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I proved one statement. Can somebody check, did I make it correct and mathematically precise?

If $S \subseteq \mathbb R$ contains one of its upper bound, show that this upper bound is the supremum of $S$.

$\underline{\textbf{Proof}}$: So let $S$ be nonempty set with $u \in S$ and $u$ is an upper bound of $S$. Then $s\leq u \ \forall s \in S$. Now let $v$ be another upper bound of $S$.

If $v \not\in S$, then $u\leq v$ $\implies$ u is a least upper bound, so $u$ is sup$(S)$.

If $v \in S$, then $s\leq v \ \forall s \in S$ because $v$ is an upper bound of $S$ and $u\leq v$. But since $u$ is also an upper bound of $S$ we have, that $v\leq u$ because $u$ is an upper bound of $S$ and $v \in S$. So we have $u\leq v, v\leq u$ thus $u=v$.

Since $v$ was arbitrary, another upper bound of $S$ and we got, that $u\leq v$ in both cases, we can conclude, that $u$ is sup$(S)$.

mathguruu
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    Everything is correct, but you can summarize it much more (you don't really need to proceed by cases): if $v$ is another upper bound of $S$, then $u\leq v$ since $u\in S$. That suffices to conclude. – Samuel M. A. Luque Jun 24 '22 at 22:10
  • Another method: If $v$ is a real number such that $v<s$, then $v$ is not an upper bound of $S$ as there exists $s \in S$ such that $s>v$. So, the upper bound $u$ must be the least upper bound of $S$. – Aman Kushwaha Jun 25 '22 at 02:38

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