Infimum

Last updated Nov 1, 2022

Let $(T \leq)$ be a Total Ordering. Let $A \subset T$. Let $T’$ be the Reverse Ordering of $T$ and let $A’ = A$, but with the Reverse Ordering. Then $\inf\limits A := \sup\limits A’$.

1. In effect this is really saying that the Infimum is the greates Lower Bound. The use of the Reverse Ordering in the definition makes it easy to carry results about Supremum over to Infimum.

1. Infimum of Sum bounds Sum of Infimum
2. Non-Decreasing Continuous Functions preserve Infimum
3. Non-Increasing Continuous Functions send Infimum to Supremum
4. Non-negative Multiplication Commutes with Infimum
5. Negative Multiplication sends Infimum to Supremum
6. Infimum of Negative is Negative of Supremum

Let $(T \leq)$ be a Total Ordering. Let $A \subset T$. Set $L = {x \in T : x \leq a \text{ } \forall a \in A}$. That is, $L$ is the Set of Lower Bounds of $A$. Then $\inf A$ (if it exists) is the Maximum of $L$.

1. This is equivalent to definition 1, because Lower Bounds become Upper Bounds and maxima become minima in the Reverse Ordering.