Talk:Axiom of power set

message: Maybe somebody should add a note to the article explaining that the "subset" referred to is the notion of subset within the theory (in an attempt to ward off the confusion you typically get when somebody first hears that ZFC has countable models). -- Cwitty 01:11, 22 November 2003 
:I added a "Limitations" section to try to address this, since I think I had this misunderstanding for a while in regard to what the power set axiom implies (and was pretty unsettled regarding ZF set theory as a result until this was resolved). Germyb2 (talk) 04:41, 11 January 2023 (UTC)
replies: text: not really, as ⊂ then needs to be defined, lengthening the statement --Henrygb 11:18, 12 October 2005 (UTC)
thread_title:  Subset in what model? 

message: The article states that the Cartesian product of X and Y is a subset of the power set of the power set of the union of X and Y. But isn't it a subset of the power set of the union of X and Y? Why the double power set? Am I missing something? 
X = {a, b} Y = {c, d}
XuY = {a, b, c, d}
P(XuY) = {{}, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}, {a, b, c, d}}
XxY = {{a, c}, {a, d}, {b, c}, {b, d}}
I probably am. But what is it, I wonder... 24.238.113.229 (talk) 05:52, 8 June 2008 (UTC)
replies: text: You are confusing unordered pairs with ordered pairs. JRSpriggs (talk) 12:03, 8 June 2008 (UTC)
replies: text: Yes. Yes I am. Is that detail too elementary to be included in the article? It clarifies a lot, at least for me. 24.238.113.229 (talk) 21:14, 8 June 2008 (UTC)
replies: text: I tried to make it clearer in the article. How do you like it now? JRSpriggs (talk) 03:57, 9 June 2008 (UTC)
replies: text: That's quite helpful, thanks! I have been browsing through Wikipedia's articles on Set Theory for some time, and while I am generally able to work through them, there are cases when I happen to be unacquainted with some basic idea (in this case (x, y) = {{x}, {x, y}}), which prevents me from following the argument. I think the problem is that the dense nature of mathematical argumentation and notation often obscures details that would, if spelled out, be heavily wikified. As a mathematical autodidact, I would love it if Wikipedia would spell such things out just a bit more, so that I could more easily find and rectify the gaps in my knowledge. 24.238.113.229 (talk) 23:26, 11 June 2008 (UTC)
thread_title:  Cartesian product defined by just one powerset operation -- I think? 

message: I mean, what about "the set of all sets". Surely "the set of all sets" cannot have a power set as the power set of "the set of all sets" would be a set larger then "the set of all sets". But "the set of all sets" is the set of all sets and therefore the largest set. So this axiom leads too absurdity. I for one would rather ditch the axiom of power sets then the axiom of self consistency, but that is just me. It's "All that is is The All and yet The All is All that is." Vs. "Well surely there is something more then The All which contains The All. That which is above is as below, ya know." I mean it really cuts deep into the hermetic controversy. --Michaelidman  —Preceding unsigned comment added by 66.31.206.34 (talk) 04:15, 5 January 2009 (UTC)
replies: text: You are assuming that there is a set of all sets. This is not true in most forms of set theory including the favorite ZFC. JRSpriggs (talk) 10:05, 5 January 2009 (UTC)
thread_title:  Why is this uncontroversial?

message: In this article, the inductive definition of the cartesian product is not associative.  So, Ax(BxC)≠(AxB)xC.  --Gallusgallus (talk) 18:54, 29 March 2010 (UTC)
replies: text: Yes, that's true. The same is true for any other definition of binary Cartesian product. The elements of Ax(BxC) are pairs the second element of which is an element of BxC; the elements of (AxB)xC are pairs the second element of which is an element of C. So Ax(BxC) cannot possibly equal (AxB)xC regardless of how the pairs are represented in set theory. — Carl (CBM · talk) 02:37, 30 March 2010 (UTC)
thread_title:  Cartesian product of finite collection