Is $\lim\limits_{n \to \infty} n$ "equal" to $\mathbb{N}$?

Question

In set theory, the natural numbers are defined by means of inductive sets and the successor operation

$S(n+1) = n \cup \{n\}$

As such, we have

$1 = \{0\}$, $2 = \{0, 1\}$, $3 = \{0, 1, 2\}$, etc.

Thus, as the natural numbers $n$ get larger and larger, they get closer to approximating the set $\{0, 1, 2, 3, \ldots\}$, which of course, is $\mathbb{N}$. So while this seems a bit under-handed, since limits are really only defined for functions, it seems true in a sense that $$ \lim_{n \to \infty} n = \mathbb{N} $$

Is there any way in which this is rigorously true? Are there any consequences of this "fact"?

Answer 1

There are notions of upper and lower limit of a sequence of sets (or of elements of a Boolean algebra), and when these coincide for a particular sequence, they are often called simply the limit. The upper limit of a sequence of sets $A_n$ consists of those elements that are in $A_n$ for infinitely many $n$. The lower limit consists of those that are in $A_n$ for all but finitely many $n$. In this sense, $\mathbb N$ is indeed the limit of the sequence of sets usually used to code the natural numbers (von Neumann's coding scheme, where $n=\{0,1,\dots,n-1\}$).

It should be emphasized, though, that (1) this set-theoretic notion of "limit" is quite different from the notion with the same name in calculus and (2) this result about $\lim_{n\to\infty}n$ depends strongly on the particular way that one chooses to code natural numbers as sets. For example, with the coding (proposed, I believe, by Zermelo) where $0$ is the empty set and $n+1=\{n\}$, we'd get that $\lim_{n\to\infty}n$ is the empty set.

Answer 2

One can make sense of assigning a numerical value to the sequence of natural numbers but what one needs is a refinement of the notion of limit rather than the limit itself. Thus, Skolem's construction of a nonstandard model of $\mathbb{N}$ involves working with sequences of natural numbers. Here the equivalence class of the sequence $(1,2,3,\ldots)$ will be an infinite integer. Note that Skolem's construction is indeed completely constructive. In particular, it relies neither on the axiom of choice nor on the existence of ultrafilters, presumably to Asaf's great joy.