I was reading through the definition of upper/lower bounds in Walter Rudin's Mathematical Analysis:

1.7 Definition

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

Wouldn't a corollary of this be that any set can only have one of its upper/lower bounds in itself? It would always be the element $x = \beta$ in $x \le \beta$ (or $x \ge \beta$ for lower bounds)?

  • $\begingroup$ What about $E=\{0\}$ in $S=\mathbb R?$ $\endgroup$ – zhw. Jan 3 '18 at 18:21
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    $\begingroup$ Sorry, my bad, I didn't read the problem carefully. $\endgroup$ – zhw. Jan 3 '18 at 18:28

What you are proposing is true. In a partially ordered set, if $\beta_1$ and $\beta_2$ are upper bounds of $E$ such that $\beta_1\in E$ and $\beta_2\in E$, then $\beta_1\le \beta_2$ (since $\beta_2$ is an upper bound of $E$) and $\beta_2\le\beta_1$ (since $\beta_1$ is an upper bound of $E$). Conclude $\beta_1=\beta_2$, by antisymmetry.


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