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P: 39
 Quote by willem2 I don't understand this at all. The proposed list will have an countably infinite number of reals, and each real will have an countably infinite number of digits (adding an infinite number of zero's if necessary). For the proof, it's only necessary that for each number N, there's an N'th row, and there's an N'th column. Since both the number of rows and the number of columns is infinite, this is obviously true, so it's always possible to find the N'th number of the diagonal and change it. Since this works for all N, we can find the complete diagonal.
Yes, I understand what you are saying. But aren't you completely tossing out everything that I had previously pointed out?

You're basically assuming a "square" situation.

You're basically saying that since it's infinitely many rows and infinitely many columns that somehow makes it doable or even "square" in the sense that it's infinity by infinity in dimension.

But you're not taking into account the observations I've given. That lists of numbers represented by numerals cannot be made into 'square' lists. Pretending that "infinity by infinity" represents a square list that can be completed is the folly.

You'd also have to ignore the limitations on the slope of any line that needs to go down these lists crossing out a digit at time.

The slope of that line cannot be made steep enough to keep up with a list that would be required to represent even the natural numbers let alone the reals.

The very assumption that this problem can be viewed as a "square" infinity-by-infinity list is indeed the folly.

That's the false assumption right there that can't hold true.