Real Number at Distance Zero from Closed Real Interval is 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 a closed real interval.
Then:
- $\map d {x, I} = 0 \implies x \in I$
Proof
From the definition of distance:
- $\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
- $\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Because $I$ is an interval, if $x \notin I$ then $x$ is either an upper bound or a lower bound for $I$.
Suppose $x$ is an upper bound for $I$.
Let $B$ be the supremum of $I$.
Then because $I$ is closed, $B \in I$.
So:
\(\ds \forall y \in I: \, \) | \(\ds \size {x - y}\) | \(=\) | \(\ds x - y\) | |||||||||||
\(\ds \) | \(=\) | \(\ds x - B + B - y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x - B + \size {B - y}\) |
Now from Infimum Plus Constant:
- $\inf_{y \mathop \in S} \size {x - y} = x - B + \inf_{y \mathop \in S} \size {B - y}$
But we also have:
- $x - B \ge 0$
- $\map d {B, S} \ge 0$
- $\map d {x, S} = 0$
So it follows that $x = B$ and so $x \in I$.
A similar argument applies if $x$ is a lower bound for $I$.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: Exercise $\S 2.13 \ (5) \ \text {(iii)}$