Genady

What do you mean, it solves them right away?

So, after taking cube of the expression in question, and simplifying, we get it (i.e., the cube) being equal to Do you see something peculiar here?

No, I don't know.

If we define "a type T set" to be "a set of numbers from 1, increasing by 1, up to a finite number", then there is no infinite set of type T.

Yes, it is. But there is no such set in your construction. IOW, there is no set in your construction "that starts at 1, increases by 1 and has infinite elements."

I understand. There is no set equal to N in your construction. There is nothing to compare to because there is no infinite set in your construction that starts at 1 and increases by 1.

There are never infinite elements and so no such n is needed. There is no such jump.

A.I. says,

It doesn't go through the entire NN every time, but rather a random subset. So, it produces different response every time you run it. I got a similar, correct answer on the 6th trial.

Any comment? Question?

This is a possibility. Or, just hanged around with a black hole for a while.

Yep. This reminded me of my physics teacher who liked to say, "When I ask them any question, they give me any answer."

You are right. +1 Here is my take:

Because it never goes from R to N.

This does not exist. It never happens. There is no such symmetry breaking. There are only finite numbers of elements in R on the left side. As has been said above, R(n) is always finite. There are only R's on the left, never N.

Yes.

R(n) does not change to infinite. R(n) and {R(n)} are different things. The former contains numbers in the range [1, n]. The latter contains sets R(n) for all n's. The former is finite, the latter is not.

Correct. R(n) is finite. {R(n) | n∈N} is not. R(n) = {1, 2, 3, and all other numbers up to n} LIST = {R(n) | n∈N} = {R(1), R(2), R(3), and all other R(n)'s}

This last example not only is not permitted, but it does not have any meaning in the set of natural numbers. Infinity is not an element of this set. This example does not make sense. Yes, each R(n) has n elements. None. Each R(n) is finite.

No. Yes.

No. Yes.

However, if n=5 there are no 5 rows/sets, but one. If you want to discuss a different mapping, then define it first.

Why did you obtain it?

No, for each n there is one and only one row, the row number n. This row has nothing to do with other rows. For each n there is one set, R(n). The list is a set of these sets. Let's call it LIST. This set, LIST is defined so that for each n∈N the set R(n)∈LIST, and for each element Q∈LIST there exists n∈N such that Q=R(n). There is no "implies" anywhere in the definitions.

Yes, each set R(n) is finite. You said something else in the previous post: This is incorrect. The list of sets is not finite. In other words, for each n, the set R(n) is finite. But the set {R(n) | n∈N} is not finite.