I am reading Rudin's book on real analysis and am stuck on a few definitions.
First, here is the definition of a limit/interior point (not word to word from Rudin) but these definitions are worded from me (an undergrad student) so please correct me if they are not rigorous. The context here is basic topology and these are metric sets with the distance function as the metric.
A point $p$ of a set $E$ is a limit point if every neighborhood of $p$
contains a point $q \neq p$ such that $ q \in E$
Also, an interior point is defined as
A point $p$ of a set $E$ is an interior point if there is a
neighborhood $N_r\{p\}$ that is contained in $E$ (ie, is a subset of
E).
I understand interior points. Ofcourse given a point $p$ you can have any radius $r$ that makes this neighborhood fit into the set. Thats how I see it, thats how I picture it.
I can't understand limit points. It seems trivial to me that lets say you have a point $p$. Then one of its neighborhood is exactly the set in which it is contained, right? ie, you can pick a radius big enough that the neighborhood fits in the set.
Ofcourse I know this is false. Our professor gave us an example of a subset being the integers. He said this subset has no limit points, but I can't see how.
Best Answer
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.