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I have attached an image of how I was visualizing a limit point, but I'm now not so sure that I have understood the concept correctly after attempting to really draw out what I was visualizing.

I'll mention the definition of Neighbourhood and Limit Point from Rudin's Analysis just for a refresher:

Definitions

Let $X$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $X$. A neighbourhood of $p$ is a set $N_{r}(p)$ consisting of all $q$ such that $d(p,q)<r$, for some $r>0$. A point $p$ is a limit point of the set $E$ if every neighbourhood of $p$ contains a point $q \neq p$ such that $q \in E$

I ran into a roadblock understanding the boundary, closure, and interior of halfspaces bounded by hyperplanes, and I think it runs back down to my misunderstanding of the limit point. Here is the figure I created: Limit Point

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EDIT: New figure for a limit point $p$ New Limit Point Figure Please let me know if this is more accurate!

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I am imagining this as happening at the "infinitesimal" level, so for example, $r_{1}$ is not actually as large as shown in the figure. The three general operations I imagine happen are:

$$d(r_{n},r_{n-1}) \rightarrow 0$$ $$\forall r \in \mathbb{R}, \exists q \in E \text{ }|\text{ } d(p,q_{n})<r_{n}, n \in \mathbb{N}$$ $$d(p,q_{1}) \rightarrow 0$$

So I think of every neighbourhood around the point $p$ as being circles of expanding radius, and have the condition that there must be some $q \neq p$ where $q \in N_{r}(p)$ and I think of this as being a condition for each radius. Is this an overcomplication? I realized that for say some $r_{1}$ that if there exists a $q \in E$ so that $d(p,q) < r_{1}$, then for any $r_{i} > r_{1}$, that same point $q$ which worked for the neighbourhood $N_{r_{1}}(p)$ will work for the neighbourhood $N_{r_{i}}(p)$. So, are these three operations sort of additional constraints on the situation that do not add any relevant information?

Main Question

My problem now, assuming that the condition is that for all $r \in \mathbb{R}$, there is some $q \in E$ so that $d(p,q) < r$, for an arbitrarily small $r>0$, then how can I visualize this? How can two points $p$ and $q$ where $p \neq q$ actually be distinct when I can shrink my radius arbitrarily small? If anyone could give me a concrete example or an explanation of where I went wrong/what could clear things up for me that would be very helpful.

Thank you!

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    $\begingroup$ The nhoods shouldn't be thought of as expanding, but rather contracting. As the radii of the nhoods tends to 0, you can still find points in $E$ distinct from $p$. Though the radius can be arbitrarily small, it is still positive, and there is "room" for the point $q$ (points take up no space). Note also that the $q$ may change depending on the nhood. $\endgroup$ – David Mitra Dec 21 '11 at 20:01
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    $\begingroup$ I don't understand all of what you're saying, but here are at least some points: a) There seems to be a mistake in the second of your displayed equations; the dummy variable $r$ of the universal quantifier doesn't occur anywhere else. b) The "Main Question" appears to suffer from quantifier reversal -- it's not "there is a $q$ such that for all $r$" but "for all $r$ there is a $q$ such that". c) It's unusual to define a neighbourhood of $p$ as only open balls containing $p$ -- usually a neighbourhood is any set containing such an open ball (or in some texts, any open such set). $\endgroup$ – joriki Dec 21 '11 at 20:06
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    $\begingroup$ I would perhaps visualize it as the point being "really close" to the set, since any neighbourhood of the point contains a point from the set. $\endgroup$ – Mikko Korhonen Dec 21 '11 at 20:07
  • $\begingroup$ I agree, quantifier order is important! $\endgroup$ – Dylan Moreland Dec 21 '11 at 20:07
  • $\begingroup$ P.S.: I just checked the book and it does have that definition of a neighbourhood -- just be aware that this is not the usual definition; see e.g. Wikipedia. $\endgroup$ – joriki Dec 21 '11 at 20:12
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The problem seems to be that you’ve got your quantifiers backwards. It isn’t that there’s a single $q\in E$ such that $d(p,q)<r$ for arbitrarily small $r$. Rather, for each positive $r$, no matter how small, there is a $q\in E$ such that $d(p,q)<r$. The choice of $q$ depends on $r$. In symbols, this is the difference between

$$\exists q\in E\setminus\{p\}\;\forall r>0 (d(p,q)<r$$ and $$\forall r>0\;\exists q\in E\setminus\{p\}(d(p,q)<r\;.$$

