I know, I know, this question has been asked several times, but I feel mine is a little bit different.
Imagine a correspondence between $[0,1]$ and natural numbers in the following sense:
$$ 0.12 \longleftrightarrow 21 $$
$$ 0.443 \longleftrightarrow 344 $$
$$0.12345 \longleftrightarrow 54321 $$
Now, before you argue that in fact I'm leaving rationals as $\frac{1}{3}$, let me explain a bit more.
What this correspondence does is: given a real numer $ r \in [0,1] $, which can be expressed as (or can not?) $ r= x_1 \cdot 10^{-1} + x_2 \cdot 10^{-2} + x_3 \cdot 10^{-3} +$ … (For example: $\frac{1}{3} = 0.33333…=0.3+0.03+0.003+…$). I now define $n_r$ as $n_r=x_1 \cdot 10^0 + x_2 \cdot 10^1 + x_3 \cdot 10^2 + …$ in a recursive way. So I will have $ n_{\frac{1}{3}} = 3 + 30+300+… $ as a limit of that recursion.
What's the problem with my reasoning? If it is about that last series, isn't the nature of $\mathbb{N}$ precisely to be defined in an inductive way? So I suppose $ n_r$ defined like that is still a natural number.
PD: just an undergrad student who doesn't like infinity
Best Answer
Incidentally, I don't think the OP has made the best case for non-duplicateness that they could, so let me do that here. Ignoring expansion issues, the point is the difference between their suggested map $$0.a_1a_2a_3...\mapsto \sum a_i\cdot 10^{i-1}$$ and the more common idea $$0.a_1a_2...a_n\mapsto \sum a_i\cdot 10^{n-i+1}.$$
Their suggestion makes much more sense - it really does look like a "limiting description" of something, the idea being that e.g. "$...333$" makes a lot more sense than "$333...$." This in turn makes their "$\mathbb{N}$ should be closed under limits of recursions" idea much more relevant, in my opinion, than it would otherwise be. I think this additional coherence is actually rather valuable, and makes this question meaningfully different - at least, from the potential duplicates I've been able to find.
J.W. Tanner's answer has gotten it exactly right: the expression "$3+30+300+...$" looks like a description of a natural number, but it isn't. There's an interesting subtlety here, though:
This is one of those things which becomes less obvious the more we think about it, so it's worth analyzing a bit. The naive response is that natural numbers are finite - that's sort of the whole point - so we obviously can't have a natural number with an infinite number of digits. While this actually is perfectly right, it's also somewhat unsatisfying and may reasonably feel circular at first.
(Incidentally, this is why in my opinion it's much better to first present the diagonal argument for powersets: that for every set $X$ there is no surjection from $X$ to $\mathcal{P}(X)$. There's no need to define anything subtle here.)
All of what follows should really be a comment on J.W. Tanner's answer, but ... it's slightly too long.
So let's look at your query on this exact point:
In the interests of brevity I'm not going to be totally formal in what follows, but I promise that no serious errors have been made.
"Defined in an inductive way" is a somewhat slippery thing to say, and it's created a crucial confusion in this case. In natural language we think of induction as a way of building more and more things, but that's not really the right picture. The rule
isn't really about induction; it's really a "closure property," and that's a much simpler sort of thing. For example, it's also true that $1$ is a real number and if $n$ is a real number then $n+1$ is also a real number, but we wouldn't say that that amounts to the real numbers satisfying any kind of induction.
Rather, induction comes in when we say that the only way to build natural numbers is by applying the above rules. Specifically, consider the following very-clearly-limitative claim:
The principle $(*)$ probably looks mysterious at first, but it's actually equivalent to the usually-phrased principle of induction.
Induction implies $(*)$: Suppose $X$ contains $1$ and is closed under $n\mapsto n+1$; we want to show $\mathbb{N}\subseteq X$. Well, consider $X\cap\mathbb{N}$. This set contains $1$ and is closed under $n\mapsto n+1$ (since both $X$ and $\mathbb{N}$ have these properties), so by applying induction in $\mathbb{N}$ we have $X\cap\mathbb{N}=\mathbb{N}$.
$(*)$ implies induction: Suppose $X\subseteq\mathbb{N}$ contains $1$ and is closed under $n\mapsto n+1$. Then by $(*)$ we have $\mathbb{N}\subseteq X$, so $X=\mathbb{N}$.
So when we say "$\mathbb{N}$ is built inductively," what we really mean is that $\mathbb{N}$ is as small as it could possibly be while satisfying some basic properties (namely, that $1\in\mathbb{N}$ and $\mathbb{N}$ is closed under $n\mapsto n+1$). Put another way:
Nothing is a natural number unless it absolutely has to be.
"OK," you might say, "but that's not what I think of natural numbers as! What if we replace $\mathbb{N}$ with some number system $\hat{\mathbb{N}}$ which does allow such infinite expressions?"
The good news is:
We can totally do this! The jargon here is "non-Archimedean discrete ordered semiring" or "nonstandard model of arithmetic" or something similar, but without diving into that let's just point out that we can totally whip up a perfectly well-behaved algebraic structure here.
The thing we get might indeed have the same cardinality as - or even be strictly bigger than - $\mathbb{R}$!
However, the drawback is that this is really mixing things up. We shouldn't compare the new $\hat{\mathbb{N}}$ to the old $\mathbb{R}$; e.g. what's the "$...3333$"th digit of $\pi$? Instead, we should compare $\hat{\mathbb{N}}$ to some $\hat{\mathbb{R}}$ which is the analogue of $\mathbb{R}$ for $\hat{\mathbb{N}}$. And once we whip up such a thing ... we'll see again that there is no surjection from $\hat{\mathbb{N}}$ to $\hat{\mathbb{R}}$.
Ultimately this takes us back to my parenthetical comment at the start of this answer: I think it's pedagogically better, most of the time, to present the fully general fact that every set is strictly smaller than its powerset before focusing on a particular example like $\mathbb{N}$ vs. $\mathbb{R}$, if for no other reason than that the general fact (once understood) makes the "goalpost-moving" issue above ("if we go from $\mathbb{N}$ to $\hat{\mathbb{N}}$ we should also go from $\mathbb{R}$ to $\hat{\mathbb{R}}$") basically unsurprising.