# 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:

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

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

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

$\blacksquare$