Creating the set of natural numbers

Question

I am not a mathematician but an engineer, so I can read some basics of the language proofs are written in. Second I am bad in probability and infinity and my question covers both. So I like to apologize for not using a mathematical language here:

Suppose you have an urn with as much black as white balls of infinite amount. Could such a set be defined without any logical conflicts?

Now start drawing a random amount of balls between 0 and infinity. Put the balls in order as you draw them and think of them as white being a zero and black being a 1. The order and the color will now represent a binary number.

If I repeat this process for an infinite number of times, will this produce the set of natural numbers? Could this be proofed?

Answer 1

Firstly, let me address your first question. Consider the following experiment:

There is an urn with one black and one white ball. We draw a ball, write down $0$ if it's white, $1$ if it's black, then put the ball back.

Repeating this experiment is clearly equivalent to taking balls from the urn in your experiment. This hopefully settles your concerns; otherwise, the concept of a multiset may interest you (see Wikipedia; it can be generalised to infinitely many copies of the same element).

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The next concern is a means to decide on the number of balls we're going to take from the urn (or, how many times the experiment above is carried out). We can't simply say "We'll take any natural number, all with equal probability" because this probability would become $0$ (suppose it were some $a >0$; then the probability of selecting a number wouldn't be $1$, but $\sum\limits_{n \in \Bbb N} a = +\infty$).

Therefore a different probability distribution has to be used, e.g. a geometric distribution (cf. Wikipedia).

For reasonable choices of distribution, there will be a nonzero chance of generating any number (because each number $n$ has a binary expansion of certain length $k$; if $k$ can be selected by our distribution, we can get $n$ by a proper selection of black and white balls).

Presuming that you mean by "repeating an infinite number of times" the limit of drawing finitely many times (otherwise it's rather hard to make mathematical sense of this statement) we will in general end up with the notion of almost surely, meaning that it occurs with probability $1$ (which isn't the same as that it necessarily occurs; see again Wikipedia).

The only result I could presently come up with is (assuming reasonable distribution in the sense above):

Let $n$ be a natural number. Then $n$ will be selected almost surely.

This is however not equivalent to saying that we will almost surely obtain all of $\Bbb N$. An analogy: If we were to generate an infinite bit-string, with $0,1$ equiprobable (both occurring with $p = \frac12$) then we would almost surely not end up with $000\ldots$, or for that matter, any specific bitstring. But we would end up with some bitstring.

A more rigorous development of these ideas leads into the realms of measure theory. I hope this addresses most of your questions.

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Addendum: As Henning Makholm pointed out in a comment, we actually have:

All of $\Bbb N$ will be generated almost surely.

This is because "All of $\Bbb N$ is generated" is the complement of the event:

Some $n$ is not generated.

whose probability is (bounded above by — ignoring the case of two or more omitted integers) $\sum\limits_{n \in \Bbb N} \mathsf{Pr}(\text{$n$ is not generated}) = \sum\limits_{n \in \Bbb N} 0 = 0$.

Answer 2

Unfortunately the process goes wrong in a couple of places.

The first place it goes wrong is when you say 'draw randomly'. From your knowledge of the uniform distribution on finite sets (i.e. you pick any one ball with the same probability as any other), it should seem intuitively obvious what this means in this context... but unfortunately it breaks down: you can't define a uniform probability distribution on a countable set. If you did, and the probability that any given ball is drawn with probability $p$, then you'd need $$p+p+p+\cdots = 1$$ but if $p=0$ then $p+p+p+\cdots = 0$ and if $p>0$ then $p+p+p+\cdots = \infty$, so this can't happen.

The second place it goes wrong is that an infinite binary string usually won't represent a natural number: for example, the infinite sequence of $1$s, $11111\dots$, isn't a natural number.