Kuratowski's Closure-Complement Problem/Interior

Theorem

Let $\R$ be the real number line with the usual (Euclidean) topology.

Let $A \subseteq \R$ be defined as:

\(\ds A\)

\(:=\)

\(\ds \openint 0 1 \cup \openint 1 2\)

Definition of Union of Adjacent Open Intervals

\(\ds \)

\(\)

\(\, \ds \cup \, \)

\(\ds \set 3\)

Definition of Singleton

\(\ds \)

\(\)

\(\, \ds \cup \, \)

\(\ds \paren {\Q \cap \openint 4 5}\)

Rational Numbers from $4$ to $5$ (not inclusive)

The interior of $A$ in $\R$ is given by:

\(\ds A^\circ\)

\(=\)

\(\ds \openint 0 1 \cup \openint 1 2\)

Union of Adjacent Open Intervals

Proof

From Interior equals Complement of Closure of Complement:

$A^\circ = A^{\prime \, - \, \prime}$

From Kuratowski's Closure-Complement Problem: Closure of Complement:

\(\ds A^{\prime \, -}\)

\(=\)

\(\ds \hointl \gets 0\)

Unbounded Closed Real Interval

\(\ds \)

\(\)

\(\, \ds \cup \, \)

\(\ds \set 1\)

Singleton

\(\ds \)

\(\)

\(\, \ds \cup \, \)

\(\ds \hointr 2 \to\)

Unbounded Closed Real Interval

It can be determined by inspection that:

$A^{\prime \, - \, \prime} = \openint 0 1 \cup \openint 1 2$

Hence the result.

$\blacksquare$