Jump to content

Talk:Boolean algebra (structure)

Page contents not supported in other languages.
From Wikipedia, the free encyclopedia

About isomorphisms

[edit]

In § 4, Homomorphisms and isomorphisms, why's isomorphisms being mentioned in the title, but not in the section itself? Are all homomorphisms also isomorphisms? That should be written then.

--Unknowledgeable (talk) 13:50, 18 July 2016 (UTC)[reply]

I've added the definition of isomorphism; not every homomorphism is an isomorphism. Andrewbt (talk) 05:45, 31 December 2017 (UTC)[reply]

Anti-absorption

[edit]

Another useful identity (theorem, not axiom) is:

a ∨ (¬ab) = ab
a ∧ (¬ab) = ab

Does this have a standard name? It follows immediately from distributivity, of course. Should it be mentioned in the article? --Macrakis (talk) 20:09, 26 March 2021 (UTC)[reply]

$1$ and $0$ are supposed to be distinct in the definition

[edit]

I think it is typical (See for instance page 10 of "Introduction to Boolean algebras" by Paul Halmos) that in the definition of the algebra, $0$ and $1$ are distinct elements. Should I make the change? PierreQuinton (talk) 13:59, 20 September 2024 (UTC)[reply]

In English, "two elements 0 and 1" means "two" not "one or two". D.Lazard (talk) 14:21, 20 September 2024 (UTC)[reply]
I'd say it's a little ambiguous. I would say "two distinct elements" if that's definitely what we mean. (I think it's a slightly controversial point whether there's a one-element Boolean algebra; in my usage there is not, but I'm not sure there's a complete consensus on it in the literature.) --Trovatore (talk) 19:16, 20 September 2024 (UTC)[reply]
We have a remark about that in section Definition: "A Boolean algebra with only one element is called a trivial Boolean algebra or a degenerate Boolean algebra. (In older works, some authors required 0 and 1 to be distinct elements in order to exclude this case.)"
Moreover, as far as I remember, it makes a big difference if the trivial one-element boolean algebra is to be excluded: an extra axiom is needed, since it is not an equation, the class is no longer a variety, hence Birkhoff's variety theorem no longer applies to it. That might be the reason why the distinctness requirement is missing in newer works. - Jochen Burghardt (talk) 05:20, 23 September 2024 (UTC)[reply]