Prove a ball is open

Question

So I have the following ball: $$B_{1}^{\infty}(0,0)=\left \{ (x,y) \in \mathbb{R}^{2}:\text{max} \left \{ |x|,|y| \right \}<1 \right \}$$ I have to prove it is open. My issue here is that the definition of open ball is

$$B_{r}(x_{o})=\left \{ x \in \mathbb{R}^{n}:\left \| x_{o}-x \right \|<r \right \}$$

and I reckon that the $<r$ part is very similar, so I don't know how to prove it is open. For me, it is already proven by definition. However, I also think I am probably mistaken. What is the difference between the definitions? How do I prove it is open?

Answer 1

The same result can be rephrased in the context of metric spaces.

Consider a metric space $(X,d_{X})$, $x\in X$ and an open ball $B_{\delta}(x)$. We shall prove it is open indeed.

If $y\in B_{\delta}(x)$, take $\varepsilon = \delta - d_{X}(x,y)$. Then, if $z\in B_{\varepsilon}(y)$, one concludes that \begin{align*} d_{X}(z,x) & \leq d_{X}(z,y) + d_{X}(y,x)\\\\ & < \varepsilon + d_{X}(y,x)\\\\ & = \delta - d_{X}(x,y) + d_{X}(x,y)\\\\ & = \delta \end{align*}

Hence we conclude that $z\in B_{\delta}(x)$, and we are done.

Hopefully this helps!

Answer 2

A set $E$ is open if (in your notation and in two dimensions) for every $(x_0,y_0) \in E$ there exists $r > 0$ with the property that $B_r(x_0,y_0) \subset E$.

Consider $E = B_1^\infty(0,0)$. Let $(x_0,y_0) \in E$ and choose $0 < r < 1 - \max(|x_0|,|y_0|)$.

If $(x,y) \in B_r(x_0,y_0)$ then $\sqrt{|x-x_0|^2 + |y - y_0|^2} < r$ so that $$|x| \le |x_0| + |x-x_0| \le |x_0| + \sqrt{|x-x_0|^2 + |y - y_0|^2} < |x_0| + r < 1$$ and likewise $|y| < 1$. Consequently $\max(|x|,|y|) < 1$ which means $(x,y) \in E$. That is, $$B_r(x_0,y_0) \subset E$$ which means $E$ is open.