An uncountable countable set
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From: Virgil on 7 Sep 2006 17:58 In article <45004abc(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > I find it disappointing, but not surprising, that you don't understand > such a simple proof, since it's contradictory to your education. I do > find it annoying that you feel the right to disagree with it without > understanding it. If you feel there is a problem with the proof, please > state the logical error I made. If the string is all finite bits, and > none of them ever can possibly achieve an infinite value, then how can > the string have an infinite value? There's nowhere in the string where > that can occur. It's that simple. Grok it. What no one else can understand is why TO thinks any string need have an "infinite value". The rest of us can deal quite well with infinitely many ( and endless sequence of) finite strings, so long as there is no finite upper limit on the finite lengths of those finite strings. If TO cannot deal with that, the more fool he.
From: Virgil on 7 Sep 2006 18:00 In article <45004b50(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > At which bit does this string attain an infinite value? At which element in an endless sequence does it end?
From: Virgil on 7 Sep 2006 18:05 In article <45004b9b$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Dik T. Winter wrote: > > In article <4500215f(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > > writes: > > > So, while ...111 may not be considered a standard natural, I see no > > > reason why it should not be considered, say, an extended natural. > > > > Why not use the proper name such numbers already have? 2-adics. > > No reason, except that 2-adics could possibly have infinite bit > positions. Not standard 2-adics. TO's version of anything can be unregulated but standard n-adics are not. > I wanted to make sure that detail was not lost, because it's > crucial to the proof of finiteness. TO has yet to provide any proof of anything, much less a proof of the finiteness of things which are not.
From: Virgil on 7 Sep 2006 18:07 In article <45004ccc(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > stephen(a)nomail.com wrote: > > Tony's objection was that because you are always adding balls, > > you can never get 0. But apparently his own math says that > > if you keep adding 1, you eventually get, ..111111111111, and > > if you add 1 to that, you get 0. So in Tony's world, if > > you just add 1 ball at a time, and never remove any balls, > > you can end up with 0 balls at noon. > > > > Stephen > > > > It's nice to see you are enjoying making fun of the number circle. Making fun of TO's idiocies is a growth industry.
From: Virgil on 7 Sep 2006 19:10
In article <450059fb(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > Virgil wrote: > >>> Then show that the set of all natural numbers does not have cardinal > >>> number > >>> aleph-0 and ordinal number w. A proof please. > > > > Tony Orlow wrote: > >>> I have already shown how the set of bit positions in the binary naturals > >>> has no cardinal or ordinal that can be assigned to it. > > > > Virgil wrote: > >>> TO has claimed it, but not proved it. > >>> TO claims a lot but never proves any of it. > >>> Since the set of digit positions in any positional notation for the > >>> members of N is indexed by N, the cardinality of the set of bit > >>> positions equals the cardinality of its index set. > > > > Tony Orlow wrote: > >> Since the bit strings are the power set of the set of bit positions, > >> each natural being a unique subset of 1 positions, the set of naturals > >> is power set to the set of bit positions. > > > > That sounds reasonable at first glance. But further study shows > > that there is a direct one-to-one correspondence between the bitstrings > > and the naturals. So the logical conclusion is that they are the same > > size. If you associate all the (finite) bitstrings with naturals, they > > are in fact the same set. > > > > So there can't be an in-between cardinality because the sets > > have the same cardinality. > > > > > >> In your concoction, the naturals are power set to nothing. > > > > Exactly. To be so, there would have to be an infinite set smaller > > than N that could not be bijected with N. But there is no such set. > > > > Specifically, like I said above, the set of bit positions bijects with > > the set of bitstrings. For every natural, there is a bit index, and > > vice versa. For every natural, there is a bitstring, and vice versa. > > So every bit index is a natural, and every bitstring is a natural. > > Hence they are exactly the same set. > > > > That's my point. You can biject the naturals with the bit posiitons, and > also with the strings, but the set of strings is the power set of the > bit positions. The set of finite strings is NOT the power set of the set of finite bit positions. It is only the set of infinite strings that can represent the entire power set of the set of finite bit positions. So TO's "point" was pointless. |