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More specifically:

What is the maximum Cardinality of the set of all diagonal new numbers producible from a single arbitrary countably infinite list comprised of either unique rational or irrational numbers ?

The original list remains static and unchanged, only new diagonally derived numbers from the original list form a new set which will be of max Cardinality - what ?

"Diagonally derived" means a number in the new set was produced by using the original list as a "seed or reference point" and systematically altering each diagonal digit across the range of the number base being used so that any number in the new set conforms to Cantor's "not on the list" claim.

Discuss.

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  • $\begingroup$ The question has to be made more precise. Under one interpretation, the answer is "1": take the diagonal number that results from the given sequence of numbers, and you are done. Under another interpretation, the answer is $\omega_1$: start in the same way as before; add the new number to the sequence somewhere; then take the diagonal again; repeat $\omega_1$ many times. $\endgroup$ – Casper Apr 26 '17 at 1:56
  • $\begingroup$ I'm afraid it's still not precise. What do you mean by the new numbers being "diagonally derived from the original list". $\endgroup$ – Casper Apr 26 '17 at 2:10
  • $\begingroup$ To me it sounds like you're asking 'how many modifications can we make to the diagonal to produce new numbers,' this is one such choice for however many elements are in the diagonal. So if you used a decimal expansion and had countably many choices, we can do $10^{\aleph_0}$ possibilities. $\endgroup$ – Alfred Yerger Apr 26 '17 at 2:14
  • $\begingroup$ It's the cardinality of the reals. Choosing each element of the diagonal is the same as choosing the decimal digits of a real number. But you can see this another way. Given a countable list of reals, there are uncountably many reals not on the list. Any one of them could be the antidiagonal. $\endgroup$ – user4894 Apr 26 '17 at 3:23
  • $\begingroup$ (user4894) Great intuitive argument. But the complaint is: The "systematically altering" procedure, which Cantor himself uses, seems to only generate more countable numbers. So don't you kind of have to "imagine" there are numbers that are not on the list that somehow explode into total uncountability ? $\endgroup$ – user439222 Apr 26 '17 at 4:42
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Assuming you have a finite number, at least $2$, of choices at each position and a countably infinite set of positions, the number is $\mathfrak c$, the power of the continuum. You have at least $2^{\aleph_0}$ possible numbers and less than $\aleph_0^{\aleph_0}$. $\mathfrak c = 2^{\aleph_0}\le \aleph_0^{\aleph_0} \le (2^{\aleph_0})^{\aleph_0}=2^{(\aleph_0^2)}=2^{\aleph_0}=\mathfrak c$

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