Genady

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.

No, it does not. What makes you think it does? It says, "each n∈N", doesn't it? Here is the definition again: "each n∈N is mapped to set R(n)= {x∈N | 1≤x≤n}" There is no limit on n.

This is needed because mapping connects elements of two sets. In your case, each element of the domain set ("from") is a natural number, and each element of the codomain set ("to") is a set of natural numbers. Very well. So, following the definition, "each n∈N is mapped to set R(n)= {x∈N | 1≤x≤n}", please define the issue that bothers you.

The lack of reason to think otherwise.

Life does not have a purpose. But people do, sometimes.

The OP question is, Is a moral free market possible. My response is, If the free market is moral already, then surely it is possible.

It is not a different topic because if it is moral then the answer to the OP question is yes.

What is immoral in the free market?

It depends on the culture, I guess. In my school, when we were given an equation to solve, there was always a possibility that the equation does not have a solution, in which case we were expected to prove that it is so. The idea of your proof is right and the proof is very close, except the issue you've mentioned about the area covered by the points. This issue can be eliminated by a modification of the proof where instead of removing points from a 1x1 square we find where to put a point. (I hope this last hint helps rather than distracts.) +1