Does the statement characterise an upper and lower bound?

I was reading Rudin's textbook on real analysis and came across the following definition of upper and lower bounds:

"Suppose $S$ is an ordered set, and $E \subset S$.

If there exists a $\beta \in S$ such that $x \le \beta$ for every $x \in E$,

we say that $E$ is bounded above, and call $\beta$ an upper bound of $E$."

Would the following be a correct logical interpretation of the above statement?

$$\exists\beta\in S : x\le\beta\space\forall x\in E$$

Solution 1:

Actually, you have to describe the details to make it logical. This logic you wrote has no clear meaning without mentioning that $S$ is an ordered set and $E \subset S$. In this case, $E$ will be bounded above in $S$ (not in every set). So this characterization works when we have the following:

1- $S$ is an ordered set.

2- $E \subset S$.

3- $\exists \beta \in S$ ($\beta$ not unique) such that $\forall x \in E \Rightarrow x \leq \beta.$

Moreover,

4- $E$ has a least upper bound (LUP) if $\exists^{1} b \in S$ such that $\forall \beta \in S$ (for every upper bound $\beta$ of $E$ in $S$) then $b \leq \beta$.