Distance from Subset of Real Numbers to Infimum/Proof 1
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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 $S$ be bounded below such that $\xi = \inf S$.
Then:
- $\map d {\xi, S} = 0$
Proof
From the definition of distance:
- $\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
- $\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Let $\xi = \inf S$.
Consider $\map d {-\xi, S'}$ where $S' = \set {-\xi: \xi \in S}$.
By Negative of Infimum is Supremum of Negatives:
- $\xi = \inf S \implies -\xi = \sup S'$
Thus from the above, $\map d {-\xi, S'} = 0$ and hence the result.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: Exercise $\S 2.13 \ (5) \ \text {(ii)}$