Wikipedia: Cantors Diagonal argument/Talk

I'm glad there is an article on this in Wikipedia. Cantor's Diagonal argument is my favourite piece of Mathematics - Andre Engels

quote

Its result can be used to show that the notion of the [set of all sets]? is an inconsistent notion; if S would be the set of all sets then P(S) would at the same time be bigger than S and a subset of S.

end quote

Can the diagonal argument really do this? If we accept the existence of uncountable infinities (and i guess you do if you accept the diagonal argument) then it is pretty clear that the set of all sets must be of uncountable size if it exists. How do you define an ordered list of elements of an uncountable set? If you can't define an ordered list, how do you apply the diagonal argument? (Maybe you can, but it's not immediately obvious how. I think a bit more needs to be written about this if the above quote is to remain.)

If you look at the second proof you can see that it makes no assumptions about S being ordered. That is in fact the main difference between the actual diagonalization argument and its generalization that is used in the second proof. I will add a remark that the second proof is not equal to the actual diagonal argument but a generalization of it. -- Jan Hidders

That sounds good to me. Thanks.

OTOH, the line:

: But since there are expansions such as 0.01999... and 0.02000... that represent the same real number,

seems like pure and utter bunk. 0.01999... and 0.02000... are only the same number to computers, who can't see off into infinity. In the mathematical world, they're completely different.

Or has it been too long since my math degree?

: Yes, it has been too long. You are mistaken. --AV

Could you be slightly more terse? :-) Seriously, please expand. Wikipedia demands proof!

: Consider the following simple explanation. If a and b are two real numbers, a<b, then there is another real number strictly between them - for example, (a+b)/2. If 0.019999... and 0.20000... were distinct, then the first would be smaller than the second, and therefore their average would be strictly between them; but a quick consideration of its decimal expansion will convince you that there can be no decimal expansion strictly between 0.01999.... and 0.200..... . Therefore, the numbers represented by these two expansions are actually equal.

: This is a rather roundabout way to demonstrate the obvious, but it might be more convincing to you. --AV

: If I understand it correctly, the property of being able to represent a rational number by two infinite decimals is a logical consequence of how the real numbers are constructed. Without this property, Bad Things would happen. Comments appreciated, I am off to contemplate real numbers for a while.

: It is a consequence, but I don't see the Bad Things you mean. Some rational numbers, for instance 1/3, can only be represented by one infinite decimal. --AxelBoldt

: Right, didn't mean to claim some couldn't be represented by 1 infinite decimal, only that some require being able to be represented by 2 infinite decimals, e.g., 0.999... = 1.000.... Some Bad Thing would *surely* happen if one decided that 0.999... != 1.000..., but I have not yet taken the time to work out the consequence. Maybe after dinner...

Here's another argument: look at the formula for geometric series, and then calculate 0.999999... = 9/10 + 9/100+ 9/1000 + 9/10000 + .... = 9 ( 1/10 + (1/10)² + (1/10)³ + ... ) = 9 ( 1 / (1 - 1/10) - 1) = 1. --AxelBoldt

  1/3      0.333...
+ 2/3    + 0.666...
-----    ----------
  3/3  =   0.999... = 1

Excellent arguments, all -- thanks! I sit corrected. -- NK