Infimum Plus Constant
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Theorem
Let $T$ be a subset of the set of real numbers.
Let $T$ be bounded below.
Let $\xi \in \R$.
Then:
- $\ds \map {\inf_{x \mathop \in T} } {x + \xi} = \xi + \map {\inf_{x \mathop \in T} } x$
where $\inf$ denotes infimum.
Proof
From Negative of Infimum is Supremum of Negatives, we have that:
- $\ds -\inf_{x \mathop \in T} x = \map {\sup_{x \mathop \in T} } {-x} \implies \inf_{x \mathop \in T} x = -\map {\sup_{x \mathop \in T} } {-x}$
Let $S = \set {x \in \R: -x \in T}$.
From Negative of Infimum is Supremum of Negatives, $S$ is bounded above.
We have:
\(\ds \map {\sup_{x \mathop \in S} } {x + \xi}\) | \(=\) | \(\ds \xi + \map {\sup_{x \mathop \in S} } x\) | Supremum Plus Constant | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {\inf_{x \mathop \in T} } {x + \xi}\) | \(=\) | \(\ds -\map {\sup_{x \mathop \in T} } {-x + \xi}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \xi - \map {\sup_{x \mathop \in T} } {-x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \xi + \map {\inf_{x \mathop \in T} } x\) |
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: Exercise $\S 2.13 \ (3)$