Connected Subsets of Real Numbers

Metric Spaces

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Connected Subsets of Real Numbers: Characterization and Key Properties

Explore the concept of connected subsets of real numbers through rigorous definitions, interval characterizations, and complete proofs that clarify their topological structure.

Interval

A subset \( I \subseteq \mathbb{R} \) is said to be an interval if for all \( x \lt z \lt y \) where \( x, y \in I \), then \( z \in I \).

Not an Interval

A subset \( I \subseteq \mathbb{R} \) is said to be not an interval if there exists \( z \in \mathbb{R} \) where \( x \lt z \lt y \) such that \( z \notin I \).

Connected Sets in R are Intervals

Let \( (\mathbb{R}, d) \) be the space of real numbers with the usual metric and let \( I \subseteq \mathbb{R} \). If \( I \) is connected, then \( I \) is an interval.

Given that \( (\mathbb{R}, d) \) is the space of real numbers with the usual metric and \( I \subseteq \mathbb{R} \).

Assume \( I \) is connected.

To prove: \( I \) is an interval.

If possible, suppose \( I \) is not an interval. Then there exists \( z \in \mathbb{R} \) such that \( x \lt z \lt y \), with \( x, y \in I \), but \( z \notin I \).

Define subsets:

\( A = I \cap (-\infty, z) \)
\( B = I \cap (z, \infty) \)

Clearly, \( I = A \cup B \) and \( A \cap B = \Phi \). Both \( A \) and \( B \) are open in the subspace topology on \( I \), and both are non-empty since \( x \in A \) and \( y \in B \).

Hence \( I \) is disconnected — a contradiction.

Therefore, \( I \) must be an interval.

Conversely: Let \( I \) be an interval.

To prove: \( I \) is connected.

Suppose \( I \) is disconnected. Then there exist two non-empty disjoint subsets \( A \) and \( B \) such that:

\( I = A \cup B \)
\( A \cap \overline{B} = \Phi \) and \( \overline{A} \cap B = \Phi \)

Choose \( x \in A \), \( y \in B \), and assume \( x \lt y \). Since \( I \) is an interval, the closed interval \([x, y] \subseteq I \).

Let \( z = \sup(A \cap [x, y]) \), which exists and lies in \([x, y]\).

For any \( \epsilon \gt 0 \), there exists \( a \in A \cap [x, y] \) such that:

z - \epsilon \lt a \lt z \Rightarrow |z - a| \lt \epsilon \Rightarrow d(z, a) \lt \epsilon \Rightarrow a \in B(z, \epsilon) \cap A

So \( z \) is a limit point of \( A \), hence \( z \in \overline{A} \).

Case A: Suppose \( z \notin A \). Then since \( I = A \cup B \), \( z \in B \). But \( z \in [x, y] \subseteq I \), so \( z \in I \), contradicting the assumption that \( I \) is not an interval. So this case leads to a contradiction.

Case B: Suppose \( z \in A \). Then:

z \in A \Rightarrow z \notin \overline{B} \Rightarrow \exists \delta \gt 0 \text{ such that } (z - \delta, z + \delta) \cap B = \Phi

Then there exists \( z_1 \in (z, y) \) such that \( z_1 \notin B \), and since \( z_1 \notin A \cup B \), it implies \( z_1 \notin I \), contradicting \( z_1 \in [x, y] \subseteq I \).

Thus, our assumption that \( I \) is disconnected leads to contradiction. Hence \( I \) is connected.