Can we raise aleph cardinals to other aleph cardinals?
in [Math]
|
Prev: Arthur Rubin's churlish notion of a person's nickname Re: Wikipedia failing to follow their own rules, and Wikipedia abuse of defamation of character
Next: solution manual of basic principle and calculation in chemical engineering by David Himmelblau
From: Dave Seaman on 30 Sep 2008 07:29 On Mon, 29 Sep 2008 22:09:33 -0700 (PDT), lite.on.beta(a)gmail.com wrote: > What would aleph0 ^ aleph0 be? aleph0 ^ aleph0 = 2^aleph0 = c = cardinality of the continuum > What would aleph1 ^ aleph1 be? aleph1 ^ aleph1 = 2^aleph1 > What would cardinality of P^k(N) be? > [That is taking power sets k times on naturals. If k=1, then it is > P(N) = 2^aleph0 P^2(N) = 2 ^ 2 ^ aleph0 = 2 ^ c p^3(N) = 2 ^ 2 ^ c, etc. > c^alpeh_i is easier since c = 2^aleph_0, therefore > c^alpeh_i = 2^aleph_i*aleph0 = 2^aleph_i > Can anyone give me a set with cardinality aleph_1? The set of countable ordinals. In fact, aleph_1 happens to be the name given to that very set. The cardinality of aleph_1 is aleph_1. > No derivations required. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. <http://www.indybay.org/newsitems/2008/03/29/18489281.php>
From: lite.on.beta on 30 Sep 2008 12:41 On Sep 30, 7:29 am, Dave Seaman <dsea...(a)no.such.host> wrote: > On Mon, 29 Sep 2008 22:09:33 -0700 (PDT), lite.on.b...(a)gmail.com wrote: > > What would aleph0 ^ aleph0 be? > > aleph0 ^ aleph0 = 2^aleph0 = c = cardinality of the continuum > > > What would aleph1 ^ aleph1 be? > > aleph1 ^ aleph1 = 2^aleph1 > Which rule(in math for cardinals) are you using to change aleph1 to 2 (or alpeh0 to 2 in first one)? Or is this just a theorem involving the actual sets whose cardinalities would be those (i.e., like the theorem that proves 2^aleph0 = c or the one that says |P(A)| = 2^|A| ) ? If it involves only cardinal math, how did you do that? > > > What would cardinality of P^k(N) be? > > [That is taking power sets k times on naturals. If k=1, then it is > > P(N) = 2^aleph0 > > P^2(N) = 2 ^ 2 ^ aleph0 = 2 ^ c p^3(N) = 2 ^ 2 ^ c, etc. > Yes makes sense. So all you can simplify is the "head"=(2^aleph0) of the tower of exponentiations. > > > c^alpeh_i is easier since c = 2^aleph_0, therefore > > c^alpeh_i = 2^aleph_i*aleph0 = 2^aleph_i > > Can anyone give me a set with cardinality aleph_1? > > The set of countable ordinals. In fact, aleph_1 happens to be the name > given to that very set. The cardinality of aleph_1 is aleph_1. > > I will get back to this once I learn more about ordinals. Thank you. > > > No derivations required. > > -- > Dave Seaman > Third Circuit ignores precedent in Mumia Abu-Jamal ruling. > <http://www.indybay.org/newsitems/2008/03/29/18489281.php>
From: Dave Seaman on 30 Sep 2008 12:57 On Tue, 30 Sep 2008 09:41:01 -0700 (PDT), lite.on.beta(a)gmail.com wrote: > On Sep 30, 7:29?am, Dave Seaman <dsea...(a)no.such.host> wrote: >> On Mon, 29 Sep 2008 22:09:33 -0700 (PDT), lite.on.b...(a)gmail.com wrote: >> > What would aleph0 ^ aleph0 be? >> >> aleph0 ^ aleph0 = 2^aleph0 = c = cardinality of the continuum >> >> > What would aleph1 ^ aleph1 be? >> >> aleph1 ^ aleph1 = 2^aleph1 >> > Which rule(in math for cardinals) are you using to change aleph1 to 2 > (or alpeh0 to 2 in first one)? The law of exponents and the Schroeder-Bernstein theorem. 