3
$\begingroup$

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"?

$\endgroup$
7
  • 1
    $\begingroup$ It depends on what you consider $\infty$. $\endgroup$
    – Jeel Shah
    Mar 30, 2014 at 4:12
  • $\begingroup$ Instead of limit, we can use union. The union of a chain is a kind of set limit. This kind of idea is useful, it simplifies the theory of ordinals. $\endgroup$ Mar 30, 2014 at 4:13
  • $\begingroup$ @AndréNicolas Union seems too heavy-handed though; a union doesn't really capture the fact that each natural numbers contains all the ones below it. $\endgroup$ Mar 30, 2014 at 4:17
  • $\begingroup$ It is the union of a chain. Such unions have particularly nice structure. $\endgroup$ Mar 30, 2014 at 4:19
  • $\begingroup$ So is $\mathbb{N}$ the smallest infinite limit ordinal $\omega$? (en.wikipedia.org/wiki/Limit_ordinal) $\endgroup$ Mar 30, 2014 at 4:33

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Great answer, especially the part about it being specific to the construction (and hence, not all that relevant). $\endgroup$ Apr 3, 2014 at 18:49
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Is seems strange to me to regard the equivalence class of $(1,2,3,\dots)$ in Skolem's model as the "limit" of the finite numbers. The model contains lots of elements strictly above all the finite numbers yet strictly below the equivalence class of $(1,2,3,\dots)$. $\endgroup$ Apr 3, 2014 at 14:43
  • $\begingroup$ I was just describing a way of making sense of the sequence as an infinite number. The OP's question did not emphasize that he is looking for a least ordinal. $\endgroup$ Apr 3, 2014 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.