Adriana Xenides Dead
in [Math]
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From: Tim Little on 8 Jun 2010 23:57 On 2010-06-09, |-|ercules <radgray123(a)yahoo.com> wrote: > What's the deduction step in ZF that says "higher infinities exist". Herc is dumb enough to think that every proof in ZF consists of a single deduction step, *the* step. - Tim
From: |-|ercules on 9 Jun 2010 00:50 "Tim Little" <tim(a)little-possums.net> wrote > On 2010-06-09, |-|ercules <radgray123(a)yahoo.com> wrote: >> Right, but statement 1 is sufficient to disprove that modifying the >> diagonal produces a new sequence of digits. > > As predicted, Herc can't understand the difference nor why it makes > nonsense of his claims. > > >> WHERE is the new sequence of digits coming from when all possible >> digit sequences are computable to all, as in an infinite amount of, >> finite lengths? > > For *all* N, the sequence differs from the Nth entry in the list at > the Nth digit (and possibly other positions as well). It is new > because for *every* sequence in the list, the question "is it the same > as this sequence" is answered "no". > So you think the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS and this does not contradict that ALL sequences of digits are on the computable list of reals up to all (an infinite amount of) digit positions? Herc
From: Tim Little on 9 Jun 2010 21:18 On 2010-06-09, |-|ercules <radgray123(a)yahoo.com> wrote: > So you think the antidiagonal comes up with an actual NEW SEQUENCE > OF DIGITS and this does not contradict that ALL sequences of digits > are on the computable list of reals up to all (an infinite amount > of) digit positions? Your question is incoherent, as it implies that every finite segment being on the list is equivalent to the entire sequence being on the list. The equivalence is false, so your question is meaningless. To be more precise: The statement "ALL sequences of digits are on the list of computable reals up to all finite digit positions" is true. The statement "ALL sequences of digits are on the list of computable reals (to infinite digit positions)" is false. Your question equates these statements. Perhaps you should focus on this issue, as it seems to be the core of your inability to understand Cantor's proof. As an aside, any list of all computable reals is not a computable list, so your use of the term "the computable list of reals" is incorrect in your context. Finally, the introduction of computable reals is a red herring: Cantor's proof had nothing to do with computability. - Tim
From: |-|ercules on 9 Jun 2010 22:06 "Tim Little" <tim(a)little-possums.net> wrote ... > On 2010-06-09, |-|ercules <radgray123(a)yahoo.com> wrote: >> So you think the antidiagonal comes up with an actual NEW SEQUENCE >> OF DIGITS and this does not contradict that ALL sequences of digits >> are on the computable list of reals up to all (an infinite amount >> of) digit positions? > > Your question is incoherent, as it implies that every finite segment > being on the list is equivalent to the entire sequence being on the > list. The equivalence is false, so your question is meaningless. > > To be more precise: > > The statement "ALL sequences of digits are on the list of computable > reals up to all finite digit positions" is true. > > The statement "ALL sequences of digits are on the list of computable > reals (to infinite digit positions)" is false. > > Your question equates these statements. Perhaps you should focus on > this issue, as it seems to be the core of your inability to understand > Cantor's proof. > I don't equate those statements, I merely substitute "all" with "infinite amount of". How many natural numbers are there in "all natural numbers"? Given that, do you still maintain you cannot parse this question? >> So you think the antidiagonal comes up with an actual NEW SEQUENCE >> OF DIGITS and this does not contradict that ALL sequences of digits >> are on the computable list of reals up to all (an infinite amount >> of) digit positions? Herc
From: Tim Little on 10 Jun 2010 01:20
On 2010-06-10, |-|ercules <radgray123(a)yahoo.com> wrote: > I don't equate those statements, I merely substitute "all" with > "infinite amount of". Incorrectly so. No sequence in the list has all its digits in common with the antidiagonal. There may be some sequences in the list that have an infinite amount of digits in common with the antidiagonal. For any finite n there may also be sequences that agree to n places. None of these descriptions are substitutable for each other. More mathematically, let A be the antidiagonal and L be the list. True: "For all n in N, there exists S in L such that S matches A to n places". False: "There exists S in L such that for all n in N, S matches A to n places". The former statement is a formalisation of "any finite number of places" and the latter formalises "infinite amount of places". You are treating the two statements as equivalent, which is why some people are saying you have quantifier dyslexia. They differ only in the ordering of the quantifiers. > Given that, do you still maintain you cannot parse this question? The question can be parsed, but it includes a counterfactual premise and so is ill-posed. You seem to like analogies, so here is one: Your repetition of your ill-posed question is a mathematical equivalent of saying "Is the elephant in your refrigerator green or yellow? STOP DODGING THE QUESTION AND ANSWER GREEN OR YELLOW!" when the truth is simply that there is no such elephant and so any question of its colour is meaningless. Likewise your substitution of different terms as if they were equivalent renders a YES/NO answer to your question meaningless. They are not equivalent, and if you want a meaningful answer you need to ask a meaningful question. - Tim |