# Open Real Interval is Open Set

## Theorem

Let $\R$ be the real number line considered as an Euclidean space.

Let $\openint a b \subset \R$ be an open interval of $\R$.

Then $\openint a b$ is an open set of $\R$.

### Corollary

Let $\R$ be the real number line considered as an Euclidean space.

Let $A := \openint a \infty \subset \R$ be an open interval of $\R$.

Let $B := \openint {-\infty} b \subset \R$ be an open interval of $\R$.

Then both $A$ and $B$ are open sets of $\R$.

## Proof

Let $c \in \R$ such that $a < c < b$.

Let $\epsilon = \min \set {b - c, c - a}$.

From the definition of positive it follows that $\epsilon \in \R_{>0}$.

Let $\map {B_\epsilon} c = \openint {c - \epsilon} {c + \epsilon}$ be the open $\epsilon$-ball of $c$.

Let $y \in \map {B_\epsilon} c$.

Then:

$\size {c - y} < \epsilon$

implies

$-\epsilon < c - y < \epsilon$

We have that:

 $\displaystyle a$ $=$ $\displaystyle c - \paren {c - a}$ $\displaystyle$ $\le$ $\displaystyle c - \epsilon$ $\displaystyle$ $<$ $\displaystyle y$ $\displaystyle$ $<$ $\displaystyle c + \epsilon$ $\displaystyle$ $\le$ $\displaystyle c + \paren {b - c}$ $\displaystyle$ $=$ $\displaystyle b$

In other words:

$y \in \map {B_\epsilon} c \implies y \in \openint a b$

Thus:

$\map {B_\epsilon} c \subseteq \openint a b$

It follows that, by definition, $\openint a b$ is a neighborhood of $c$.

The result follows by definition of open set.

$\blacksquare$