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:

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

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:

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

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 2.13 \ (3)$