Showing the Closed unit disk is not open but is closed and perfect – metric spaces

metric-spacesreal-analysis

I'm trying to show that the closed unit disk i.e. $D = \{z \in \mathbb{C} : |z| \leq 1\}$ is closed and perfect but it is not open.

I have managed to show that it is closed (I think) but am unsure of my final step.

Showing it is Closed:

Consider $D^{c} = \{z \in \mathbb{C} : |z| >1\}$ and let $z_{0} \in D^{c}, |z_{0}| = 1+r, r > 0.$ Then if we consider $B_{r}(z_{0}) = \{w \in \mathbb{C} : |z_{0} – w| < r\}$

By the triangle inequality I get $|z_{0}| \leq |z_{0} – w| + |w| \Rightarrow |w| \geq |z_{0}| – |z_{0} – w| = 1+r – r = 1$. Therefore $|w| \in D^{c}.$ Can I then say that $B_{r}(z_{0}) \subseteq D^{c}$. Hence $D^{c}$ is open and $D$ is closed.

Questions on the above:

  1. I believe that equality at the end of the triangle inequality argument by $ >$ but I can't see why?
  2. Can I make the argument that just because $|w| \in D^{c}$ then the ball is contained in $D^{c}$

Showing it is not Open:

I was trying to do this directly. Take $(v,w) \in D$ and let $\epsilon = 1 – \sqrt{v^{2} + w^{2}}$ then if $(s,t) \in B_{\epsilon}((v,w))$ we have that $d((s,t), (0,0)) \leq d((s,t), (v,w)) + d((v,w), (0,0)).$ But this doesn't seem correct to me, as after doing the calculations I get that the set is open.

Showing it is Perfect:

Rudin defines a Perfect set as one that is closed and every point in the set is a limit point. I only need to check that every point is a limit point, as I've shown the set is closed.

I am stuck here as well. I choose a point $w \in D$ and let $r > 0.$ Now we need to show if we draw a ball with radius $r$ center $w$ that we get a point that is different to $w.$

$B_{r}(w) = \{v \in \mathbb{C} : |w – v| \leq r\}$

Best Answer

You are close to proving $D^c$ is open. Just notice there that you have mistakenly put the equal sign here $|w|\ge |z_0|-|z_0-w|\gt 1+r-r=1$ since $(|z_0-w|\lt r)$

Showing it is not open is trivial. Take a point on the boundary,let us say $z_0$, then $|z_0|=1$ . For every $\epsilon \gt 0$, consider the neighbourhood $B(z_0,\epsilon)$ .Then $B(z_0,\epsilon)\cap D^c\neq \emptyset$ ( By definition of the boundary point ).

This shows $B(z_0,\epsilon) \nsubseteq D$.Hence $z_0$ is not interior point.

Now $D\subset D'$(the set of limit points of $D$) , since the points in $D$ are either interior points or boundary points. To show it is perfect, we just have to show there is no limit point of $D$ in $D^c$, which is easy since every point in $D^c$ is interior and $ D\cap D^c=\emptyset$. So no neighbourhood of a point in $D^c$ contains a point of $D$. Thus proving the claim.

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