Talk:Stone's representation theorem for Boolean algebras

message: ----
Talk moved from former "Stone's duality". --Markus Krötzsch 14:39, 18 Apr 2004 (UTC)
replies: text: Stone's duality generalises to infinite sets of propositions the use of truth tables to characterise elements of finite Boolean algebras.
I think of a truth table as a device to define logical operators, i.e. to define functions {T,F}n→{T,F}, and of the elements of a Boolean algebras as logical propositions that can be combined with the operators and, or, not. So I don't quite understand how truth tables are used to characterise elements of finite Boolean algebras. Could that be explained a bit more?
Also, a prominent Stone space is the Cantor set; does it correspond to an interesting Boolean algebra? AxelBoldt 16:06, 13 Feb 2004 (UTC)

message: Why was this article moved from Stone's duality, when that is a term much more likely to be used for the subject of this article (and Stone duality more so)?  --- Charles Stewart 19:51, 9 Jun 2005 (UTC)
thread_title: Move

message: The simplicity of the properly stated theorem and its elegant consequences are over shadowed by the interjection of categorical language. Also, at what point in the history of Stone's representation theorem, has category theory had any place?  -- Differentiablef  — Preceding unsigned comment added by Differentiablef (talk • contribs) 06:57, 20 January 2011 (UTC)
thread_title:  Simpler statements of the theorem 

message: It's quite a statement to say that the Stone duality was the 'first' example of a non-trivial duality of categories - there are plenty of other great classical theorems far predating it which may also be phrased as giving (non-trivial) dualities of categories. Unless there is a qualification given to this, I'm taking it out.  —Preceding unsigned comment added by 137.158.152.207 (talk) 07:11, 11 November 2010 (UTC)
thread_title:  First Example? 

message: "Note that B embeds in S(B) by sending each element b to the principal ultrafilter generated by b." — What could it mean?? Elements of B are, basically, clopen subsets of S(B), not points of S(B). Boris Tsirelson (talk) 17:47, 1 September 2011 (UTC)
replies: text: Well, this unfortunate sentence means "B can be embedded into S(B)" (by some injective mapping defined immediately after), rather than "B is a subset of S(B)". What do you suggest? Would it be better to write: "Note that B can be embedded..."? Kope (talk) 20:39, 9 September 2011 (UTC)
replies: text: No, I wonder about the essence of that claim, not just the form of the phrase. Let us take the simplest case: a finite Boolean algebra. It contains 2n elements, where n is the number of points in S(B). In which sense can the 2n-element set be embedded into the n-element set? Boris Tsirelson (talk) 16:40, 10 September 2011 (UTC)
replies: text: The other way around.An n-element set into a 2n-element set. Kope (talk) 13:59, 11 September 2011 (UTC)
replies: text: That is, "S(B) can be embedded into B"? First, this is not the claim in the article. Second (and worse), it works for the simplest case (of finite Boolean algebras), but fails in general. Take for instance an atomless countable Boolean algebra (it is unique up to isomorphism). Its Stone space is the Cantor set (up to homeomorphism). Which way do you embed the continuum of points of the Cantor set into the countable Boolean algebra? Boris Tsirelson (talk) 19:11, 11 September 2011 (UTC)
replies: text: Nope. B has n elements and it embeds in the 2n-element S(B). Kope (talk) 21:19, 11 September 2011 (UTC)
replies: text:  :-( Boris Tsirelson (talk) 05:19, 12 September 2011 (UTC)

text: And then it should be also explained that a principal filter is related to a subset of the Stone space... Boris Tsirelson (talk) 11:32, 12 September 2011 (UTC)
replies: text: We crossed edits. I added a sentence just now about how the isomorphism is defined, is that what you were thinking of? — Carl (CBM · talk) 11:34, 12 September 2011 (UTC)
replies: text:  :-) Boris Tsirelson (talk) 11:39, 12 September 2011 (UTC)
thread_title:  Strange phrase 

message: It says on this page
{x \in S(B) : b \in x }
Where b is an element of $B$
When defining the basis. If x \in S(B), then it is an ultrafilter on B, meaning it should be a set of subsets of B. Thus, b \in x notationally implies b is a subset of B rather than an element of B.  — Preceding unsigned comment added by 131.215.220.161 (talk) 21:34, 14 April 2016 (UTC)
thread_title:  Singleton or element