Regarding the basic topology Ex, 2.21 , 3rd edition of Principles of Mathematical Analysis by Rudin I am firstly confused about the example 2.21 (c) when as a special case, the nonempty finite set has discrete point elements and hence about (d) too:
' If I've no limit point in my metric space corresponding to a given metric, does it make sense to talk about a set being 'closed'? 
Secondly, (f): How can the set of all complex numbers be both open and  closed (assuming the canonical Euclidean norm to be the metric)...I think that it should be closed (going as per the definition that all the limit points are in the set).
And (g) : I don't understand how the segment $(a,b)$  when thought about as the segment $(a,0)$ till $(b,0)$ (which seems confusing to me in the first place, as any neighborhood in $R^2$  must be a circle and not a segment) is closed (thinking in terms of the canonical Euclidean norm as the metric).
Also, another confusion is regarding the (e) where it says that $0$ is a limit point of $E$ ...but I was able to prove the contrary.
Any help would be greatly appreciated. Thanks
 A: Concerning (c): Let $x \in \mathbb R ^2$. Then $\{x\}$ is the complement of $\mathbb R ^2 \setminus \{x\}$ which is open; therefore $\{x\}$ must be closed. Now, if your finite set is $\{x_1, \cdots, x_n\}$, then it is a finite union of closed sets: $\bigcup \limits _{k=1} ^n \{x_k\}$. Since finite unions of closed sets are closed themselves, then $\{x_1, \cdots, x_n\}$ is closed.
Concerning (d): The argument above can no longer be used, because $\mathbb Z$ is infinite. Let us show that the complementary of $\mathbb Z$ is open. Pick some arbitrary $x \in \mathbb R ^2 \setminus \mathbb Z$; surround it by an open ball $B$ centered in $x$. Now, if you make the radius of $B$ small enough it will not intersect $\mathbb Z$; therefore, for every $x \in \mathbb R ^2 \setminus \mathbb Z$ you have found an open set $B$ such that $x \in B \subset \mathbb R ^2 \setminus \mathbb Z$, which shows that $\mathbb R ^2 \setminus \mathbb Z$ is open and therefore its complement must be closed. This reasoning could also be used for (c).
Concerning (f): $\mathbb R ^2$ is both open and closed because this is how the concept of "topology" is defined. A topology on some arbitrary space $X$ is a collection of subsets of $X$ (that we shall call "open subsets") with the properties, among others, that $X$ and $\emptyset$ belong to it (i.e are open). But, since $\emptyset$ is open, this means that its complementary $X \setminus \emptyset = X$ is closed, therefore $X$ is both open and closed. The same is true for $\emptyset$. If a space is connected, then these two subsets are the only subsets both open and closed. This may be counterintuitive, but you will get used to this (this happens because $\emptyset$ itself is a counterintuitive set, with strange properties).
Concerning (g): The interval $(a,b)$ is not closed in $\mathbb R ^2$ because the points $(a,0)$ and $(b,0)$ are limit points that do not belong to it. It is not open either, because its complement is not closed (use a similar argument).
Concerning (e): Yes, $0$ is a limit point of $E$. What does the concept of limit point mean? It is a point that is the limit of some eventually nonconstant sequence of points in $E$. Well, consider the sequence $1, \frac 1 2, \frac 1 3, \cdots, \frac 1 n, \cdots$; this sequence lives in $E$ (in fact it is the whole $E$, but this is not important) and tends to $0$, so $0$ is a limit point of $E$. Nevertheless, $0 \notin E$, so $E$ cannot be closed (since it does not contain all its limit points).
