Talk:Passing to the infimum

Yes, and moreover:
If $\forall a\in A\exists b\in B[b\le a]$, then every lower bound of $B$ is a lower bound of $A$. Thus, $\inf B$ (if exists) is a lower bound of $A$. Thus, if also $\inf A$ exists, then $\inf B\le\inf A$.
No need to assume that the two are comparable. Boris (talk) 08:02, 17 April 2016 (UTC)