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:

\(\ds \map {\sup_{x \mathop \in S} } {x + \xi}\)

\(=\)

\(\ds \xi + \map {\sup_{x \mathop \in S} } x\)

Supremum Plus Constant

\(\ds \leadsto \ \ \)

\(\ds \map {\inf_{x \mathop \in T} } {x + \xi}\)

\(=\)

\(\ds -\map {\sup_{x \mathop \in T} } {-x + \xi}\)

\(\ds \)

\(=\)

\(\ds \xi - \map {\sup_{x \mathop \in T} } {-x}\)

\(\ds \)

\(=\)

\(\ds \xi + \map {\inf_{x \mathop \in T} } x\)

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: Exercise $\S 2.13 \ (3)$