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Supremum of a Cofinal Subset is the Supremum of the Set

Last updated Nov 6, 2022

# Statement

Let $(X, \leq)$ be a Total Ordering and let $A \subset X$. Let $S \subset A$ be a Cofinal Subset. If $\sup\limits S \in X$, then $\sup\limits A = \sup\limits S$.

# Proof

$\sup\limits S$ is an Upper Bound for $A$ since for every $a \in A$, there exists $b \in S$ so that $b \geq a$ ($S$ is a Cofinal Subset of $A$). Because $\sup\limits S \geq b$ for all $b \in S$, we have that $\sup\limits S \geq a$ for all $a \in A$. On the other hand, any Upper Bound for $A$ must also be an Upper Bound for $S$ since $S \subset A$. Thus, any Upper Bound for $A$ is at least $\sup\limits S$. This means $\sup\limits S$ is the least upper bound of $A$ (i.e. $\sup\limits A = \sup\limits S$). $\blacksquare$

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