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

## 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 closure of the interior of $A$ in $\R$ is given by:

\(\displaystyle A^{\circ \, -}\) | \(=\) | \(\displaystyle \left[{0 \,.\,.\, 2}\right]\) | Definition of Closed Real Interval |

## Proof

From Kuratowski's Closure-Complement Problem: Interior:

- $A^\circ = \left({0 \,.\,.\, 1}\right) \cup \left({1 \,.\,.\, 2}\right)$

From Closure of Union of Adjacent Open Intervals:

- $A^{\circ \, -} = \left[{0 \,.\,.\, 2}\right]$

$\blacksquare$