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## Main Question or Discussion Point

I wrote a long response hoping to get to the root of AlienRender's confusion, but the thread closed before I posted it. So I'm putting it here.

let ##f(n) = 2n##, this is a one-one and onto mapping of the natural numbers onto the even natural numbers.

Let ##L## be the enumeration/list ##2, 4, 6, 8, ...## . And lets call ##L## a thing, a complete object.

Now suppose you want to take ##L## and concatenate some more numbers, say ##3k##, for all ##k## in ##N##, to make an extended list, ##L'## . (For now lets ignore the problems with the idea of concatenating to an infinite list. )

Well, wait a minute, isn't there a problem? Because if we add these new numbers, we need new rows/natural numbers to correspond them to, but we've used them all up. It seams there is a contradiction since ##L'## is a subset of ##N##, yet we can't seam to construct ##L'## from ##L## in a way that preserves a bijection.

It would seam that ##g(n)=L'_n## is not a bijection with ##\{L'_i \mid i \in N\}## and ##N##, so does this imply that ##|\{L'_i \mid i \in N\}| > |N|##, a contradiction since it's a subset of ##N##? What in the world is ##g(n) = L'_n## if not ##f(n)## which it cannot always be? Conclusion? Well, you can't concatenate to the end of an infinite list can you?

Actually, IMO, infinite enumerations should be considered as a processes. What does it mean to concatenate processes? Not the same thing as what one imagines when they think of concatenating two lists. In this view, diagionlization would perhaps be considered a coupled and synchronized process in which for each step of process 1, there is a step of process 2, yet there is no final step.

Lastly, on a tangent, we can't do diagonalization on an enumeration of ##N## in unary, because we have only one symbol. If we use binary, we have the problem that the number of digits grows slower than the number of rows. We could however do something similar, just one by one, add the numbers on the "rows" construct the new one, call it ##x##, through the application of ##g(n)=\sum_{i=1}^nn## at step ##n##. ##L_n=g(n)## is fine, but the problem is that ##x## is infinite and thus not a natural number, so it was never supposed to be on the list in the first place. This is why the proof would not work, meaning that diagonalization in this way doesn't prove ##|N|<|N|##.

In the case of diagonalization of an assumed list of all of the reals, we don't have the limitation, since we are working in the fractional parts of the numbers, thus the constructed number is still finite in size, and reals are allowed to require an infinite length representation.

It seams you're argument is something like this:You know very well what digits and rows. The diagonal uses it for goodness' sake. Please stop this nonsense.

When you ASSUME that there are as many real numbers as there are digits in a single real number, this isn't true for N either. It's a given that it isn't true. If it's not true for N, what does it matter that it's not true for R?

let ##f(n) = 2n##, this is a one-one and onto mapping of the natural numbers onto the even natural numbers.

Let ##L## be the enumeration/list ##2, 4, 6, 8, ...## . And lets call ##L## a thing, a complete object.

Now suppose you want to take ##L## and concatenate some more numbers, say ##3k##, for all ##k## in ##N##, to make an extended list, ##L'## . (For now lets ignore the problems with the idea of concatenating to an infinite list. )

Well, wait a minute, isn't there a problem? Because if we add these new numbers, we need new rows/natural numbers to correspond them to, but we've used them all up. It seams there is a contradiction since ##L'## is a subset of ##N##, yet we can't seam to construct ##L'## from ##L## in a way that preserves a bijection.

It would seam that ##g(n)=L'_n## is not a bijection with ##\{L'_i \mid i \in N\}## and ##N##, so does this imply that ##|\{L'_i \mid i \in N\}| > |N|##, a contradiction since it's a subset of ##N##? What in the world is ##g(n) = L'_n## if not ##f(n)## which it cannot always be? Conclusion? Well, you can't concatenate to the end of an infinite list can you?

Actually, IMO, infinite enumerations should be considered as a processes. What does it mean to concatenate processes? Not the same thing as what one imagines when they think of concatenating two lists. In this view, diagionlization would perhaps be considered a coupled and synchronized process in which for each step of process 1, there is a step of process 2, yet there is no final step.

Lastly, on a tangent, we can't do diagonalization on an enumeration of ##N## in unary, because we have only one symbol. If we use binary, we have the problem that the number of digits grows slower than the number of rows. We could however do something similar, just one by one, add the numbers on the "rows" construct the new one, call it ##x##, through the application of ##g(n)=\sum_{i=1}^nn## at step ##n##. ##L_n=g(n)## is fine, but the problem is that ##x## is infinite and thus not a natural number, so it was never supposed to be on the list in the first place. This is why the proof would not work, meaning that diagonalization in this way doesn't prove ##|N|<|N|##.

In the case of diagonalization of an assumed list of all of the reals, we don't have the limitation, since we are working in the fractional parts of the numbers, thus the constructed number is still finite in size, and reals are allowed to require an infinite length representation.

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