For a simple example, let $$E=\left\{\frac1n:n\in\mathbb{Z}^+\right\}\;,$$ and let $p=0$. For $r=0.1$, you can take for $q$ any $1/n$ with $n>10$. With $r=0.01$, on the other hand, you’ll need to choose a $1/n$ with $n>100$. And so on.

I’d also say that your picture is inside-out: you should think of circles of decreasing radius squeezing in closer and closer to $p$. Then $p$ is a limit point of $E$ if within each of those circles, no matter how close to $p$, there is at least one point of $E$ different from $p$ itself.

Added:

Now let’s take a look at your three ‘general operations’.

$$d(r_{n},r_{n-1}) \to 0$$

If you choose a sequence $\langle r_n:n\in\mathbb{N}\rangle$ converging to $0$, then it will automatically be the case that $d(r_{n},r_{n-1}) \to 0$ as $n\to\infty$, but this is a side-effect, not something on which you should focus. What’s important is that $r_n\to 0$ as $n\to\infty$; if that’s the case, and if for each $n\in\mathbb{N}$ you have a $q_n\in E$ such that $q_n\ne p$ and $d(p,q_n)<r_n$, then $p$ is a limit point of $E$.


$$\forall r \in \mathbb{R} \exists q \in E \big(d(p,q_{n})<r_{n}, n \in \mathbb{N}\big)$$

As written, this doesn’t make sense: how are the single $r$ and $q$ in the quantifiers related to the $r_n$ and $q_n$ in the quantified statement? You could correctly write any of the following, since all of them say that $p$ is a limit point of $E$:

$$\begin{align*} &\forall r>0 \exists q(r)\in E\big(q(r)\ne p\text{ and }d(p,q(r))<r\big)\\ &\forall n\in\mathbb{Z}^+\exists q_n\in E\left(q(r)\ne p\text{ and }d(p,q_n)<\frac1n\right)\\ &\forall n\in\mathbb{N}\exists q_n\in E\left(q(r)\ne p\text{ and }d(p,q_n) < \frac1{2^n}\right) \end{align*}$$

Indeed, if $\langle r_n:n\in\mathbb{N}\rangle$ is any sequence of positive real numbers converging to $0$, you could take

$$\forall n\in\mathbb{N}\exists q_n\in E\big(q(r)\ne p\text{ and }d(p,q_n) < r_n\big)$$

as your definition of ‘$p$ is a limit point of $E$’.


$$d(p,q_{1}) \to 0$$

As written this makes no sense, since $p$ and $q_1$ are single, fixed points: there is no sequence here. Did you mean $d(p,q_n)\to 0$? That isn’t enough as it stands, because it says nothing about the nature of the $q_n$. What does work is this:

A point $p$ is a limit point of a set $E$ if and only if there is a sequence $\langle q_n:n\in\mathbb{N}\rangle$ of points of $E\setminus\{p\}$ such that $d(p,q_n)\to 0$ as $n\to\infty$.

(This of course assumes that there is a metric $d$; this definition doesn’t work for topological spaces in general.)

