MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I found in Awodey's "Theory Category", second edition p34, a set of poset axioms to define Boolean algebras, whereas I don't see how they can be sufficient.

Here are the axioms:

A Boolean algebra is a poset B equipped with distinguished elements 0, 1, binary operations $a \vee b$ of "join" and $a \wedge b$ of "meet", and a unary operation $\neg b$ of "complementation". These are required to satisfy the conditions $$0 \le a$$ $$a \le 1$$ $$a \le c \text{ and } b > \le c \text{ iff } a\vee b \le c $$ $$c \le a \text{ and } c \le b \text{ > iff } c \le a \wedge b$$ $$a \le \neg > b \text{ iff } a \wedge b = 0$$ $$\neg\neg a =a$$

I can see that from these axioms I can deduce several of the common properties of boolean algebras, so for example I can prove $\forall c, c\wedge\neg c =0$, $\forall (a,c), a\wedge c \le c$, etc. but there are some properties I cannot see how they are provable in this axiomatics.

Here one of which I tried so hard (and failed) to prove from the axioms above: $\forall (a,b), a \le a \vee b$

Could somebody guide me? I am trying to make something impossible? Is my mistake on things I think that are required to be true in a Boolean algebra and are not?

share|cite|improve this question

closed as off-topic by Ricardo Andrade, David White, Carlo Beenakker, Daniel Moskovich, Ryan Budney Oct 19 '13 at 13:25

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, David White, Carlo Beenakker, Daniel Moskovich, Ryan Budney
If this question can be reworded to fit the rules in the help center, please edit the question.

In your second-last axiom, you probably want to say $a\wedge b=0$ rather than $a\vee b=0$. – Joel David Hamkins Mar 19 '13 at 2:44
Yes, thank you. Corrected. I confused three times "vee" and "wedge" while redacting... I hope all is corrected now. – Almeo Maus Mar 19 '13 at 2:48
The claim that $a\leq a\vee b$ seems to follow immediately from the third axiom, by taking $c=a\vee b$, since $c\leq c$. – Joel David Hamkins Mar 19 '13 at 2:52
Exact. Now I feel so stupid that I'd like to close my question... Thank you so much. – Almeo Maus Mar 19 '13 at 2:56
Hey, don't worry about it! We all miss easy things sometimes... – Joel David Hamkins Mar 19 '13 at 3:02
up vote 0 down vote accepted

As remarked by Joel, this is indeed very direct starting from $a\vee b \le a \vee b$ and using the third axiom. Many thanks to have unstucked me.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.