Projective Dimension and Supremum

Question

Here is a lemma that appears in A Course in Ring Theory by Passman. In the last section of the proof the writer shows that, $\mbox{pd }A_i\leq n\iff \mbox{pd }A\leq n$ and finishes the proof. I don't understand how this implies that $\mbox{pd }A=\mbox{sup }\{\mbox{pd }A_i\mid i\in I\}$. I hope one of you can shed some light on this matter.

Answer 1

A supremum is an upper bound so $\operatorname{pd}A_i \leq \sup\{\operatorname{pd}A_i \ | \ i\}$. Letting $n = \sup\{\operatorname{pd}A_i \ | \ i\}$ then gives $\operatorname{pd}A \leq \sup\{\operatorname{pd}A_i \ | \ i\}$. Conversely $\operatorname{pd}A \leq \operatorname{pd}A$ so letting $n = \operatorname{pd}A$ gives $\operatorname{pd}A_i \leq \operatorname{pd}A$ for all $i$. As supremum is a least upper bound $\sup\{\operatorname{pd}A_i \ | \ i\} \leq \operatorname{pd}A$.