Infimum of Singleton

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Then for all $a \in S$:


 * $\inf \left\{{a}\right\} = a$

where $\inf$ denotes infimum.

Proof
Since $a \preceq a$, $a$ is a lower bound for $\left\{{a}\right\}$.

Let $b$ be another lower bound for $\left\{{a}\right\}$.

Then necessarily $b \preceq a$.

It follows that indeed:


 * $\inf \left\{{a}\right\} = a$

as desired.

Also see

 * Supremum of Singleton