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  • $\begingroup$ Can you please explain your notation a little bit for the "$\ {p}$"? I'm not sure if those are just typos but I think I understand what you are trying to say. I understand that quantifiers matter, I fixed the order in the "main question". My question still stands, for if those 3 procedures I mentioned are what is happening or not? $\endgroup$ – Samuel Reid Dec 21 '11 at 20:41
  • $\begingroup$ @Samuel: Something didn’t display in your comment; are you asking about the notation $E\setminus\{p\}$? $A\setminus B$ is the set difference of $A$ and $B$, the set of all things that are in $A$ but not $B$, so saying $\exists q\in E\setminus\{p\}$ is just saying that there is a point $q\in E$ with $q\ne p$. $\endgroup$ – Brian M. Scott Dec 21 '11 at 20:48
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    $\begingroup$ @Samuel: You’ve fixed the order of the written quantifiers, but not the way you’re thinking about them when you ask how two points can be distinct when you shrink the radius arbitrarily small. If it were always the same two points, obviously they couldn’t be distinct, but they’re not the same two points: as you shrink $r$, $q$ changes as well. $\endgroup$ – Brian M. Scott Dec 21 '11 at 20:53
  • $\begingroup$ I see, I hadn't seen that notation before. I thought $E - {p}$ was customary for set difference. Thank you for the response it was very helpful. I am currently fixing up my figure and I will post an edit on the main question when I am finished. I would appreciate if you could comment on if the edited version is a more accurate visual representation. $\endgroup$ – Samuel Reid Dec 21 '11 at 20:54
  • $\begingroup$ Would you mind to explain how to show limit point of $E = {\frac1n:n\in\mathbb{Z}}$ is 0? Do I have to divide it into two cases? positve and negative? $\endgroup$ – Timothy Leung May 19 '13 at 11:33
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Let me give a different attempt to explain why you are misunderstanding. Yes, many people have told you, and likely convinced you, that there is something wrong. However, a very simple example would just tell the exact directionality of the wrongness.

You should see the link between limit points and Rudin's 2nd proof of the irrationality of sqrt 2. (As a sidenote, I just discovered that that particular proof is just plain wrong.)

Remember, Rudin was saying that sqrt 2 is irrational because you can get a sequence of points in the rationals, always getting closer to sqrt 2, but never reaching it? If you now consider the set of points made by any such sequence, that set of points has as a limit point the sqrt of 2, but sqrt 2 is NOT in the set, nor is it even in the set of rationals.

Let us look even more into depth. Rudin gave, in Example 1.1, an algorithm. That is, for any p in (positive) rationals you pick, he tells you how to find a q that is closer to sqrt 2, but never reaching. Thus, for any such p you pick, you now have an infinite sequence that will give you better and better approximations to sqrt 2, always going monotonically closer, and from one direction (never overshooting). Just repeatedly apply the algorithm to find the new q.

This sequence of points in rationals, when seen as elements of a set, is a countable set. If you stop the sequence at any integer n, it would be a finite set of isolated points. Isolated, in the same sense as the set of integers are made of isolated points in the rationals / reals. PROOF: For any integer you pick, just pick a radius less than 1, and suddenly you have a neighbourhood (in rationals / reals) around this integer, where there is no other point in the set of integers in this neighbourhood. Hence isolated. QED.

Similarly, the finite set of isolated points that make up a truncated sequence for sqrt 2, are isolated because you can pick the distance between the two closest points as a radius, and suddenly your neighbourhood with any point is isolated to just that one point.

Hence why the name ``limit point''---the idea is to convey the limit of a sequence. You need to even to talk about Cauchy sequences and so forth.

That is, look at the fully infinite sequence. Then sqrt 2 is a limit point, because no matter how small (as long as non-zero) you pick a radius for the neighbourhood around sqrt 2, you will find that the neighbourhood around sqrt 2 will include not just one other point in the set, but infinitely many (that is part of a theorem quite close after Rudin defined limit points). This captures the entire meaning of limit points, and is the original motivation anyway. Rudin just kept doing this bait-and-switch, telling you things, but not why.

As should be now obvious, limit points may not be in the set itself. Namely boundary points of open sets are clearly limit points of the set. That gives the sensible generalisation of the concept of boundaries away from the colloquial 2D and 3D simple cases. The intuition that is built by Venn diagrams is made more rigorous and generalised.

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  • $\begingroup$ I forgot to add: Consider a perfect set, like an interval. Every point is a limit point and vice versa. That should give you the idea of it being a continuum, and it is precisely what that is. $\endgroup$ – xiaokj Sep 13 '16 at 17:09
  • $\begingroup$ My reference of the mistake, is now online at math.stackexchange.com/questions/1925602/… $\endgroup$ – xiaokj Sep 13 '16 at 18:48

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