\mathbf{X}}) \) and \( (\mathbf{Y}, d_{\mathbf{Y}}) \) be two metric spaces, and let \( f: \mathbf{X} \to \mathbf{Y} \) be a function. Then \( f \) is continuous at every point \( a \in \mathbf{X} \) (i.e., continuous on \( \mathbf{X} \)) if and only if for every open set \( G \subseteq \mathbf{Y} \), the preimage \( f^{-1}(G) \) is open in \( \mathbf{X} \).

We have \( (\mathbf{X}, d_{\mathbf{X}}) \) and \( (\mathbf{Y}, d_{\mathbf{Y}}) \) as metric spaces, and \( f: \mathbf{X} \to \mathbf{Y} \).

Let \( f \) be continuous at every point \( a \in \mathbf{X} \), and let \( G \subseteq \mathbf{Y} \) be open.

To prove: \( f^{-1}(G) \) is open in \( \mathbf{X} \).

Let \( x \in f^{-1}(G) \). Then,

f(x) \in G.

Since \( G \) is open, there exists \( \epsilon \gt 0 \) such that

B(f(x), \epsilon) \subseteq G.

Since \( f \) is continuous at \( x \), there exists \( \delta \gt 0 \) such that for all \( z \in \mathbf{X} \),

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

Equivalently, \( f(z) \in B(f(x), \epsilon) \) whenever \( z \in B(x, \delta) \).

Hence, for all \( z \in B(x, \delta) \):

\begin{align*} & f(z) \in B(f(x), \epsilon) \\ \Rightarrow\ & f(z) \in G \quad \text{since} \quad B(f(x), \epsilon) \subseteq G \\ \Rightarrow\ & z \in f^{-1}(G) \end{align*}

So, \( B(x, \delta) \subseteq f^{-1}(G) \), implying \( f^{-1}(G) \) is open.

Conversely

Assume for every open set \( G \subseteq \mathbf{Y} \), the set \( f^{-1}(G) \) is open in \( \mathbf{X} \).

To prove: \( f \) is continuous at every point \( a \in \mathbf{X} \).

Let \( a \in \mathbf{X} \) and let \( \epsilon \gt 0 \) be arbitrary. Then \( B(f(a), \epsilon) \) is open in \( \mathbf{Y} \), and so \( f^{-1}(B(f(a), \epsilon)) \) is open in \( \mathbf{X} \).

Also, \( a \in f^{-1}(B(f(a), \epsilon)) \). So, there exists \( \delta \gt 0 \) such that

B(a, \delta) \subseteq f^{-1}(B(f(a), \epsilon)).

Thus, for all \( x \in \mathbf{X} \) such that \( d_{\mathbf{X}}(x, a) \lt \delta \), we have

\begin{align*} & d_{\mathbf{X}}(x, a) \lt \delta \\ \Rightarrow\ & x \in B(a, \delta) \\ \Rightarrow\ & x \in f^{-1}(B(f(a), \epsilon)) \\ \Rightarrow\ & f(x) \in B(f(a), \epsilon) \\ \Rightarrow\ & d_{\mathbf{Y}}(f(x), f(a)) \lt \epsilon \end{align*}

Hence, \( f \) is continuous at \( a \in \mathbf{X} \). Since \( a \) was arbitrary, \( f \) is continuous on \( \mathbf{X} \).

Closed Set Characterization of Continuity

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

Then \( f \) is continuous on \( \mathbf{X} \) (i.e., continuous at every point \( a \in \mathbf{X} \)) if and only if for every closed set \( G \subseteq \mathbf{Y} \), the preimage \( f^{-1}(G) \subseteq \mathbf{X} \) is closed.

We have \( (\mathbf{X}, d_{\mathbf{X}}) \) and \( (\mathbf{Y}, d_{\mathbf{Y}}) \) as metric spaces, and \( f: \mathbf{X} \to \mathbf{Y} \).

Assume \( f \) is continuous on \( \mathbf{X} \), and let \( G \subseteq \mathbf{Y} \) be closed.

To prove: \( f^{-1}(G) \) is closed in \( \mathbf{X} \).

Since \( G \) is closed, its complement \( G^c = \mathbf{Y} \setminus G \) is open in \( \mathbf{Y} \).

By continuity of \( f \), the preimage \( f^{-1}(G^c) \) is open in \( \mathbf{X} \). But,

f^{-1}(G^c) = \mathbf{X} \setminus f^{-1}(G)

Since the complement of \( f^{-1}(G) \) is open, it follows that \( f^{-1}(G) \) is closed in \( \mathbf{X} \).

Conversely

Assume that for every closed set \( G \subseteq \mathbf{Y} \), the set \( f^{-1}(G) \) is closed in \( \mathbf{X} \).

To prove: \( f \) is continuous on \( \mathbf{X} \).

Let \( U \subseteq \mathbf{Y} \) be open. Then \( U^c = \mathbf{Y} \setminus U \) is closed.

By assumption, \( f^{-1}(U^c) \) is closed. But,

f^{-1}(U^c) = \mathbf{X} \setminus f^{-1}(U)

Hence, \( f^{-1}(U) \) is open in \( \mathbf{X} \), showing that \( f \) is continuous.

