# Kuratowski's Closure-Complement Problem/Interior of Complement of Interior

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## Theorem

Let $\R$ be the real number space under the usual (Euclidean) topology.

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

\(\displaystyle A\) | \(:=\) | \(\displaystyle \left({0 \,.\,.\, 1}\right) \cup \left({1 \,.\,.\, 2}\right)\) | Definition of Union of Adjacent Open Intervals | ||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle \cup \, \) | \(\displaystyle \left\{ {3} \right\}\) | Definition of Singleton | |||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle \cup \, \) | \(\displaystyle \left({\Q \cap \left({4 \,.\,.\, 5}\right)}\right)\) | Rational Numbers from $4$ to $5$ (not inclusive) |

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

\(\displaystyle A^{\circ \, \prime \, \circ}\) | \(=\) | \(\displaystyle \left({\gets \,.\,.\, 0}\right)\) | Definition of Unbounded Open Real Interval | ||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle \cup \, \) | \(\displaystyle \left({2 \,.\,.\, \to}\right)\) | Definition of Unbounded Open Real Interval |

## Proof

From Complement of Interior equals Closure of Complement:

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

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

\(\displaystyle A^{\prime \, -}\) | \(=\) | \(\displaystyle \left({\gets \,.\,.\, 0}\right]\) | Definition of Unbounded Closed Real Interval | ||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle \cup \, \) | \(\displaystyle \left\{ {1} \right\}\) | Definition of Singleton | |||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle \cup \, \) | \(\displaystyle \left[{2 \,.\,.\, \to}\right)\) | Definition of Unbounded Closed Real Interval |

From Interior of Closed Real Interval is Open Real Interval:

- $\left({\gets \,.\,.\, 0}\right]^\circ = \left({\gets \,.\,.\, 0}\right)$

and:

- $\left[{2 \,.\,.\, \to}\right)^\circ = \left({2 \,.\,.\, \to}\right)$

From Interior of Singleton in Real Number Space is Empty:

- $\left\{ {1} \right\}^\circ = \varnothing$

$\blacksquare$