# Kuratowski's Closure-Complement Problem/Closure

## 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 $A$ in $\R$ is given by:

 $\displaystyle A^-$ $=$ $\displaystyle \left[{0 \,.\,.\, 2}\right]$ Definition of Closed Real Interval $\displaystyle$  $\, \displaystyle \cup \,$ $\displaystyle \left\{ {3} \right\}$ Definition of Singleton $\displaystyle$  $\, \displaystyle \cup \,$ $\displaystyle \left[{4 \,.\,.\, 5}\right]$ Definition of Closed Real Interval ## Proof

$\left({\left({0 \,.\,.\, 1}\right) \cup \left({1 \,.\,.\, 2}\right)}\right)^- = \left[{0 \,.\,.\, 2}\right]$
$\left\{ {3} \right\}$ is closed in $\R$
$\left\{ {3} \right\}^- = \left\{ {3} \right\}$
$\left({\Q \cap \left({4 \,.\,.\, 5}\right)}\right)^- = \left[{4 \,.\,.\, 5}\right]$

The result follows from Closure of Finite Union equals Union of Closures.

$\blacksquare$