# Kuratowski's Closure-Complement Problem/Closure

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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 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

From Closure of Union of Adjacent Open Intervals:

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

From Real Number is Closed in Real Number Space:

- $\left\{ {3} \right\}$ is closed in $\R$

From Set is Closed iff Equals Topological Closure:

- $\left\{ {3} \right\}^- = \left\{ {3} \right\}$

From Closure of Rational Interval is Closed Real Interval:

- $\left({\Q \cap \left({4 \,.\,.\, 5}\right)}\right)^- = \left[{4 \,.\,.\, 5}\right]$

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

$\blacksquare$