This is rather a philosophical question. Although it uses topological notions, it isn't any precise mathematics, so maybe one cannot take it very seriously.

Sometimes I try to picture an infinite set of points. Assuming that the space $E$ in which my visual imagination takes place is a simply connected bounded subspace of the plane $\mathbb{R}^2$, I can only visualize at most uncountably many points. But since $E$ is separable, so is every subspace and I can't really explicitly imagine uncountably many points, e.g. having a circle in mind: I couldn't decide whether there only lie denumerably points on it or not. I could decide for myself that there are uncountably many, but this I would consider cheating.

But I sometimes try to explicity imagine a countable set, viz as discrete, which isn't possible for uncountable sets, again because $E$ is separable. Since it's also bounded such a subset cannot be closed therein, for otherwise it would be compact, and therefore finite. One example would be the set $\{\tfrac{1}{n};\; n\in\mathbb{N}\}$, but here again it's hard for me to tell if its visualization is indeed denumerable. This might be due to the feeling that visualized points have an actual "size" in my mind. And it seems that any bounded countable discrete set must locally look like this particular set because it must have an accumulation point outside of it. Or maybe one can determine that such a visualization is in fact denumerable?

The problem for me seems to be the phenomenon of an accumulation point, and I feel that there's no hope if "too many" of such points are actually in the set. I just don't know how many "too many" exactly is. Maybe one is already too much. And maybe it's enough if there's any accumulation point at all, i.e. not even in the set. In that case, it would be impossible (if my argumentation isn't wrong or an assumption too strict).

So my question would be if this is the end of this discussion or not? Is there any, maybe completely alternative, way of visualizing a countable set knowing it isn't uncountable? I don't know, the question might be very weird (I feel a bit ashamed posting it), but maybe someone has a satisfactory answer.

  • $\begingroup$ +1 for pointing out several times that this is a very weird question and not exaggerating at all. $\endgroup$ – DonAntonio Sep 23 '12 at 16:05
  • $\begingroup$ Related (possibly duplicated?): math.stackexchange.com/questions/89353 $\endgroup$ – Asaf Karagila Sep 23 '12 at 16:18
  • $\begingroup$ Please let me know if the linked answer is helpful, otherwise I will try to come up with a new one. $\endgroup$ – Asaf Karagila Sep 23 '12 at 18:43
  • $\begingroup$ @AsafKaragila Oh, sorry. It doesn't help me much while it gives some perspective. First of all, I know nothing about model theory, secondly I can't see why one can distinguish an $\aleph_0$-line from an $\aleph_1$-line when put next to each other. Also, I don't understand how one can scale the cardinality of a set. Maybe I misunderstood your point, though. Anyway, I found your post hard to follow which is why I haven't given feedback yet – I feel like I can't really judge if it answers my questions, but so far it hasn't helped much. (But thanks.) $\endgroup$ – k.stm Sep 23 '12 at 19:04

I'll try to answer the question of how to visualize a countable set $C$ in, e.g., $\mathbb{R}^2$, although I'm not sure if this was your main question.

If $C$ is dense then I'm not sure how to visualize it differently than the ambient space. Instead let's consider the special case that $C$ is closed. In this case the usual proof of the Cantor–Bendixson theorem (see https://en.wikipedia.org/wiki/Perfect_set_property) maybe gives a way to "visualize" $C$ in terms of countable discrete sets, which you suggested that you already can visualize.

The proof of this theorem shows that if we transfinitely iterate the process of removing the isolated points of a countable set $C$, taking intersections at limit stages, we end up with the empty set after countable many stages. So running this process in reverse, it says that $C$ is "constructed" in countably many stages by starting with a countable discrete set, and at each successor stage, adding another countable discrete set whose limit points are the current set. (Well, what happens at the limit steps if you go in reverse isn't really a "construction", but a fuzzy question gets a fuzzy answer.)

Edit by questioner: This answer actually didn't satisfy me, but the discussion in the comments of this answer eventually convinced me to believe that one only can visualize (in the sense I meant) finite unions of bounded connected subspaces of the plane, why the answer to my question would be "no, there is no way". This is why I accepted the answer.

  • $\begingroup$ I want to visualize $\aleph_0$ as the cardinality of a bounded subspace of $\mathbb{R}^2$ since I can't imagine unbounded sets without "leaving something out". Although I find interesting what you said, it doesn't answer my question: It's more about visualizing a countable discrete bounded set, without being an visualization of an uncountable one at the same time. But as you said, it is a fuzzy question, so I apologize and many thanks for your effort anyway. $\endgroup$ – k.stm Sep 24 '12 at 7:49
  • $\begingroup$ @K.Stm. Okay, but I definitely don't understand the question then. What uncountable set do you visualize when you try to visualize the countable discrete bounded set $\{1/(n+1) : n \in\mathbb{N}\}$? $\endgroup$ – Trevor Wilson Sep 24 '12 at 15:34
  • $\begingroup$ I'm tempted to say "that very set", but I guess a more elaborate one would be that I either fail to visualize it or that I wouldn't call my visualization a set (if that's what you meant). For example the visualization of a line isn't $\mathbb{R}$ oder $\mathbb{Q}$, but rather can represent both. As I said, my whole question is about finding a visualization of a countable set which doesn't also represent an uncountable one. $\endgroup$ – k.stm Sep 24 '12 at 17:03
  • $\begingroup$ Could you explain why you can't visualize the set $\{1/(n+1) : n \in \mathbb{N}\}$? Also, what about $\mathbb{N}$ itself? Or is the problem that you can visualize these sets but that the look the same as some uncountable sets? If so, which uncountable sets? $\endgroup$ – Trevor Wilson Sep 25 '12 at 3:03
  • $\begingroup$ I can't visualize $\mathbb{N}$ because it is unbounded. Well, of course I DO somehow visualize $\mathbb{N}$ and $\{ 1/n ;\; n \in \mathbb{N}\}$. But for $\mathbb{N}$ one pictures some natural numbers and then thinks "and so on". What one really visualizes are finitely many points. And with the other set, one visualizes finitely many points which become closer and closer, eventually becoming indistuingishibly close such that they form some sort of line – then I wouldn't know an argument that such a representation doesn't look like an uncountable set. But it get's blurry down there. $\endgroup$ – k.stm Sep 25 '12 at 6:17

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.