# Closed Real Interval is Closed in Real Number Line

Jump to navigation
Jump to search

## Theorem

Let $\left({\R, \tau_d}\right)$ be the real number line considered as a topological space under the usual (Euclidean) topology.

Let $\left[{a \,.\,.\, b}\right]$ be a closed interval of $\R$.

Then $\left[{a \,.\,.\, b}\right]$ is closed (in the topological sense) in $\left({\R, \tau_d}\right)$.

## Proof

From Open Sets in Real Number Line:

- $U := \left({-\infty \,.\,.\, a}\right) \cup \left({b \,.\,.\, +\infty}\right)$

is an open set in $\left({\R, \tau_d}\right)$.

Consider the complement relative to $\R$:

- $V := \complement_\R \left({U}\right) = \left[{a \,.\,.\, b}\right]$

By definition, $\complement_\R \left({U}\right)$ is closed (in the topological sense) in $\left({\R, \tau_d}\right)$.

But by construction:

- $\complement_\R \left({U}\right) = \left[{a \,.\,.\, b}\right]$

Hence the result.

$\blacksquare$