Limit points and interior points

Solution 1:

I understand in your comment above to Jonas' answer that you would like these things to be broken down into simpler terms.

Think about limit points visually. Suppose you have a point $p$ that is a limit point of a set $E$. What does this mean? In plain terms (sans quantifiers) this means no matter what ball you draw about $p$, that ball will always contain a point of $E$ different from $p.$

For example, look at Jonas' first example above. What you should do wherever you are now is draw the number line, the point $0$, and then points of the set that Jonas described above. Namely draw $1, 1/2, 1/3,$ etc (of course it would not be possible to draw all of them!!).

Now an open ball in the metric space $\mathbb{R}$ with the usual Euclidean metric is just an open interval of the form $(-a,a)$ where $a\in \mathbb{R}$. Now we claim that $0$ is a limit point. How?

Given me an open interval about $0$. For now let it be $(-0.5343, 0.5343)$, a random interval I plucked out of the air. The question now is does this interval contain a point $p$ of the set $\{\frac{1}{n}\}_{n=1}^{\infty}$ different from $0$? Well sure, because by the archimedean property of the reals given any $\epsilon > 0$, we can find $n \in N$ such that

$$0 < \frac{1}{n} < \epsilon.$$

In fact you should be able to see from this immediately that whether or not I picked the open interval $(-0.5343,0.5343)$, $(-\sqrt{2},\sqrt{2})$ or any open interval.

Now let us look at the set $\mathbb{Z}$ as a subset of the reals. What you do now is get a paper, draw the number line and draw some dots on there to represent the integers. Can you see why you are able to draw a ball around an integer that does not contain any other integer?

Having understood this, looks at the following definition below:

$\textbf{Definition:}$ Let $E \subset X$ a metric space. We say that $p$ is a limit point of $E$ if for all $\epsilon > 0$, $B_{\epsilon} (p)$ contains a point of $E$ different from $p$.

$\textbf{The negation:} $ A point $p$ is not a limit point of $E$ if there exists some $\epsilon > 0$ such that $B_{\epsilon} (p)$ contains no point of $E$ different from $p$.

From the negation above, can you see now why every point of $\mathbb{Z}$ satisfies the negation? You already know that you are able to draw a ball around an integer that does not contain any other integer.

Solution 2:

Consider the set $\{0\}\cup\{\frac{1}{n}\}_{n \in \mathbb{N}}$ as a subset of the real line. Let's consider 2 different points in this set. First, let's consider the point $1$. Is it a limit point?

No. In fact, if we choose a ball of radius less than $\frac{1}{2}$, then no other point will be contained in it. So it's not a limit point.

Consider the point $0$. For any radius ball, there is a point $\frac{1}{n}$ less than that radius (Archimedean principle and all). So for every neighborhood of that point, it contains other points in that set. Thus it is a limit point.

Does that make sense?

Solution 3:

The definition of limit point is not quite correct, because $p$ need not be in $E$ to be a limit point of $E$.

Note that for $p$ to be a limit point of $E$, every neighborhood of $p$, no matter how small, must intersect $E$ in points other than $p$. So if there is a small enough ball at $p$ so that it misses $E$ entirely (unless $p$ happens to be in $E$), then $p$ is not a limit point. When $p$ is a limit point, there are points from $E$ arbitrarily close to $p$.

Examples:

• In $\mathbb R$, $0$ is a limit point of $\left\{\frac{1}{n}:n\in\mathbb Z^{>0}\right\}$, but $-1$ is not.
• In $\mathbb R$, $\mathbb Z$ has no limit points. For each $p\in\mathbb R$, there is a closest integer $n\neq p$, and the ball of radius $|p-n|$ centered at $p$ does not intersect $\mathbb Z$ (except perhaps at $p$).

If $p$ is a not a limit point of $E$ and $p\in E$, then $p$ is called an isolated point of $E$. This is good terminology, because $p$ is "isolated" from the rest of $E$ by some sufficiently small neighborhood (whereas limit points always have fellow neighbors from $E$). If $p$ is not in $E$, then not being a limit point of $E$ is equivalent to being in the interior of the complement of $E$.