About dense subalgebras of boolean algebras.

Question

I started studying about Boolean algebras and got stuck doing an exercise. Maybe it's trivial but really I can't do it. First, a definition

A subset $S$ of $B$, where $B$ is a Boolean algebra, is said to be dense in $B$ if for $0<b\in B$, there is some $a\in S$ such that $0<a\leq b$.

Now, the exercise:

Let $A$ be a dense subalgebra of a Boolean algebra $B$, $a\in A$ and $b\in B$. Prove:

(a) If $a<b$ there is some $a_1\in A$ such that $a<a_1\leq b$.

(b) If $\emptyset\neq S\subseteq A$ and $a=\bigvee S$ (here $\bigvee$ denotes the least upper bound, i.e., the supremum ) exists in $A$ then $\bigvee S$ exists in $B$ and equals $a$.

(c) If $0<b<1$ and $S=\{a\in A\mid 0<a\leq b \}$ then $b=\bigvee S$.

I think that I have the solution for (a) and is the next:

As $0\leq a<b$ then $b>0$. By hypothesis $A$ is dense in $B$ and $b\in B$, then, there is $c\in A$ such that $0<c\leq b$. Let $a_1=a\vee c$. Since $A$ is a subalgebra then $a_1\in A$ and moreover, since $b$ is an upper bound of $a$ and $c$ holds $a_1\leq b$ and $a_1>a$. Therefore $a<a_1\leq b$. Is it correct?

For (b), I only need to prove that $a$ is the least upper bound of $S$ in $B$. But, I don't know how. Maybe using (a). If we take $b\in B$ another upper bound of $S$ such that $b<a$ then, by (a), there is $a_1\in A$ such that $b\leq a_1<a$. But, here, I can't see a contradiction because we don't know if $a_1\in S$ or $a_1\notin S$. Another way is proving directly that $a\leq b$, but, how?

For (c), take $c$ an upper bound for $S$ such that $0<c<b$. Then $b\wedge c'\neq 0$ (here $c'$ is the complement of $c$) because, by De Morgan Laws, if $b\wedge c'=0$ then $c'\leq b'$ and therefore $b\leq c$ but this doesn't happens by hyphotesis. Then $0<b\wedge c' \in B$. Since $A$ is dense then there is $a_1\in A$ such that $0<a_1\leq b\wedge c'\leq c'$. Therefore $c\leq a'$ and $A$ is a subalgebra, then $a'\in A$. But, how can I conclude?

I really appreciate any help you can provide me. Thanks.

Answer 1

As $0\leq a<b$ then $b>0$. By hypothesis $A$ is dense in $B$ and $b\in B$, then, there is $c\in A$ such that $0<c\leq b$. Let $a_1=a\vee c$. Since $A$ is a subalgebra then $a_1\in A$ and moreover, since $b$ is an upper bound of $a$ and $c$ holds $a_1\leq b$ and $a_1>a$. Therefore $a<a_1\leq b$. Is it correct?

No. You can conclude that $a_1\geq a$ but not $a_1>a$. For instance, the $c$ that you chose could have just been $a$ itself, in which case $a_1=a\vee a=a$.

To get $a_1$ that is strictly greater than $a$, you'll need to apply the density property to something other than $b$, something that guarantees $c$ will not just be contained in $a$.

A stronger hint is hidden below:

Apply density to $b\wedge a'$.

For (b), I only need to prove that $a$ is the least upper bound of $S$ in $B$. But, I don't know how. Maybe using (a). If we take $b\in B$ another upper bound of $S$ such that $b<a$ then, by (a), there is $a_1\in A$ such that $b\leq a_1<a$. But, here, I can't see a contradiction because we don't know if $a_1\in S$ or $a_1\notin S$. Another way is proving directly that $a\leq b$, but, how?

I'm not sure why you think it is relevant whether $a_1\in S$. To get a contradiction, you just have to prove that $a$ fails to be the least upper bound of $S$ in $A$. Can you show that $a_1$ is a smaller upper bound of $S$?

(Incidentally, I don't know why the statement of (b) assumes that $S\neq \emptyset$. This assumption is unnecessary and has no effect on the proof.)

For (c), take $c$ an upper bound for $S$ such that $0<c<b$. Then $b\wedge c'\neq 0$ (here $c'$ is the complement of $c$) because, by De Morgan Laws, if $b\wedge c'=0$ then $c'\leq b'$ and therefore $b\leq c$ but this doesn't happens by hyphotesis. Then $0<b\wedge c' \in B$. Since $A$ is dense then there is $a_1\in A$ such that $0<a_1\leq b\wedge c'\leq c'$. Therefore $c\leq a'$ and $A$ is a subalgebra, then $a'\in A$. But, how can I conclude?

Notice that $a_1\in S$ (why?). Now can you reach a contradiction?

More details are hidden below.

Since $c$ is an upper bound for $S$, $a_1\leq c$. Thus since $a_1\leq c$ and $a_1\leq c'$, we have $a_1\leq c\wedge c'=0$, a contradiction.