Continuity of Composition of Functions

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

Let \( f: \mathbf{X} \to \mathbf{Y} \) and \( g: \mathbf{Y} \to \mathbf{Z} \) be functions.

If \( f \) is continuous at \( a \in \mathbf{X} \) and \( g \) is continuous at \( f(a) \in \mathbf{Y} \), then the composition \( g \circ f: \mathbf{X} \to \mathbf{Z} \), defined by \( (g \circ f)(x) = g(f(x)) \), is continuous at \( a \in \mathbf{X} \).

Let \( f: \mathbf{X} \to \mathbf{Y} \) and \( g: \mathbf{Y} \to \mathbf{Z} \), and let \( a \in \mathbf{X} \).

To prove: \( g \circ f \) is continuous at \( a \).

Let \( \epsilon \gt 0 \). Since \( g \) is continuous at \( f(a) \), there exists \( \delta_1 \gt 0 \) such that for all \( y \in \mathbf{Y} \):

d_{\mathbf{Z}}(g(y), g(f(a))) \lt \epsilon \quad \text{whenever} \quad d_{\mathbf{Y}}(y, f(a)) \lt \delta_1

Since \( f \) is continuous at \( a \), there exists \( \delta_2 \gt 0 \) such that for all \( x \in \mathbf{X} \):

d_{\mathbf{Y}}(f(x), f(a)) \lt \delta_1 \quad \text{whenever} \quad d_{\mathbf{X}}(x, a) \lt \delta_2

Therefore, for all \( x \in \mathbf{X} \) such that \( d_{\mathbf{X}}(x, a) \lt \delta_2 \):

\begin{align*} & d_{\mathbf{Y}}(f(x), f(a)) \lt \delta_1 \\ \Rightarrow\ & d_{\mathbf{Z}}(g(f(x)), g(f(a))) \lt \epsilon \\ \Rightarrow\ & d_{\mathbf{Z}}((g \circ f)(x), (g \circ f)(a)) \lt \epsilon \end{align*}

Hence, \( g \circ f \) is continuous at \( a \in \mathbf{X} \).

Closure and Continuity: Preimage Containment

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

If \( f \) is continuous on \( \mathbf{X} \), then for any subset \( B \subseteq \mathbf{Y} \), we have:

\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})

We have \( (\mathbf{X}, d_{\mathbf{X}}) \) and \( (\mathbf{Y}, d_{\mathbf{Y}}) \) are two metric spaces, and let \( f: \mathbf{X} \mapsto \mathbf{Y} \) be a function.

Let \( f \) be continuous on \( \mathbf{X} \), and \( B \subseteq \mathbf{Y} \).

We aim to prove \(\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})\).

Since \( \overline{B} \subseteq \mathbf{Y} \) is closed and \( f \) is continuous, \( f^{-1}(\overline{B}) \subseteq \mathbf{X} \) is closed. Therefore:

f^{-1}(\overline{B}) = \overline{f^{-1}(\overline{B})}

\begin{align*} & B \subseteq \overline{B} \\ \Rightarrow & f^{-1}(B) \subseteq f^{-1}(\overline{B}) \\ \Rightarrow & \overline{f^{-1}(B)} \subseteq \overline{f^{-1}(\overline{B})} \\ \Rightarrow & \overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B}) \end{align*}

Hence, the result follows.

Closure Implies Image Preimage Inclusion

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

If for any subset \( B \subseteq \mathbf{Y} \),

\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})

then for any subset \( A \subseteq \mathbf{X} \), we have:

f(\overline{A}) \subseteq \overline{f(A)}

Let \( B = f(A) \subseteq \mathbf{Y} \). Then:

\begin{align*} & A \subseteq f^{-1}(B) \\ \Rightarrow & \overline{A} \subseteq \overline{f^{-1}(B)} \\ \Rightarrow & \overline{A} \subseteq f^{-1}(\overline{B}) \\ \Rightarrow & f(\overline{A}) \subseteq f(f^{-1}(\overline{B})) \\ \Rightarrow & f(\overline{A}) \subseteq \overline{B} = \overline{f(A)} \end{align*}

Continuity via Closure of Preimages

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

If for any subset \( A \subseteq \mathbf{X} \), \( f(\overline{A}) \subseteq \overline{f(A)} \), then \( f \) is continuous on \( \mathbf{X} \).

Let \( B \subseteq \mathbf{Y} \) be closed and \( A = f^{-1}(B) \subseteq \mathbf{X} \). We aim to prove \( A \) is closed.

\begin{align*} & f(\overline{A}) \subseteq \overline{f(A)} = \overline{f(f^{-1}(B))} \\ \Rightarrow & f(\overline{A}) \subseteq \overline{(f \circ f^{-1})(B)} \subseteq \overline{B} = B \\ \Rightarrow & \overline{A} \subseteq f^{-1}(B) = A \end{align*}

Since \( A \subseteq \overline{A} \), we conclude \( A = \overline{A} \), i.e., \( A \) is closed. Hence, \( f \) is continuous.