In a Boolean algebra B and $Y\subseteq B$, $p$ is an upper bound for $Y$ but not the supremum. Is $qI don't think that this is the case. I am reading over one of my professor's proof, and he seems to use this fact. Here is the proof:
Let $B$ be a Boolean algebra, and suppose that $X$ is a dense subset of $B$ in the sense that every nonzero element of $B$ is above a nonzero element of $X$. Let $p$ be an element in $B$. The proof is to show that $p$ is the supremum of the set of all elements in $X$ that are below $p$. Let $Y$ be the set of elements in $X$ that are below $p$. It is to be shown that $p$ is the supremum of $Y$. Clearly, $p$ is an upper bound of $Y$. If $p$ is not the least upper bound of $Y$, then there must be an element $q\in B$ such that $q<p$ and $q$ is an upper bound for $Y$ ...etc.
I do not see how this last sentence follows. I do see that if $p$ is not the least upper bound of $Y$, then there is some upper bound $q$ of $Y$ such that $p$ is NOT less than or equal to $q$. But, since we have only a partial order, and our algebra is not necessarily complete, I do not see how we can show anything else.
So, is my professor's proof wrong, or am I just missing something fundamental?

Let $p$ and $q$ be upper bounds of $Y$. Then $p\wedge q$ is an upper bound of $Y$ and $p\wedge q\leq p$ and $p\wedge q\leq q$. Now if for all upper bounds $q$ of $Y$, $p\leq p\wedge q\leq q$, $p$ must be the least upper bound. Otherwise $p\wedge q<p$ for some $q$ and $p\wedge q$ is a strictly lower upper bound of $Y$. 
