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:

$\displaystyle \map {\inf_{x \mathop \in T} } {x + \xi} = \xi + \map {\inf_{x \mathop \in T} } x$

where $\inf$ denotes infimum.


Proof

From Negative of Infimum is Supremum of Negatives, we have that:

$\displaystyle -\inf_{x \mathop \in T} x = \map {\sup_{x \mathop \in T} } {-x} \implies \inf_{x \mathop \in T} x = -\map {\sup_{x \mathop \in T} } {-x}$

Let $S = \set {x \in \R: -x \in T}$.

From Negative of Infimum is Supremum of Negatives, $S$ is bounded above.

We have:

\(\displaystyle \map {\sup_{x \mathop \in S} } {x + \xi}\) \(=\) \(\displaystyle \xi + \map {\sup_{x \mathop \in S} } x\) Supremum Plus Constant
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {\inf_{x \mathop \in T} } {x + \xi}\) \(=\) \(\displaystyle -\map {\sup_{x \mathop \in T} } {-x + \xi}\)
\(\displaystyle \) \(=\) \(\displaystyle \xi - \map {\sup_{x \mathop \in T} } {-x}\)
\(\displaystyle \) \(=\) \(\displaystyle \xi + \map {\inf_{x \mathop \in T} } x\)

$\blacksquare$


Sources