0
$\begingroup$

The Cartesian product of a finite number of countable sets (each with cardinality > 1) is countable. However, the Cartesian product of a countably infinite number of countable sets (each with cardinality > 1) is uncountable. For example, see here and here.

Let $\{E_n\}_{n\in\mathbb{N}}$ be a sequence of countable sets (each with cardinality > 1) and let $$S=E_1\times\cdots\times E_n\times\cdots $$

So $S$ is uncountable. Now let $$T_n=E_1 \times E_2 \times \cdots E_n $$

So loosely speaking, as $n$ approaches infinity, $T_n$ becomes uncountable. A friend of mine has trouble grasping this, as in calculus, for example, the limit of an infinite series can be approximated by finite partial sums. My explanation is that Cantor's diagonal argument used to prove the uncountability doesn't work for finite sets, and "countability" is not something defined using epsilon-delta.

1.Can anyone think of a better and more intuitive explanation? Or an explanation that is more to the point?
2.What are some other (preferably simple) examples, where "the limit in infinity is not consistent with the finite processes that lead up to it"?

$\endgroup$
  • 1
    $\begingroup$ What exactly do you mean by the "limit" of a sequence of sets? There isn't an obvious definition that will behave the way you describe. $\endgroup$ – Nate Eldredge May 17 '15 at 2:57
  • $\begingroup$ $S$ is really defined in the first line as the Cartesian product of a countably infinite number of sets: $S=E_1\times\cdots\times E_n\times\cdots $. I intended the limit as a consequence of the definitions of $S$ and $T_n$. I suppose this does not fall within the usual definitions of set limits? But the results based on the definitions are correct, isn't that so? $\endgroup$ – FreshAir May 17 '15 at 3:03
  • 1
    $\begingroup$ No, there is no definition of limit for sets like this. Sometimes, people use $\lim_n T_n$ to mean $\bigcup_n T_n$, but that happens generally when the sets $T_n$ are nested, and there is generally clarification about this when mentioned. In this case the $T_n$ are not nested, so it doesn't really make sense to me to call it a "limit." $\endgroup$ – Shalop May 17 '15 at 3:06
  • $\begingroup$ I see - thanks. So then, how would I express that idea that as $n$ approaches infinity, $T_n$ would be the same as $S$? $\endgroup$ – FreshAir May 17 '15 at 3:09
  • $\begingroup$ There's something called an inverse limit that works kind of like this. Look up for example how the (uncountable) $p$-adics are the inverse limit of finite fields. $\endgroup$ – Gregory Grant May 17 '15 at 3:12
1
$\begingroup$

I think the concept you are after is pretty well captured in the concept of an Inverse Limit

Take a look at the examples on that wiki page. Indeed the limit of finite objects can be uncountable. I'm not sure if in your specific example that $S$ is the limit of the $T_n$, that would be good to check rigorously. It does say "Let $(I, =)$ be the trivial order (not directed). The inverse limit of any corresponding inverse system is just the product." which is encouraging.

$\endgroup$
  • 2
    $\begingroup$ Yes, I think that it is the case that the inverse limit of the $T_n$ is indeed $S$. For $m \geq n$, we have the projection $\pi_{n,m}: T_m \to T_n$ sent by $(t_1,...,t_m) \mapsto (t_1,...,t_n)$. And with respect to this family of functions, we have an imbedded copy of $S$ within $\prod_nT_n$, i.e, $(t_1,t_2,t_3,...) \mapsto ((t_1),(t_1,t_2),(t_1,t_2,t_3),...)$. And this imbedded copy of $S$ is precisely the inverse limit as it's defined on the Wiki page... $\endgroup$ – Shalop May 17 '15 at 3:51
  • 1
    $\begingroup$ That's great then he can say exactly what he wanted to say, he just needs to reverse the arrow $$\underset{{\leftarrow}}\lim T_n=S$$ $\endgroup$ – Gregory Grant May 17 '15 at 4:00
  • $\begingroup$ Thank you, Gregory Grant and Shalop! Looks like I'll have something to chew on for a while. $\endgroup$ – FreshAir May 17 '15 at 19:30
1
$\begingroup$

Here is perhaps another example of where finite processes and infinite processes differ. Consider the sum of rational numbers. Any finite sum of rational numbers is itself rational. However, if you take an infinite sum of rational numbers, you can obtain an irrational result. For example, $$ \sum_{k=1}^\infty\frac{1}{k^2}=\frac{\pi^2}{6}. $$ I know this isn't quite related to sets, but hopefully it is somewhat related to what you were looking for.

$\endgroup$
  • $\begingroup$ Yes! This is a beautiful example. All the partial sums are rational, but the limit is not. The Liouville numbers are examples of this as well. $\endgroup$ – FreshAir May 17 '15 at 19:29

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.