Cantor's Diagonal?
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From: Virgil on 4 Feb 2010 15:45 In article <4a86efc3-0a32-4e3e-85dc-3f65b764564a(a)q4g2000yqm.googlegroups.com>, zuhair <zaljohar(a)gmail.com> wrote: > > > It seems to me that a modification of this argument can actually work > > > for every well orderable set, however I don't know if a modification > > > of this argument can be made general enough to prove that the power of > > > every non well orderable set is bigger than it. > > > > The standard proof that the cardinality of a power set is greater than > > that of the base set in no way requires that either of the sets be > > ordered, much less well ordered. > > Agreed, provided what you mean by the standard proof the one in which > prove that *every* function from a set to its power is not surjective. It is enough to show that ANY such function fails to be surjective, which one can do by considering for any function a set determined by that function which is not in its range.
From: zuhair on 4 Feb 2010 18:59 On Feb 4, 3:45 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <4a86efc3-0a32-4e3e-85dc-3f65b7645...(a)q4g2000yqm.googlegroups.com>, > > zuhair <zaljo...(a)gmail.com> wrote: > > > > It seems to me that a modification of this argument can actually work > > > > for every well orderable set, however I don't know if a modification > > > > of this argument can be made general enough to prove that the power of > > > > every non well orderable set is bigger than it. > > > > The standard proof that the cardinality of a power set is greater than > > > that of the base set in no way requires that either of the sets be > > > ordered, much less well ordered. > > > Agreed, provided what you mean by the standard proof the one in which > > prove that *every* function from a set to its power is not surjective. > > It is enough to show that ANY such function fails to be surjective, > which one can do by considering for any function a set determined by > that function which is not in its range. ANY is EVERY. Zuhair
From: zuhair on 4 Feb 2010 19:02 On Feb 4, 1:57 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > Zuhair, did you read my post? What do you get when you consider the > very simple situation I described? Hi Moe, yes I read it, but I wanted more details. Thanks Zuhair
From: zuhair on 4 Feb 2010 19:16 On Feb 4, 1:57 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > Zuhair, did you read my post? What do you get when you consider the > very simple situation I described? So that you don't misread my last reply. The details that I wanted was already provided by another discusser in this thread. Thanks again. Zuhair
From: MoeBlee on 4 Feb 2010 19:31
On Feb 4, 6:16 pm, zuhair <zaljo...(a)gmail.com> wrote: > The details that I wanted > was already provided by another discusser in this thread. Okay, so at this point do you have any remaining questions or doubts that the axioms of Z set theory prove that there is no bijection between the set of natural numbers (w) and {f | f: w -> {0 1}}? MoeBlee |