In a general topology exercise I have to prove the following:
Prove that the disk $D = \{(x,y) \in \mathbb R ^2: x^2 + y^2 < 1\}$ is opened in the euclidean topology.
This reminded me in how in multi-variable calculus we approximates the regions in the plane with little rectangles in order to integrate over that region. But I have no idea how to approach the problem. How should I prove this? Any tips?
Edit: This exercise is in the beginning of the book, At this point the only concepts that I'm allowed to use in the proof are the following definitions and concepts:
- Definition of topology on a set
- Definition of basis of a topology on a set
- The basis for the euclidean topology is $B=\{(x,y)\in \mathbb R ^2:a < x < b \wedge c < y < d \}$
I have not yet reached the chapter about metric spaces, so I'm not allowed do define the disk as an opened ball using some metric $d$. I think that the objective of the exercise is to prove that it is opened using the basis of the euclidean topology.
Best Answer
You can use the more general fact that, if $V\subset\mathbb{R}$ is an open set and $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is a continuous function, then $f^{-1}(V)=\{x\in \mathbb{R}^2:f(x)\in V\}$ is an open set in $\mathbb{R}^2$. In your case, take $V=(-\infty,1)$.
Edit: a more direct approach is as follows. Let $D$ be the open unit disk and $x\in D$. If we take $\delta=1-|x|>0$, where $|x|$ denotes the norm, then you can check that the open disk centered at $x$ with radius $\delta>0$ is completely contained in $D$, i.e. $$B_{\delta}(x)\overset{\text{def}}{=}\{y\in\mathbb{R}^2:|x-y|<\delta\}\subset D.$$ Then you draw a suitable open rectangle (that's part of the basis) inside $B_{\delta}(x)$, let's call it $R_x$. Then observe that $$D=\bigcup_{x\in D}R_x.$$ Therefore, $D$ is an union of basis sets, so $D$ is open.