I was reading chapter 2 (Basic Topology) of Baby Rudin and found it a bit confusing
The following definitions were listed in the book: $$$$Let $X$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $X$ $$$$
A point $q$ is a limit point of the set $E$ if every neighborhood of $p$ contains a point $q\ne p$ such that $q\in E$
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$E$ is closed if every limit point of $E$ is a point of $E$.
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E is open if every point of $E$ is an interior point of $E$.
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$E$ is perfect if $E$ is closed and if every point of $E$ is a limit point of $E$.
$$$$ The author then goes on subsequently to take a few examples of subsets of $R^2$ and states whether these are Open or Closed. A few of the examples taken were
1) the set of all complex $z$ such that $|z|<1$$$$$ 2)the set of all complex $z$ such that $z\le 1$
I had the following doubts: $$$$ $\text{Doubt 1:}$ From my understanding of the definition of a limit point, it seems that a point is considered to be a limit point of a set if every neighborhood of the point of radius $r$ (where $0<r<\infty$) contains another point $q\ne p$ such that $q\in E$.
Now consider the interval $(1,2)$. According to my understanding, every $x$ for $x\in (1,2)$ should be a limit point regardless of the radius of the neighborhood chosen since there would always be another point $y\ne x$ "to the right" or "to the left" of $x$ such that $y=x\pm h$ where $h\to 0$. Is my understanding correct? If not, where am I going wrong? $$$$ $\text{Doubt 2: }$ Consider the author's first example of the set $S_1=\{z\in C:|z|< 1\}$. Again, in this set, according to me, every point $z$ would have another point $z_0\ne z$ such that the distance between $z$ and $z_0$ tends to 0 (but is not equal to 0). Would this imply that every point of $S_1$ is a limit point of $S_1$, and is in turn $in$ $S_1$? Hence, should $S_1$ not be closed? If not, why? $$$$ $\text{Doubt 3: }$ Consider again the author's first example of the set $S_1=\{z\in C:|z|< 1\}$. The author states that this set is open, and thus evidently cannot be perfect. However, is every point of $S_1$ a limit point of $S_1$? According to me, it should be since every point $z$ of $S_1$ has a point $z_0\ne z$ towards the interior of $S_1$ such that the distance between $z$ and $z_0$ tends to 0. If not, where have I gone wrong? $$$$
$\text{Doubt 4: }$ Consider the author's second example of the set $S_2=\{z\in C:|z|\le 1\}$. This set would (according to me) be closed, and it is indeed mentioned to be closed. My confusion lies with a point $z_0$ for which $|z_0|=1$. For any neighborhood of $z_0$, there would always be a point $z_1\ne z_0$ in the interior of the set $S_2$ such that the distance between $z_0$ and $z_1$ tends to $0$. Thus, it would seem that $z_0$ would also be a limit point of $S_2$. Am I correct? If not, why?
As to 1 : yes, every point in $(1,2)$ is a limit point of $(1,2)$, and also $1$ and $2$ are, so the set of all limit points is $[1,2]$.
As to 2: $S_1$ consists only of limit points, indeed, but all points $z$ with $z=1$ are also limit points, and no points $z$ with $|z| > 1$ are. So the closure ($S_1$ plus its limit points) is $\{z: |z| \le 1\}$, which is thus a closed set. (fact: the set of limit points of a set $A$ is always a closed set (in a metric space)).
As to 3:$S_1$ cannot be perfect, because it does not contain its limit point $z=1$ say. $A$ perfect means : all points of $A$ are limit points of $A$ and all limit points of $A$ are already in $A$, (in short $A'=A$), and here $1 \in A'$ while $1 \notin A$.
As to 4: the closed disk is indeed closed, and all its points are limit points. $A$ closed means that a set contains all its limit points, $A' \subseteq A$, and this is clearly the case here, your point $1$ is a limit point of $A$ and in $A$, so no problem.