Every $p$-norm ($p \in [0,\infty]$) generates the same class of open sets on $\mathbb{R}^n$ The following claim has been made in my multivariable analysis class, and I think I have the idea of the proof but I can't quite seem to get down to the rigorous proof the instructor wants:
Every $p$-norm ($p \in [0,\infty]$) generates the same class of open sets on $\mathbb{R}^n$.
I'll also give the definitions of the terms my instructor is using:
Let $B(X,r) = \{ Y \in \mathbb{R}^n \text{ such that } ||Y-X||<r\}$ where $|| \bullet ||$ is the standard Euclidean norm. Then a set $A \subseteq \mathbb{R}^n$ is open iff for all $X\in A$, there exists $r>0$ such that $B(X,r)\subseteq A$. 
So my idea is that since all the $p$-norms are equivalent, and if there is a ball of radius $r$ in the $2$-norm, we should be able to find a ball in any $p$-norm that contains it and remains inside of $A$, but I can't quite see how we can do that.
Any ideas? Thanks!
 A: One way of proving that norms give the same topology is to find inequalities 
$$c||x||_1\leq ||x||_2\leq C||x||_1\ \ \ \ \text{ with }c>0.$$
$$\text{If }\ \ q>p\geq1\ \ \ \ \ \ \ \ \ \text{ then }\ \ \ \ \ \ \ \ \ \frac{n^{1/q}}{n^{1/p}}||x||_p\leq||x||_q\leq||x||_p.$$
The first inequality is the generalized mean inequality and the second is because 
$$\frac{||x||_p}{||x||_q}=\underbrace{\left|\left|\frac{1}{||x||_q}x\right|\right|_p\geq\left(\sum_i\frac{|x_i|^q}{||x||_q^q}\right)^{1/p}}_{q>p\implies a^q\leq a^p\text{ for }|a|<1.}=1^{1/p}=1\ \ \ \ \text{ because }\ \ \ \ \frac{|x_i|}{||x||_q}\leq1.$$
Now, such a pair of inequalities allows you to, for every open ball in one norm to put an open ball from the other norm inside. Suppose $B_p(a,r)$ is a ball of radius $r$ in the $p$-norm, then $$B_p(a,\frac{n^{1/p}}{n^{1/q}}r)\supset B_q(a,r)\supset B_p(a,r).$$
A: For $0<p<1$ the $p$-norm is only a quasinorm. Nevertheless, it is not hard to check that all of them give the same topology as the $\infty$-norm
$$||(x_1, \ldots, x_n)||_{\infty} = \max|x_i|$$
Let $x$ with $||x||_{\infty} < \epsilon$. Then $\sum_{i=1}^n |x_i|^p \le n \epsilon^p$ and so $||x||_{p} < n^{\frac{1}{p}}\cdot \epsilon$. Therefore
$$B_{\infty}( \epsilon) \subset B_{p}( n^{\frac{1}{p}}\epsilon) $$
(balls with same center) and because of degree $1$ positive homogeneity of the (quasi-)norms we get
$$B_{\infty}( n^{-\frac{1}{p}}\epsilon) \subset B_{p}( \epsilon) $$
On the other hand
$$||x||_p = (\sum|x_i|^p)^{\frac{1}{p}}\ge (\max|x_i|^p)^{\frac{1}{p}}= \max |x_i| = ||x||$$
 Therefore, if $||x||_p < \epsilon $ then $||x||_{\infty} < \epsilon$, and we get
$$  B_{p}( \epsilon) \subset B_{\infty}( \epsilon)$$
Therefore 
$$B_{\infty}( n^{-\frac{1}{p}}\epsilon) \subset B_{p}( \epsilon) \subset B_{\infty}( \epsilon)$$
and so $||\cdot ||_p$, $||\cdot||_{\infty}$ give the same topology. 
