Transition from countable to uncountable

Question

The Cantor set $\mathcal C$ is created by iteratively deleting the open middle third from a set of line segments. Let $\mathcal C_n$ be the set after n iteration. The common definition of Cantor set is the intersection of $\mathcal C_n$ for all n.

1. Can we define the Cantor set as S=limit $\mathcal C_n$ when n goes to infinity? If not, what is the difference of S and $\mathcal C$? Is S countable?

2. For each n, $\mathcal C_n$ is composed of countably many closed sets. How can a countable intersection of these sets (for all n) become uncountably many closed sets (points)?

3. The complement of $\mathcal C_n$ is the union of countably many open intervals. Same question can be asked about the complement of $\mathcal C$. How can the complement of $\mathcal C$ becomes union of uncountably many open intervals?

What are the general conclusions? Is the countable union of countable sets countable? (Consider the set of the midpoints of intervals removed from $\mathcal C_n$.)

Answer 1

Can we define the Cantor set as $S = \lim C_n$ when n goes to infinity?

Yes, but for somewhat trivial reasons: the way the limit of a sequence of nested sets $C_1 \supseteq C_2 \supseteq C_3 \supseteq \cdots$ is defined is as the intersection $$ \lim_{n\to \infty} C_n = \bigcap_{n \in \mathbb N} C_n, $$ so you can define the Cantor set this way, but it's the exact same definition with different notation.

How can the intersection of these sets (for all n) become uncountably many closed sets (points)?

This is a question that's somewhat impossible to answer. How? Just look at the construction of the Cantor set, and you can see it happening. There is something of an implication in this question that it is weird that an intersection of countably many sets which are unions of finitely many closed sets would give a closed set with uncountably many components. Perhaps that is weird, but it's certainly not in contradiction with anything (as this example shows).

How can the complement of $C$ becomes union of uncountably many open intervals?

In fact, the complement of $C$ is also the union of countably many open intervals; this despite the fact that there appear to be "uncountably many gaps in $C$". Perhaps this is causing you confusion: it's similar to the fact that in between any two (of uncountably many) irrationals, there is a rational (out of countably many).

Answer 2

1. Limits are typically defined on metric spaces, most often on numbers (usually real or complex). If $\mathcal C_n$ is a set for every $n\in\mathbb N$, then speaking of $$\lim_{n\to\infty} \mathcal C_n$$ doesn't have a standard definition. Without telling us what you mean by that limit, we cannot say if this is the same as the Cantor set.
2. Each $\mathcal C_n$ his a union of the singletons it contains, so it is also an uncountable union of closed sets.
3. Just like 2, the complement of $\mathcal C_n$ can also be written as an unciontable union of open intervals, if you want to do it.

Answer 3

1. Yes, we can. Given a sequence $(A_n)_{n\in\mathbb N}$ of subets of a given set $X$, it makes sense to say that it converges to a set $A\subset X$ if$$\limsup_nA_n=\liminf_nA_n,$$in which case $A$ is this set. Since the sequence $(\mathcal{C}_n)_{n\in\mathbb N}$ is decreasing,$$\limsup_n\mathcal{C}_n=\liminf_n\mathcal{C}_n=\bigcap_{n\in\mathbb N}\mathcal{C}_n=\mathcal{C}.$$
2. That's like asking “How can the limit of a sequence of rational numbers be irrational?” A sequence of unions of finitely many closed intervals can have uncountably many connected components, and the cantor set is a proof of that.
3. The same answer as above.

Answer 4

1. For an infinite sequence of sets $A_n$, one can make a construction which in some sense deserves the name $\lim_{n\to\infty}A_n$. There are two special cases where this is extra easy, however. That's when $A_n\subseteq A_{n+1}$ and you can take the union, and when $A_n\supseteq A_{n+1}$ and you can take their intersection. So not only is $S$ equal to $\mathcal C$, but the only way you can make sense of your definition of $S$ is as some form of a rewriting of your definition of $\mathcal C$.

2. For $\mathcal C_n$, each closed interval corresponds to a binary sequence which is $n$ long (either as sequences of $0$ and $1$, or as sequences of "left" and "right", or any of a number of different other natural options). There are only finitely such sequences for each $n$. But each element of $\mathcal C$ corresponds to an infinite such sequence, and there are uncountably many of those.

3. The complement of $\mathcal C$ only has countably many open intervals (you cannot actually have an uncountable collection of pairwise disjoint open intervals on the real line). Each open interval in the complement (except $(-\infty, 0)$ and $(1, \infty)$) was removed at some finite iteration. This means that you can index the intervals you remove at each iteration, and that way any open interval in the complement of $\mathcal C$ has an index, and the collection of all of them is countable.

And yes, a countable union of countable sets is countable. It's not even that difficult to find a way to count them all, or to find an injection into $\Bbb N\times \Bbb N$.