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Consider the set of integers, $\Bbb{Z}$. Now consider the sequence of sets which we get as we divide each of the integers by $2, 3, 4, \ldots$.

Obviously, as we increase the divisor, the elements of the resulting sets will get closer and closer.

Question: In the limit as $\text{divisor}\to\infty$, what will the "limiting" set be? (I don't think it could be $\Bbb{R}$.)

Asaf Karagila
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Atom
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    In order to define limits, you need a topology. Which topology on the set of rational numbers are you considering? – J.-E. Pin Nov 11 '19 at 08:44
  • @J.-E.Pin I don't know. Can't I say 'the usual one'? Which we encounter the most in introductory calculus? – Atom Nov 11 '19 at 09:44
  • What is "the usual one"? The topology on the real numbers (that you encounter in introductory calculus) is on real numbers, here you're talking about sets of integers. If you're thinking about the order topology, note that it's not even Hausdorff if the order is not linear, and since this is not a linear order, the topology is certainly not "the usual one". So to reiterate @J.-E.Pin, you need to specify a topology if you want to talk about convergence. – Asaf Karagila Nov 11 '19 at 15:18

5 Answers5

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The typical way to define limits of sets is via $$\liminf_{n\to\infty} A_n = \bigcup_{n\geq 1} \bigcap_{k \geq n} A_k \\ \limsup_{n\to\infty} A_n = \bigcap_{n\geq 1} \bigcup_{k\geq n} A_k$$

Using these and $A_n = f_n(\mathbb{Z})$ where $f_n(x) = x/n,$ we have $$\liminf_{n\to\infty} A_n = \mathbb{Z} \\ \limsup_{n\to\infty} A_n = \mathbb{Q} $$ In particular, the limit doesn't exist.

Henry
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Brian Moehring
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Let $A_n=\{x/n:x\in\mathbb{Z}\}$ with $n$ any integer greater than $1$ then it is easy to see that $\mathbb{Z}\subset A_n\subset \mathbb{Q}$. We claim that $$\limsup_n A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)=\mathbb{Q}$$ and $$\liminf_n A_n = \bigcup_{N=1}^\infty \left(\bigcap_{n \ge N} A_n\right)=\mathbb{Z}.$$ See set-theoretic limit for more details about the limit of a sequence of sets.

Robert Z
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1) Let $S_n = \{ \frac{z}{n} \ | \ z \in \mathbb{Z} \}$ and $p_i$ be $i$-th prime integer.

2) It has no limit! Because since $(n,n+1)=1$ we have $S_n \cap S_{n+1} = \mathbb{Z}$, so always new set miss any non integer rationals included in previous one and get some new ones.

3) But $\limsup$, exists. If you consider $a_n=\prod_{i=1}^n p_i^n$, then set sequence $S_{a_n}$ is an strict increasing sequence, with respect to inclusion order, that for every $m$ there is a $k$ that $m | a_k$, so $S_m \subseteq S_{a_k}$, Therefore it tends to $\mathbb{Q}$.

4) Also $\liminf$, exists. As we see $S_n \cap S_{n+1} = \mathbb{Z}$, so it tends to $\mathbb{Z}$.

Ali Ashja'
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The three answers thus far assume by limits of the sets you mean the common value of the set-theoretic $\liminf$ and $\limsup$ (where convergence means they agree). This is a highly reasonable assumption, given that you did not specify a meaning for the limit of sets yourself.

However, I want to point out that there are other possibilities for defining a limit of sets. For example, given a sequence of sets $(S_n)$ with $\forall n, S_n \subseteq X$ for some topological space $X$, you could define

$$\lim_n S_n = \{x\in X\mid \exists (s_n) \subset X, s_n \to x \wedge \forall n, s_n \in S_n\}$$

By this definition with $X = \Bbb R$, the limit of your sets is indeed $\Bbb R$.

Paul Sinclair
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    Why not $\Bbb Q$? – Asaf Karagila Nov 11 '19 at 18:20
  • Answer accepted! – Atom Nov 11 '19 at 18:35
  • No problem with the basic reasoning, but there is a technical vagueness with using the former axiom of unrestricted comprehension to define the limit. To fix it, we would need to define $$\lim_n S_n = {x \in Y \mid \exists (s_n) \subset X, s_n \to x \wedge \forall n, s_n \in S_n}$$ for some $Y \supseteq X$ but then it depends on this choice of $X$ and $Y.$ For instance, $X=Y=\mathbb{Q}$ gives the limit $\mathbb{Q},$ $\mathbb{Q} \subseteq X \subseteq Y = \mathbb{R}$ gives the limit $\mathbb{R},$ but also we could add a point at infinity in $Y$ and then it could be in the limit too. – Brian Moehring Nov 11 '19 at 19:58
  • @BrianMoehring - I was in a hurry when I wrote this, and just forgot to include $\in X$ (it is not even sensible to use any larger set, since it requires the topology on $X$). It is now fixed. – Paul Sinclair Nov 12 '19 at 00:04
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    @AsafKaragila - true, one could take $X = \Bbb Q$ and therefore the limit would be $\Bbb Q$, but it seemed apparent to me that Atom was working in the reals, and used that space without comment. As I said to Brian, I wrote this in a hurry, and did not spare time for specifics. This too is now fixed. – Paul Sinclair Nov 12 '19 at 00:08
  • I'm not sure what you mean by "it is not even sensible to use any larger set, since it requires the topology on $X$". I would be plenty sensible to have $X=\mathbb{Q}$ and $Y = \mathbb{R}$ as the completion of $\mathbb{Q}$. Similarly, if $X = \mathbb{R}$ then $Y = \mathbb{R} \cup {\infty}$ might be the one-point compactification. In both, the topology of the larger set is induced by that of the smaller. Ultimately, however, I agree your edit fixes the issues of vagueness. – Brian Moehring Nov 12 '19 at 07:34
  • @BrianMoehring - In the definition for the set-limit, I introduced a topological space $X$. The definition is based on convergence of sequences in that space $X$. It is the topology of $X$ that gives meaning to that convergence. Why in the world would one suppose that I somehow meant convergence in some other completely unmentioned space?? – Paul Sinclair Nov 12 '19 at 15:28
  • My whole point was to give you the most general scheme for defining $\lim_n S_n$ based on your original post. You specified that you wanted $Y=X,$ which I'm fine with, but you proceeded to say that $X \subsetneq Y$ is not sensible, which is the only thing I'm contending. For another concrete example, consider for instance the extension $\mathbb{R} \subset \mathbb{R} \cup {-\infty, \infty}$ of the reals inside the extended reals. This is the common extension with which we introduce limits. All elements in the set are finite, but the limits may be infinite. – Brian Moehring Nov 12 '19 at 21:11
  • In direct answer to your question "Why in the world would one suppose that I somehow meant convergence in some other completely unmentioned space??" -- I don't. I never did. All I ever said is that you need to specify the space in which your limits exist and due to the existence of certain extensions of $X$, you are allowed to use a larger space for the limits on which the topology is induced by the topology on $X$. – Brian Moehring Nov 12 '19 at 21:19
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If you are physicist or applied mathematician, if you define $S_n = \{ z_i/n, z_i=i, i = 1..\infty \in \mathbb{Z}\}$, then you can just say that $\lim_{n\rightarrow\infty} S = \{ \lim S_n \}$, which will yield you $\lim S = \{0, 0, 0, 0,...\}$. This will happen because $\lim$ is just working on every fixed element, pretty similar to partial differential.

sanaris
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