1

The Cantor set, as showed in books and Wikipedia, is defined in terms of $C_k$, the finite Cantor set of level $k$:

$$ \mathcal{C} = \bigcap_{k=1}^\infty C_k $$

But after intersection only the "last" remains, so, why not to define it by a limit?

$$\mathcal{C} = \lim_{k \to \infty} C_k$$

It is perhaps a naive intuition, but I not see a good justification.

(adding here a note after first answer, only for comment the comments)

NOTE: if it is not only a question of choice of notation, but also about context and semantics.
Can I tell an engineer that the intersection is a kind of specification, something like a project to explain "what I need", and the limit is a "what I get", the end result?    ...Or perhaps the inverse, as suggested by @HansLundmark (thanks the comment! also thnaks @SangchulLee!). I am supposing that $C_k$ is a "decreasing sequence", $C_1 \supseteq C_2 \supseteq C_3 \supseteq \dotsb$,
so using your anser we can say
 "it's natural to define the limit as the intersection: $C_n \to \mathcal{C}$ as $n\to\infty$",
 where $\mathcal{C}$ is defined by intersection.

About comment of @LordShark: its is possible to use "limits" notation in the context of set sequences without "develop a theory" for it? @HansLundmark's link is a satisfactory answer for it?

Matt Samuel
  • 58,164
Peter Krauss
  • 163
  • 8
  • 3
    Why develop a theory of "limits" of sets to define $\cal C$ when it's simply an intersection? – Angina Seng Mar 24 '19 at 18:53
  • 1
    The intersection is the limit in this case (when you have a decreasing sequence of sets). See here: https://math.stackexchange.com/questions/107931/lim-sup-and-lim-inf-of-sequence-of-sets/1912079#1912079 – Hans Lundmark Mar 24 '19 at 18:54
  • 1
    Of course we can perfectly make sense of $\mathcal{C} = \lim_{k\to\infty} C_k$. On the other hand, we need some works to put that notion to a rigorous mathematical framework. And that is sort of asking too much for this very specific case, as pointed out by others. – Sangchul Lee Mar 24 '19 at 18:56
  • Hi @SangchulLee, I edited a note, can you check if it is ok? – Peter Krauss Mar 24 '19 at 20:48
  • 1
    The notion of limit explained in HansLundmark's answer is sufficient for this case, which is an instance of limit notion in lattice theory. It definitely works for your sequence. Since $\mathcal{C}$ can be realized as a bona-fide limit, you can interpret $\mathcal{C}$ as the ideal target and $C_n$ as approximations of $\mathcal{C}$ whose accuracy improves progressively in $n$, just as for limits in $\mathbb{R}$. – Sangchul Lee Mar 24 '19 at 21:43

1 Answers1

5

Essentially, the intersection is the limit. There is no final set, and we don't want to appeal to any concept of convergence of sets or categorical limits when this is usually introduced in an undergraduate course because this would not be understood by the students. But you're allowed to take arbitrary intersections, and this is an elementary way to get the limit of a nested decreasing sequence of sets.

Matt Samuel
  • 58,164
  • Can I tell an engineer that the intersection is a specification, something like a project to explain "what I need", and the limit is a "what I get", the end result? – Peter Krauss Mar 24 '19 at 19:17
  • 1
    @PeterKrauss: No, that is totally wrong. The intersection is equal to the limit (in the sense as discussed in the comments), and they are both the end result. – Eric Wofsey Mar 24 '19 at 23:34