I'm going through Rudin chapter two on Topology, and have been working on open sets, closed sets and material related to that. I understand the concepts, but am struggling to apply it to some examples. In the book it has the following:
The set of all complex numbers $z$ such that $|z| < 1$ is Open and Bounded, but not Closed and not Perfect. I'm trying to prove each to get practice with the definitions.
But am stuck (so far I've only tried proving it is open)
Open:
Let $D = \{(x,y) \in \mathbb{R}^{2} : \sqrt{x^{2} + y^{2}} < 1\}.$ I need to show that every point in $D$ is an interior point i.e. if I draw a ball around a point in $D$ that ball lies completely inside $D$.
My idea was to take $\epsilon = 1 – \sqrt{(s)^{2} + (t)^{2}}$ i.e. the segment that connects the point $(s,t)$ to the boundary of $D$.
Let $x = (x_{1}, y_{1}) \in D$ and let $a = (a, b) \in D.$ Then $B_{\epsilon}(a) = \{x \in D: d(x,a) < \epsilon\}.$ I get a bit stuck now. I need to apply the definition of the ball and probably the triangle inequality (I tried the 1D case a simplified example).
$ \sqrt{x_{1}^{2} – a^{2}} + \sqrt{y_{1}^{2} – b^{2}} < \epsilon$
Any help appreciated.
Thanks.
Best Answer
Take $(x,y)\in D$ and let $\varepsilon=1-\sqrt{x^2+y^2}$. If $(s,t)\in B_\varepsilon\bigl((x,y)\bigr)$, then\begin{align}d\bigl((s,t),(0,0)\bigr)&\leqslant d\bigl((s,t),(x,y)\bigr)+d\bigl((x,y),(0,0)\bigr)\\&<1-\sqrt{x^2+y^2}+\sqrt{x^2+y^2}\\&=1.\end{align}In other words, $(s,t)\in D$.