2^aleph1 <= aleph1 ^ aleph1 <= (2^aleph0) ^ aleph1 = 2 ^ (aleph0 * aleph1) = 2 ^ aleph1. > Or is this just a theorem involving the actual sets whose > cardinalities would be those (i.e., like the theorem that proves > 2^aleph0 = c or the one that says |P(A)| = 2^|A| ) ? > If it involves only cardinal math, how did you do that? The law of exponents can be derived from the definitions of multiplication and exponentiation in terms of sets. If A and B are cardinals, then A*B = |AxB| (Cartesian product) and A^B = |{ f : f : B -> A }| (set of mappings from B into A). >> >> > What would cardinality of P^k(N) be? >> > [That is taking power sets k times on naturals. ?If k=1, then it is >> > P(N) = 2^aleph0 >> >> P^2(N) = 2 ^ 2 ^ aleph0 = 2 ^ c p^3(N) = 2 ^ 2 ^ c, etc. >> > Yes makes sense. So all you can simplify is the "head"=(2^aleph0) of > the tower of exponentiations. There is even a name for these cardinals. We have beth_0 = aleph_0, and beth_(n+1) = 2 ^ beth_n. Therefore |P^k(N)| = beth_k. >> >> > c^alpeh_i ? is easier since c = 2^aleph_0, therefore >> > c^alpeh_i = 2^aleph_i*aleph0 = 2^aleph_i >> > Can anyone give me a set with cardinality aleph_1? >> >> The set of countable ordinals. ?In fact, aleph_1 happens to be the name >> given to that very set. ?The cardinality of aleph_1 is aleph_1. >> >> > I will get back to this once I learn more about ordinals. Thank you. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. <http://www.indybay.org/newsitems/2008/03/29/18489281.php>
From: Arturo Magidin on 30 Sep 2008 14:13
In article <08d8da7f-4fb2-48fa-b3e5-d0224a6e52ee(a)u57g2000hsf.googlegroups.com>, <lite.on.beta(a)gmail.com> wrote: >On Sep 30, 7:29=A0am, Dave Seaman <dsea...(a)no.such.host> wrote: >> On Mon, 29 Sep 2008 22:09:33 -0700 (PDT), lite.on.b...(a)gmail.com wrote: >> > What would aleph0 ^ aleph0 be? >> >> aleph0 ^ aleph0 =3D 2^aleph0 =3D c =3D cardinality of the continuum >> >> > What would aleph1 ^ aleph1 be? >> >> aleph1 ^ aleph1 =3D 2^aleph1 >> > >Which rule(in math for cardinals) are you using to change aleph1 to 2 >(or alpeh0 to 2 in first one)? >Or is this just a theorem involving the actual sets whose >cardinalities would be those (i.e., like the theorem that proves >2^aleph0 =3D c or the one that says |P(A)| =3D 2^|A| ) ? > >If it involves only cardinal math, how did you do that? It's a theorem. kappa^lambda is the cardinality of the set of all functions from lambda to kappa. Since 2 < aleph_0, it follows that 2^{aleph_0} <= {aleph_0}^{aleph_0}. But aleph_0 = aleph_0*aleph_0, so 2^{aleph_0} = 2^{aleph_0*aleph_0} = (2^{aleph_0})^{aleph_0}. Since aleph_0 < 2^{aleph_0}, it follows that (aleph_0)^{aleph_0} <= (2^{aleph_0})^{aleph_0} = 2^{aleph_0} So we have 2^{aleph_0} <= (aleph_0)^{aleph_0} <= 2^{aleph_0} from which equality follows. A similar argument shows that 2^{aleph_1} = (aleph_1)^{aleph_1}. In fact, that 2^{aleph_n} = (kappa)^{aleph_n} for every cardinal kappa with 2<=kappa<=aleph_{n+1}, for each ordinal n (since aleph_{n+1}<=2^{aleph_n} always holds). -- ====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes" by Bill Watterson) ====================================================================== Arturo Magidin magidin-at-member-ams-org |