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

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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 $b$.

$\blacksquare$


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