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:
 * $\ds \map {\inf_{x \mathop \in T} } {x + \xi} = \xi + \map {\inf_{x \mathop \in T} } x$

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: