Supremum of an Increasing Function is Function of Supremum
# Statement
Let $(X, \leq)$, $(Y, \leq)$ be Total Orderings equipped with their respective Order Topologys and let $f: T \to S$ be a continuous Non-Decreasing Function. Let $A \subset T$ and suppose $\sup\limits A$ exists. Then $$f(\sup\limits A) = \sup\limits f(A)$$
# Proof
Since $A$ is a Total Ordering, it is also a Directed Partial Ordering. Thus, we can view $A$ as the non-decreasing Net $(a_{a}){a \in A}$. Since $\sup\limits A$ exists, we know $a{a} \to \sup\limits A$ because Order-Preserving Nets Converge to their Supremum. Because A Function is Continuous iff it preserves Net Convergence, $(f(a_{a})){a \in A} \to f(\sup\limits A)$. Since $f$ is non-decreasing, if $a \leq b \in A$, then $f(a{a}) \leq f(b_{b})$. Therefore, $(f(a_{a})){a \in A}$ is a non-decreasing Net. Since Order-Preserving Nets Converge to their Supremum, we know $f(\sup\limits A) = \sup\limits{a \in A} f(a_{a}) = \sup\limits f(A)$. $\blacksquare$