Product of natural numbers (self-referentiality).

Question

Assume that $\mathbb{R}$ is defined axiomatically as a complete ordered field, and let us define the set of natural numbers $\mathbb{N}$ as the smallest inductive set of $\mathbb{R}$. In other words, $$\mathbb{N}=\{1,1+1,1+1+1,\ldots\} .$$

This is the standard quick (top-down) approach in Analysis (like in Baby Rudin) to introduce the set of real numbers and its distinguished subsets.

Question How can one write a proof that $\mathbb{N}$ is closed under multiplication, i.e., that given two real numbers $x,y\in\mathbb{R}$, if $x,y\in \mathbb{N}$ then $x\cdot y$ in $\mathbb{N}$.

The "problem" is that I have to show the proof to first-year calculus students, and I cannot write the proof without expecting philosophical questions about some kind of "self-referentiality" from them. I would like to give concise proof that convinces students there is no problem with that.

Any help?

PS This "problem" is also present if one wants to show that $\mathbb{N}$ is closed under addition, but it is less evident, and usually, I don't get philosophical questions on that. I want to avoid explaining the difference between theory, metatheory, etc.

Answer 1

Let $$S=\{x\in\mathbb{N}:(\forall y\in \mathbb{N})(y+x\in\mathbb{N})\}$$ and $$M=\{x\in\mathbb{N}:(\forall y\in\mathbb{N})(yx\in\mathbb{N})\}.$$

Then $S$ and $M$ are inductive subsets of $\mathbb{N}$, and are therefore equal to $\mathbb{N}$.

By definition of inductive, $y+1\in \mathbb{N}$ for all $y\in\mathbb{N}$. For $x\in S$ and $y\in\mathbb{N}$, by definition of $S$, associativity of addition, and definition of inductive, $y+(x+1)=(y+x)+1\in\mathbb{N}$. So $x+1\in S$ and $S=\mathbb{N}$.

For all $y\in\mathbb{N}$, $y\cdot 1=y\in \mathbb{N}$, so $1\in M$. For $y\in\mathbb{N}$ and $x\in M$, $$y(x+1)=yx+y\in \mathbb{N}+\mathbb{N}\subset\mathbb{N},$$ using the fact that $S=\mathbb{N}$.

Since you asked about multiplication but not addition, you may have already established that $S=\mathbb{N}$, in which case the first half of this solution can just be recalled.

Answer 2

Intuitive proof:

We know that the naturals are closed under addition. Let $a = 1 + 1 + 1 + \ldots$ and let $b = 1 + 1 + 1 + \ldots$. We know that $ab = a_1 + a_2 + a_3 + \ldots + a_b$. Hence, we get

$$(1+1+\ldots)(1+1+\ldots) = (1+1+1+\ldots)_1 + (1 + 1 + 1 + \ldots)_2 + \ldots + (1 + 1 + 1 + \ldots)_b = 1 + 1 + 1 + 1 + 1 + 1 + 1 + \ldots$$

Since this is addition, and the naturals are closed under addition, the naturals must also be closed under multiplication.

Hence, $x + y \in \mathbb N \implies xy \in \mathbb N$.

I don’t know how much rigor you need, but this is what I think is an intuitive proof.

You can just rewrite this proof with the correct symbols.

No idea how this would relate to philosophical doubts.