Suppose $S\subseteq{\mathbb{R}}$ is nonempty. Show that $u$ is an upper bound of $S$ if and only if $t\in\mathbb{R}, u<t$ imply $t\notin S$

Here's my attempt.

$\Rightarrow$ Let $u$ be an upper bound of $S$. Thus, $s\le u, \forall s\in S$. Now suppose that $\exists t\in \mathbb{R}$ such that $u<t$. Therefore $$s\le u<t \rightarrow s<t$$

Thus, $t$ is strictly greater than $s \forall s\in S$, which implies $t\notin S$.

$\Leftarrow$ Let $t,u \in \mathbb{R}$ such that $u<t$. Assume that $u$ is a lower bound of S. By definition of lower bound, $u\le s, \forall s \in S$. Thus $$u<t\le s \text{ or } u\le s<t$$ Thus it is possible for $t$ to be an element of $S$. However, this is impossible as we stated $t\notin{S}$. Thus by contradiction, our assumption that $u$ is a lower bound must be wrong. Therefore, $u$ is an upper bound of $S$.

Therefore i have shown that $u$ is an upper bound of $S$ if and only if $t\in\mathbb{R}, u<t$ imply $t\notin S$

As I am working on my proof skills, I feel as though I have done my work properly. I feel as though, logically this is a good proof, but I was hoping to have it critiqued. If it is wrong, where are the holes in the argument?

  • 1
    $\begingroup$ Another idea is to rewrite $\forall s\in S\colon s\le u$ as $\forall s\in \mathbb R\colon(s\in S\to s\le u)$ and use contraposition $\forall s\in \mathbb R\colon(\neg(s\le u)\to \neg(s\in S))$, i.e. $\forall s\in \mathbb R\colon(s>u\to s\notin S)$ $\endgroup$ – Hagen von Eitzen Feb 23 '14 at 15:13

Lower bounds have nothing to do with it, instead ...

If $u$ is an upper bound then there exists no $t$ in $S$ and $u < t$. Since for any $u\in \mathbb R$ there exists $t > u$ (lots of them in fact) it follows that such a $t$ is not in $S$.

Conversely for $t \in \mathbb R$ if $u < t$ implies that $t$ is not in $S$, then there is no $t$ in $S$ greater than $u$ so $u$ is an upper bound.


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