A solved example in Rudin's text discusses two sets. $|z|<1 \text{ and } |z|\le 1$. The first set is open and not closed while the latter is not open and closed.
Say, $A$ contains all elements $z$ such that$|z|\le 1$
All the limit points in a closed set are within the set. Therefore, for $|z|\le 1$, all points of $A$ must be limit points. Say, $p\in A$. Then every neighborhood $N_{\varepsilon}(p)$ must contain a point $q\in A,q\ne p$. How do I actually prove that such a $q$ exists?
I have the intuitive feeling that such a $q$ exists but how do I formalize it? I can draw an arbitrarily small radius around any point in the $2-D$ ball and I will have an uncountable set which will always contain a point arbitrarily close to the center. Also why doesn't this hold true in the case of $|z|<1$?
$B$ contains all $z$ such that $|z|<1$
I can find a point $p$ such that neighborhood of $N(p)\subset B$. That is again more intuitive and I cannot write it down. However, if such a neighborhood exists for $|z|<1$ then intuitively such a neighborhood also exists for $|z|\le 1$.
What am I missing? How can I prove the openness and closure of these two sets?
EDIT: I realized for a set to be closed, all limits points must be inside the set. In the case of $|z|<1, 1$ is a limit point but not inside $A$. Hence $A$ is not closed. So all I need to know now is why is $B$ open and $A$ not?
Best Answer
That is not true. For example, $A = \{1\}$ is closed (in $\mathbb{R}$) but $1$ is not a limit point of $A$. On the other hand, consider $B = (0, 1) \subset \mathbb{R}.$ Every point of $B$ is a limit point of $B$ but $B$ is not closed (in $\mathbb{R}$).
The statement that is true that a closed set contains all of its limit points. Thus, $A$ above was vacuously closed (it had no limit points) and $B$ was not closed as $0$ is a limit point of $B$ which is not in $B$.
(In fact, a closed set such that all of its points are limit points is called perfect.)
Another equivalent definition of a set being closed (in $X$) is that its complement (in $X$) is open.
Now, if you want to show that your $B = \{z \in \mathbb{C}: |z| < 1\}$ is not closed, then you must produce a point $z \in \mathbb{C}$ such that $z$ is a limit point of $B$ but not in $B$. (For example, one such point is $1 + 0\iota$, can you find all such points? What does it mean geometrically?)
Once again, note that all points of $B$ are indeed limit points of $B$ but $B$ is still not closed.
To show that $B$ is open, you must show that given any $p \in B$, there exists $\epsilon > 0$ such that $N_\epsilon(p) \subset B.$
If you think of it geometrically, any $\epsilon > 0$ which is at most the distance of $p$ from the boundary will work. So, a likely candidate is $1 - |p|.$ Now, you must show that this actually works. (The triangle inequality will be helpful.)
However, observe that we use the fact that $|p|$ is strictly less than $1$. This is why $\epsilon = 1 - |p|$ is indeed positive.
Now, to show that your $A$ is closed, you could try to argue that no point of $\mathbb{C}\setminus A$ is a limit point of $A$. Or you could try to show that $\mathbb{C}\setminus A$ is open. The latter seems easier to do with an almost identical proof as before. (Consider $\epsilon = |p| - 1 > 0$.)
To show that $A$ is not open, simply consider the point $p = 1 + 0\iota$ and show that every neighbourhood of $p$ contains a point outside $A.$ (Observe that $1+\frac{\epsilon}{2} \in N_\epsilon(p)$ for every $\epsilon > 0.$)