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What happens when you try to set n to an infinite ordinal? (where n is the iterations)

WIP!

Part 1 - Something's not right...

Now, let's start with a simple ruler, the one from 0 to ω.

If we set n to ω, there are no problems. Just that the ruler is now infinite. It should look something like this (where the dots represent the space inbetween, and let the elements be separated by commas)

0,...........................,1,.........,2,...,3,...,ω

However, if we set n to ω+1, the problems start.

0,...........................,1,.........,2,...,3,...,ω

Wait, what? There are 2 ω's?

See, when we set n to ω+1, the ω at the end will correspond to n, so it will correspond to ω+1, and the element at the end, n-1, will correspond to ω+1-1 = ω.

And let's see what happens if we try to put n as ω+2:

0,...........................,1,.........,2,...,3,...,ω, ,ω+1

Great. Now we have 2 ω's and one ω+1 that's out of the range. The same logic applies as before, but while trying to find the second to last element, we needed to substract 1 from the last entry, getting us ω.

And how's about ω2 as n?

0,.........,1,...,2,...,ω,.........,ω+1,...,ω+2,...,ω

Now we have the whole set of ω+n inserted in the ruler. Using ω3 as n doesn't help:

0,...,1,...,ω,...,ω+1,...,ω2,...,ω2+1,...,ω

Now we have the whole set of ω+n PLUS ω2+n inserted in the ruler. Let's use ω2.

0,...,ω,...,ω2,...,ω3,...,ω4,...,ω5,...,ω

Now we have the whole set of ωx+y. In fact, for any n of the form ωm and smaller than ωω, it gets us the whole set of ωm-1a+ωm-2b+ωm-3c+...+ω2x+ωy+z.

But what happens when n=ωω?

Part 2: Twisted Arithmetic

Part 3: It's literally cancer

Part 4: More twisted arithmetic

Example

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