12
$\begingroup$

There are several methods to say whether sum of series is finite or not. Can we say whether sum of series is countable or not.

For example $S_n=\Sigma_{0 \leq i \leq n}{2^i}= 2^{n+1}-1$ So for $n=\aleph_0$

$ S_{\aleph_0}=2^{\aleph_0}-1$

So can we say that S has a value which is not countable. But wouldn't that mean that sum of integers turns out be a non integer, as all integers form a set of countables. Or can we say that the series $S_i$ has a limit which lies outside set of inetgers so $\{S_1, S_2, S_3 ...\} \subset I$ has a limit $p$, such that $p \notin I$

Edit: I understood that $\{S_1, S_2, S_3 ...\} \subset I$ has a limit $p$ is incorrect as neither $\aleph_0$ nor $2^{\aleph_0}$ lie in $\mathbf{R}$ but I am still confused about other things.

Edit: Suppose we have sets $A_1, A_2, ... A_i...$ each having $2^i$ elements for $i=1,2 ...$ i.e. $\#(A_i)=2^i$. Suppose all these sets are disjoint can we say whether the set $A=\bigcup_{1 \leq i \leq \infty} A_i$ is countable or not.

$\endgroup$
  • $\begingroup$ A countable union of countable sets is countable. Since your index set is countable and each term is countable (finite, actually), the sum should be $\aleph_0$. If your index set is uncountable...I think it depends on the terms. For instance, it seems like $\sum_{x \in \mathbb{R}} 1 = 2^{\aleph_0}$. But as the answer below states, I'm not sure writing such a thing makes sense. Maybe look at this thread, too. $\endgroup$ – André 3000 Dec 5 '16 at 8:24
  • 1
    $\begingroup$ The infinity of the extended real numbers is not a cardinal number, so it does not directly correspond to either "countable" or "uncountable" infinity. $\endgroup$ – hardmath Dec 5 '16 at 16:23
  • $\begingroup$ Thanks for the answer. I have added some details to the question. $\endgroup$ – Curious Dec 5 '16 at 16:35
  • 1
    $\begingroup$ You have defined $S_n$ for $n\in\mathbb{N}$. Since $\aleph_0\not\in\mathbb{N}$, the term $S_{\aleph_0}$ is meaningless (exactly as $S_{\text{red}}$ or $S_{\text{car}}$ would be if you wrote any of them). $\endgroup$ – barak manos Dec 5 '16 at 16:51
  • 1
    $\begingroup$ With a little interpretation your 2nd edit does make sense. ($\infty \implies \aleph_0$. Note what Alfred Yerger said about confusing concepts of infinity. You are counting here, so the set-theoretic notion applies.) However, $A$ is uncountable because one of the sets in the union, $A_{\aleph_0}$, is itself uncountable. If you limit $i < \aleph_0$, then $A$ would be a countable union of countable sets, and thus would be countable itself. $\endgroup$ – Paul Sinclair Dec 5 '16 at 18:08
39
$\begingroup$

This question confuses two notions of infinity. The notion of infinity in set theory, that describes how big a set is, is not the same as the infinite of analysis. In short, the infinity of analysis means simply "grows without bound." On the other hand, the infinity of set theory talks about what kind of set you can put your set in one-to-one correspondance with. A set has cardinality $\aleph_0$ when it can be put in such a pairing with the natural numbers.

The slightly longer story: A series is a limit. In particular, it is a limit of the sequence of partial sums of its terms. We say that the series converges when this sequence approaches some finite number. In particular, the sequence converges to the limit $S$ if for every positive real number $\epsilon$, there is an index of the sequence $N$, so that for every index larger than $N$, the terms of the sequence never get more than $\epsilon$ away from $S$. What this means is that the terms get and stay close to a particular number, and as we take $\epsilon \to 0$, we can get closer and closer and closer to this number.

When this doesn't happen, it can do one of a few things. It could bounce around and never settle down. The prototypical example is $(-1)^n$. Bounces back and forth, but never settles. Another possibility is that it blows up, with terms who's (absolute) values will eventually exceed any possible number. In this case, we say that the sequence 'diverges.' Since the infinite series is just a special kind of sequence limit, here too we say the series diverges when the partial sums are getting larger, and will grow larger than any finite number.

So while the infinity of analysis deals with a kind of growth and change, the infinity of set theory is rather stagnant, and measures a kind of size. Neither is better or worse, they just capture different notions.

$\endgroup$
14
$\begingroup$

Recall that the statement $S_n = 2^{n+1} - 1$ is proven by induction. Induction is not magic - it cannot apply to things that aren't in its domain. For example, just because $S_n = 2^{n+1} - 1$ doesn't mean $S_{\mathrm{apple}} = 2^{\mathrm{apple}+1}-1$. Induction operates in two steps: first, the base case shows that the claim holds at $n = 0$. Second, the induction step shows that if the claim holds for $n$ then it holds for $n+1$. Now, to show that (for example) it works for $10$, you notice that the induction step says that if it works for $9$ then it works for $10$. If it works for $8$, then it works for $9$. And so on and so forth, back to $0$, which we already know works.

The thing is that for apples or $\aleph_0$, there isn't a way to step back to $0$ one by one. The argument by induction doesn't work, and so the final result has no reason to work either.

EDIT: You mentioned in a comment the equation $(x - 1)(x^n + x^{n - 1} + \ldots + 1) = x^{n+1} - 1$. This equation is proven by induction on $n$. It therefore holds for all natural numbers, but $\aleph_0$ is not covered by the induction. I'll draw your attention, for example, to the case where $n$ is negative, or $n = \pi$. The equation doesn't even make sense for those values of $n$, because those values are too different from positive natural numbers. The same is true of $\aleph_0$.

As I mentioned in a comment below, virtually nothing that is true of finite $n$ is also true of $\aleph_0$. Even the simple statement $n < n + 1$ is false when $n = \aleph_0$.

$\endgroup$
  • $\begingroup$ Not because it takes an indeterminate form, but yes, it becomes invalid. Anything that is proven by induction breaks down for $n = \aleph_0$; as a good rule of thumb, literally everything that's true for integers is not true for $\aleph_0$. Not even the statement $n < n + 1$ is true when $n = \aleph_0$. $\endgroup$ – Reese Dec 5 '16 at 8:38
  • $\begingroup$ But is $(x-1)(x^n+x^{n-1}+ ...+1)= x^{n+1}-1$ does not apply here? $\endgroup$ – Curious Dec 5 '16 at 8:39
  • $\begingroup$ @Curious I've added an elaboration to my answer. $\endgroup$ – Reese Dec 5 '16 at 8:46
  • $\begingroup$ Thanks for detailed answer $\endgroup$ – Curious Dec 5 '16 at 8:51

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.