How to find limit point?

Question

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?

Answer 1

All the points in a closed set are limit points.

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.

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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.

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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 $\epsilon > 0.$)

Answer 2

To show that $B = \{ z\mid |z| < 1\}$ is open choose open ball centered at some $p$ with radius $r = 1-|p|$. Now, you need to show that all the points inside this ball are indeed contained in $B$. Use triangle inequality to show this (and draw a picture to visualize what is going on).

To show that $A = \{ z\mid |z| \leq 1\}$ is closed, you can show that $A^c = \{ z\mid |z| > 1 \}$ is open just like above. I'll let you figure out what radius works.

It is not true that in the case of $|z|\leq 1$ you can always find such a neighborhood. The problem is with the boundary points, $|z|=1$. Any neighborhood of such a point will contain points in both $|z|<1$ and $|z|>1$.

Answer 3

$B$ does not contain all of its limit points. You can construct a sequence of points in $B$ converging to $1\notin B$.

To prove $A$ is closed, you could show that its complement is open.

To prove $B$ is open, see the proof here that open balls are open.