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 $\map d {x, S}$ denote 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:
- $\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
As $I \ne \O$ and $I \ne \R$ it follows that one of the following applies:
\(\ds \exists a, b \in \R: \, \) | \(\ds I\) | \(=\) | \(\ds \openint a b\) | |||||||||||
\(\ds \exists a \in \R: \, \) | \(\ds I\) | \(=\) | \(\ds \openint a \to\) | |||||||||||
\(\ds \exists b \in \R: \, \) | \(\ds I\) | \(=\) | \(\ds \openint \gets b\) |
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 $b$.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: Exercise $\S 2.13 \ (5) \ \text {(iii)}$