Suppose that every nonempty subset $X$ bounded from above has a supremum. Then every nonempty subset $Y$ bounded from below has an infimum

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.

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