- #1

Anixx

- 80

- 12

My impression that after we do countably infinite number of steps, the set will have countably infinite cardinality.

But what happens if we add exponential number of elements each step?

For instance, on step 0 we add 1 element, on step 1 we add 2 elements, on step 2 we add 4 elements, on step 3 we add 8 element, on step n we add 2^n elements?

For instance, by adding a binary register after the fractional point in a number we add 2^n elements:

One register after point, 2 elements:

0.0

0.1

Two registers after point, 4 elements:

0.00

0.01

0.10

0.11

and so on. After countably infinite steps we will have all numbers from the range [0,1].

But it has cardinality of continuum.

So, the question is, after adding 2^n elements on n-th step countably infinite times, will we have a set of uncountable cardinality?

Or, in other words, can we say that

**polynomial functions have germ at infinity, in a sense corresponding to countable infinity while an exponential function has a germ, corresponding to uncountable infinity**?

Notice, the cardinality of continuum is often designated as ##2^{\aleph_0}## which hints this maybe the case.

*I mean, yes, if we count all these numbers in the order of the steps, we would have only rational numbers, which is countable, but after we ALREADY COMPLETED the infinite number of steps, we not necessarily should count the numbers in the order of steps and we will have numbers with infinite expansions as well by that time.*

P.S.

**Some people say that even after infinite number of steps the numbers with infinitely long binary representations are still not included. I cannot see, why.**

PPS. Some other people said this is by definition. In that case I am interested to know what is the justification of such definition.