Understanding Limit points.

Question

I am trying to learn about limit points.

$\underline{\mathrm{Definition}}$ : If $X$ is a metric space then $ x \in X$ is a limit point of the set $E \subset X$ if every neighborhood $N_{r}(x)$ of $x$ contains a point $x ≠ y, y \in E$.

I don't understand this definition well. Even after doing examples I can't get a intuition for limit points.

Let's say, we need to find the limit of an interval $(1,4)$. We claim that $1$ is the limit of the set.

Proof : Choose $\epsilon > 0$ then the neighborhood $N_{r}(1) = (1 - \epsilon,1 + \epsilon)$ must contain a point $y \in (1,4)$ such that $y\neq x$.

So, from the definition I get that we are basically talking about the intersection here, i.e. $N_{r}(1) \cap (1,4)$, which is not empty here for both cases $0 < \epsilon < 1$ and $\epsilon > 1 > 0$ and doesn't contain $1$, thus $1$ is a limit point of the interval $(1,4). \square$

Am I doing it right? Even if yes, I get stuck at higher dimensions, for example I need to prove that every point of $E = \{x \in \mathbb R^{3} : |x| < r\}$ is a limit point i.e. the set is closed. It's bugging me quite a lot, due to my shallow understanding of limit points, the understanding of isolated points is being tough as well.

To summarize, my question is that what are limit points intuitively? I tried to draw a figure, but to no avail, I will be thankful if you can use a picture to aid my understanding, I think I need it a lot. Thank you.

Answer 1

Intuitively, limit points cannot be separated from your set. Let $E=(1,4)$ as in your problem and consider the point $0$. If we take an interval of radius $\frac{1}{2}$ around $0$ we get the interval $(-\frac{1}{2}, \frac{1}{2})$ and $$(-\frac{1}{2}, \frac{1}{2})\cap(1,4)=\emptyset$$ so the point $0$ can be isolated from $E$ and is not a limit point.

Consider your example of $x=1$. For any interval around $1$ say $(1-\epsilon, 1+\epsilon)$, we get $$(1-\epsilon, 1+\epsilon)\cap (1,4)=(1,1+\epsilon)\ne\emptyset$$. This says that every interval around $1$ intersects $(1,4)$ so $1$ is a limit point of $E$.

Answer 2

It makes no sense to say that “$1$ is the limit of the set”. I suppose that you meant that $1$ is a limit point of the set $(1,4)$. What this means is that, for every $r>0$, $(1-r,1+r)\cap(1,4)$ neither is $\emptyset$ nor $\{1\}$. Well, since $1\notin(1-r,1+r)\cap(1,4)$, it's clear that $(1-r,1+r)\cap(1,4)\ne\{1\}$. And $(1-r,1+r)\cap(1,4)\ne\emptyset$ because if $s=\min\left\{1+\frac r2,2\right\}$, then $s\in(1-r,1+r)\cap(1,4)$.

Concerning the set $E=\left\{x\in\Bbb R^3\mid|x|<r\right\}$, asserting that every element of $E$ is a limit point does not mean that $E$ is closed (a set $M$ is closed if and only if every limit point of $M$ belongs to $M$). And if $x\in E$, in order to prove that $x$ is a limit point of $E$ you can take $\varepsilon>0$ and then proceed as follows:

• if $x=(0,0,0)$, then take $y=\left(\min\left\{\frac\varepsilon2,\frac s2\right\},0,0\right)$;
• otherwise, consider the line segment $l$ joining $(0,0,0)$ to $x$. If $y\in l$, then $\|y-x\|$ can take any value from $0$ to $\|x\|$. Just take some $y\in l$ such that $0<\|y-x\|<\varepsilon$.

In both cases, you get an element $y$ from $N_\varepsilon(x)\cap E$ distinct from $x$.

Answer 3

Since you are looking for a picture to aid your understanding, let me give you an intuitive explanation of the definition. Let $X$ be a metric space, so there is a notion of distance and nearness. Let $E\subset X$ be a subset of your metric space.

Any point $x$ in the metric space $X$ is said to be a limit point of $E$ if you are able to find points in the set $E$ that are as "near" to $x$ as you want while not being equal to $x$.

That is, you will be find elements of $E$ inside every ball you draw around $x$, and the points you find are distinct from $x$ (in case $x$ is an element of $E$).