Let $S$ be an ordered set and $S \supset E $. Let $\alpha = \sup E \in E$. If $u \notin E$, then we have $ \sup ( E \cup \{u\} ) = \sup\{ \alpha, u \} $.
Try:
I know this result follows easy by using On the supremum of the union of two bounded sets :
$$\sup ( E \cup \{ u \} ) = \sup \{ \sup E , \sup \{ u \} \} = \sup \{ \alpha, u \}$$ since the supremum of a singleton { x } is always $x$ trivially.
Now, I would like to solve this problem without using the previous problem: This is what I have tried: Since $u \notin E$, then it follows that $ \alpha \neq u $. Let $A = E \cup \{ u \} $.
If $\alpha < u $, then $\sup\{ \alpha, u \} = u $ trivially. Since $x \leq \alpha $ for all $x \in E$, then we have $x < u $ for all $x \in A$ and since $ u \in A$, it follows at once that $u = \sup A = \sup ( E \cup \{u \} ) $.
If $u < \alpha $, then $\alpha = \sup\{u, \alpha\} $ since $\alpha $ is an upper bound inside the set $\{ u , \alpha \}$. The result follows by the same line of thought of the previous paragraph.
I would like to get feedback. Thanks.