Infimum Plus Constant

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

Let $$T$$ be bounded below.

Let $$\xi \in \mathbb{R}$$.

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

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

Let $$S = \left\{{x \in \mathbb{R}: -x \in T}\right\}$$. From Negative of Infimum, $$S$$ is bounded above.

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

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