I've yet another question about the cardinality of sets. Apologies, but I just can't seem to fully grasp it. For what it's worth, I have tried searching the site for a solution to this problem.

Let $S_n$ be the set of positive multiples of $n$:

$$S_n = \{n, 2n, 3n, 4n,\ldots\}$$

I know that this set has a cardinality of $\aleph_0$ for any positive integer $n$ because the above is the bijective function between $S_n$ and the set of natural numbers.

So what happens to the cardinality of $S_n$ as $n \to +\infty$?

Since there is a bijective function for any arbitrarily large $n$, and since the cardinality doesn't appear to change as $n$ gets arbitrarily large, my intuition tells me that the cardinality should remain $\aleph_0$.

If such is the case, would $\bigcap_{n=1}^\infty S_n$ also have a cardinality of $\aleph_0$?

Thanks so much in advance.

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    $\begingroup$ Well the intersection is empty... $\endgroup$ – Mikhail Katz Jul 31 '14 at 16:18
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    $\begingroup$ What do you mean "as $n\to\infty$"? Are you implying some kind of limit construction? $\endgroup$ – Jack M Jul 31 '14 at 16:19
  • $\begingroup$ The cardinality of a set is independent of the set's elements. Whether $n$ is a natural number or any other number, it doesn't change the number of elements in the set. $\endgroup$ – Jam Jul 31 '14 at 16:19
  • $\begingroup$ The answer to your question is yes (hint: take a look at a generic proof that rationals are countable and adapt it to your particular problem). $\endgroup$ – user132181 Jul 31 '14 at 16:20
  • $\begingroup$ Well, I suppose cardinals are orderered, so you could theoretically talk about limiting operations on them. Still, any reasonable definition of limit of a sequence of cardinals will have a constant sequence having limit equal to the generic term, and since all $|S_n|=\aleph_0$ a limit of such should still be $\aleph_0$. $\endgroup$ – Adam Hughes Jul 31 '14 at 16:23

Cardinality is not continuous. Just because a sequence of sets have the same cardinality all through the sequence doesn't mean that the limit of these sets will have the same cardinality as well.

It's easy to observe, indeed, that $|S_n|=\aleph_0$ for all $n$. So the sequence of cardinals is constant. The limit of the sets, however, is $\bigcap S_n=\{x\in\Bbb N\mid \forall n(x\in S_n)\}$. Namely all the numbers which are divisible by all the integers.

How many are there? Well, if $0\in\Bbb N$ then $0$ is such number. If you don't consider $0$ to be a natural number, then the intersection is empty. No positive integer is divisible by its successor.

Therefore the cardinality of the limit of the sequence of set, is not the limit of the cardinals of the sets in the sequence.


Certainly $|S_n|=\aleph_0$ for every $n$, so you could reasonably say $\lim\limits_{n\to\infty} |S_n| = \aleph_0$.

But that doesn't mean you've shown that some set approached by the sequence $\{S_n\}_{n=1}^\infty$ has any particular cardinality.

The set $\bigcap\limits_{n=1}^\infty S_n$ has cardinality $0$, since there is no number that is a member of $S_n$ for every value of $n$.

Notice that the set $\bigcap\limits_{n=1}^\infty S_n$ is not defined as any sort of limit. Rather it is defined by saying $m$ is a member of that set if and only if for every $n\in\{1,2,3,\ldots\}$ we have $m\in S_n$.


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