Discrete and Disconnectedness in Metric Spaces

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Discrete and Disconnectedness in Metric Spaces: Continuity Characterizations Explained

This post explores Discrete and Disconnectedness in Metric Spaces through the lens of continuity. Learn how discrete two-point metric spaces help characterize when a metric space is connected or disconnected. Discover the role of continuous mappings in identifying topological structure and gain insight into how function behavior reflects the connectedness of spaces.

Discrete Two-Point Metric Space

Let us consider \( \mathbf{X_0} = \{a, b\} \).

Define a function \( d_0 : X_0 \times X_0 \to \mathbb{R} \) by

d_0(x, y) = \begin{cases} 0 & \text{if } x = y, \\ 1 & \text{if } x \neq y. \end{cases}

Then \( (\mathbf{X_0}, d_0) \) is called a discrete two-point metric space.

Disconnection and Continuous Mapping to Two-Point Space

Let \( (\mathbf{X}, d) \) be a metric space. Then \( \mathbf{X} \) is disconnected if and only if there exists a continuous mapping of \( (\mathbf{X}, d) \) onto a discrete two-point metric space \( (\mathbf{X_0}, d_0) \).

Given that \( (\mathbf{X}, d) \) is a metric space and \( \mathbf{X} \) is disconnected.

To prove: there exists a continuous mapping from \( \mathbf{X} \) onto a discrete two-point metric space.

Since \( \mathbf{X} \) is disconnected, there exist two non-empty disjoint open subsets \( A \) and \( B \) such that \( \mathbf{X} = A \cup B \).

Define a function \( f: \mathbf{X} \to \mathbf{X_0} \) as:

f(x) = \begin{cases} 0 & \text{when } x \in A \\ 1 & \text{when } x \in B \end{cases}

Clearly, \( f \) is onto.

To prove: \( f \) is continuous.

The only open sets in \( \mathbf{X_0} \) are \( \Phi, \{0\}, \{1\}, \{0,1\} \). We compute the preimages:

\( f^{-1}(\Phi) = \Phi \) — open in \( \mathbf{X} \)
\( f^{-1}(\{0\}) = A \) — open in \( \mathbf{X} \)
\( f^{-1}(\{1\}) = B \) — open in \( \mathbf{X} \)
\( f^{-1}(\{0,1\}) = \mathbf{X} \) — open in \( \mathbf{X} \)

Thus, the preimage of every open set in \( \mathbf{X}_0 \) is open in \( \mathbf{X} \), so \( f \) is continuous.

Conversely, assume a continuous and onto function \( f : \mathbf{X} \to \mathbf{X_0} = \{0,1\} \) exists.

Let \( A = f^{-1}(\{0\}) \), \( B = f^{-1}(\{1\}) \). These are non-empty, disjoint, open subsets such that \( \mathbf{X} = A \cup B \). Hence, \( \mathbf{X} \) is disconnected.

Characterization of Connectedness via Constant Mappings

Let \( (\mathbf{X}, d) \) be a metric space. Then \( \mathbf{X} \) is connected if and only if the only continuous mappings from \( \mathbf{X} \) into a discrete two-point metric space \( (\mathbf{X}_0, d_0) \) are constant functions:

f(x) = 1 \text{ for all } x \in \mathbf{X}, \quad \text{or} \quad f(x) = 0 \text{ for all } x \in \mathbf{X}.

Suppose \( \mathbf{X} \) is connected and let \( f: \mathbf{X} \to \mathbf{X}_0 = \{0,1\} \) be continuous.

To prove: \( f \) is constant.

Since \( \{0\}, \{1\} \) are open in \( \mathbf{X}_0 \), and \( f \) is continuous, the preimages \( A = f^{-1}(\{0\}) \) and \( B = f^{-1}(\{1\}) \) are open.

Also, \( \mathbf{X} = A \cup B \), \( A \cap B = \Phi \).

If both \( A \) and \( B \) are non-empty, \( \mathbf{X} \) is disconnected — a contradiction. So one of them must be empty.

Case 1: \( A = \Phi \) implies \( f(x) = 1 \) for all \( x \in \mathbf{X} \).

Case 2: \( B = \Phi \) implies \( f(x) = 0 \) for all \( x \in \mathbf{X} \).

Hence, \( f \) is constant.

Conversely, assume all continuous maps \( f: \mathbf{X} \to \{0,1\} \) are constant.

Suppose for contradiction that \( \mathbf{X} \) is disconnected. Then \( \mathbf{X} = A \cup B \) where \( A \) and \( B \) are disjoint, non-empty, open.

Define:

f(x) = \begin{cases} 0 & \text{if } x \in A, \\ 1 & \text{if } x \in B. \end{cases}

This \( f \) is continuous and not constant — contradiction.

Therefore, \( \mathbf{X} \) must be connected.