Infimum Plus Constant

Theorem
Let $T$ be a subset of the set of real numbers.

Let $T$ be bounded below.

Let $\xi \in \R$.

Then:
 * $\displaystyle \inf_{x \in T} \left({x + \xi}\right) = \xi + \inf_{x \in T} \left({x}\right)$

where $\inf$ denotes infimum.

Proof
From Negative of Infimum, we have that:
 * $\displaystyle -\inf_{x \in T} x = \sup_{x \in T} \left({-x}\right) \implies \inf_{x \in T} x = -\sup_{x \in T} \left({-x}\right)$

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

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

From Supremum Plus Constant we have:
 * $\displaystyle \sup_{x \in S} \left({x + \xi}\right) = \xi + \sup_{x \in S} \left({x}\right)$

Hence:
 * $\displaystyle \inf_{x \in T} \left({x + \xi}\right) = -\sup_{x \in T} \left({-x + \xi}\right) = \xi - \sup_{x \in T} \left({-x}\right) = \xi + \inf_{x \in T} \left({x}\right)$