Existence of Real Number at Distance Zero from Open Real Interval not in Interval

Theorem

Let $S$ be a subset of the set of real numbers $\R$.

Let $x \in \R$ be a real number.

Let $\map d {x, S}$ be the distance between $x$ and $S$.

Let $I \subseteq \R$ be an open real interval such that $I \ne \O$ and $I \ne \R$.

Then:

$\exists x \notin I: \map d {x, I} = 0$

Proof

From the definition of distance:

$\forall x, y \in \R: \map d {x, y} = \size {x - y}$

Thus:

$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$

As $I \ne \O$ and $I \ne \R$ it follows that one of the following applies:

 $\, \displaystyle \exists a, b \in \R: \,$ $\displaystyle I$ $=$ $\displaystyle \openint a b$ $\, \displaystyle \exists a \in \R: \,$ $\displaystyle I$ $=$ $\displaystyle \openint a \to$ $\, \displaystyle \exists b \in \R: \,$ $\displaystyle I$ $=$ $\displaystyle \openint \gets b$

It follows by the definition of open real interval that $I$ has either an infimum $a$, or a supremum $b$, or both.

Thus the required value of $x$ is either $a$ or $b$.

$\blacksquare$