Topology question in Rudin book about example 2.21 I am taking a math class where the book used is Walter Rudin.
I don't understand how the author explain that in $R^2$ the complex number such $|z|<1$ is 


*

*not closed

*open

*not perfect

*bounded
My other question is: how does the distance function change the way we define the neighborhood on the set, and for subset? 
How do we apply the definitons of: open, closed, complement, perfect, dense... on the topological space $(X,d)$ with a particular $d$ or a new definition of $d$ that is different of the usual distance we operate with?
 A: Please give your progress on your first question. I think you will be able to get it without much help.
I can offer some help on your second question right away. For any distance function $d$, you always have the open ball $B(x,\delta)=\{y\in X\mid d(x,y)<\delta\}$ (We will always use $\delta>0$ to make it nondegenerate.) Different $d$'s may give different shapes of the ball, but we can think of them all as ball-shaped regions around $x$.
Perhaps you've seen $d_1(x,y)=\max({|x|,|y|})$ used in $\mathbb{R}^2$, which results in a square-shaped ball around a point, or $d_2(x,y)=|x|+|y|$ resulting in a diamond shaped ball around a point. 
The notion of complement, of course, will never change. It is just a set theoretic definition, and doesn't depend on any topology!
The notion of "open set" associated with this metric $d$ is entirely dependent upon these ball-regions. A set $\mathcal{O}$ is open in that topology if each each point of $\mathcal{O}$ is the center of an open ball contained in $\mathcal{O}$.
Once you feel you understand the open sets, closed sets will be exactly the complements of open sets.
You can convince yourself that in a metric space, a set is dense (intersects every nonempty open set nontrivially) iff it intersects every ball (with positive radius of course) nontrivially. If you get this, I think you should already understand how this works for any metric.
The same goes for perfectness: a set $Y$ will be perfect iff $B(x,\delta)\setminus\{x\}\cap Y\neq\emptyset$ for every $x\in Y$.
Again, the shape of the ball may change with the metric, but you use the ball uniformly in all of these concepts.
A: Put $D = \{z\in \mathbb{C}|\; |z| < 1\}.$
The set $D$ is not closed.  Notice that $1\not\in D$ and that ${n\over n+1}\in D$ for all
$n$.  You can see that ${n\over n+1}\to 1$ as $n\to\infty$, so 1 is in the closure of $D$ but it is not an element of $D$.  
This set is open. Choose $z\in D$.  Put $\epsilon = 1 - |z|$; from the definition of $D$ we see that $\epsilon > 0$. You can use the triangle inequality to see that if $w\in {\mathbb{C}}$ and $|z - w| < \epsilon$, $z\in D$.  
Perfect sets must be closed.  So we are done here
It is bounded.  I doubt you had difficulty with that.
