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Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!


Let $(A,\le)$ be an ordered set whose every nonempty subset $X$ bounded from above has a supremum. Then every nonempty subset $Y\subseteq A$ bounded from below has an infimum.

Let $Z$ be the set of lower bounds of $Y$. It's clear that $Z$ is bounded above by every elements in Y. Thus $Z$ has a supremum, which is denoted by $\sup Z$. Next we prove $\sup Z$ is the infimum of $Y$.

$\sup Z$ is the least upper bound of $Z$, and $Y$ consists of upper bounds of $Z \implies$ $\sup Z\le y$ for all $y\in Y$.

If $z$ is a lower bound of $Y$, then $z\in Z$ and consequently $z\le \sup Z$.

Thus $\sup Z$ is the infimum of $Y$.


Update: Since $Y$ is bounded from below, $Z$ is nonempty.

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  • 1
    $\begingroup$ Absolutely correct proof. $\endgroup$ – Mark Sep 24 '18 at 16:25
  • 3
    $\begingroup$ You should begin by emphasizing that $Z\ne\emptyset$. In general, when writing a proof, you should highlight where the assumptions are used. $\endgroup$ – Andrés E. Caicedo Sep 24 '18 at 17:59
  • $\begingroup$ Thank you @AndrésE.Caicedo! I have updated my post to reflect your suggestion. $\endgroup$ – Le Anh Dung Sep 24 '18 at 23:13

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