# Showing a set is open with respect to multiple metrics?

I've proven that for $1 \leq p < q \leq \infty$, then $\|x\|_q \leq \|x\|_p \leq n^\frac{1}{p}\|x\|_q$. How can I use this to show that if a set $A\subseteq\mathbb{R}^n$ is open with respect to the $d_2$ metric if and only if it is open with respect to the $d_p$ metric for $1 \leq p \leq \infty$?

One idea I have is to compare two balls in $A$: $B^2_r(a) = \{x\in A|d(x,a) < r_2\}$ (for the $d_2$ metric), and $B^p_r(a) = \{y\in A|d(y,a) < r_p\}$ (for the $d_p$ metric). I'm not sure where to go from here, though; should I try to show that for $p > 2, B^2_r(a) \subseteq B^p_r(a)$ and vice-versa for $p \leq 2$? Or perhaps I should compare the radii $r_2$ and $r_p$? I'm starting to think it's not so helpful to compare two balls centered around the same point...

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Your idea is right, and you're really almost there. Keep at it! –  Jesse Madnick Oct 31 '11 at 23:53
Maybe you could think about how the definition of "open set" relates to balls. –  Jesse Madnick Oct 31 '11 at 23:54
I played around with the inequalities for a while. Since $\|x\|_q \leq \|x\|_p$ when $p < q$, I considered the case where $2 < p$ to find that $\|y - a\|_p \leq \|x - a\|_2 < r_2$. This seems like a good development. In the case where $p \leq 2$, this means that $\|x - a\|_2 \leq \|y - a\|_p \leq n^\frac{1}{p}\|x - a\|_2 \rightarrow \|y - a\|_p \leq n^\frac{1}{p} \|x - a\|_2 \leq n^\frac{1}{p}r^2$. I'm not sure how to deal with this one though... –  James Hines Nov 1 '11 at 1:20

My proof is for normed spaces although you claim this is a metric space problem, you seem to be working with norms.

Let $1 \le p < q < \infty$. And we assume to be familiar with $\lVert x \rVert_q \le \lVert x \rVert_p \le n^{\frac{1}{p}} \lVert x \rVert_q$

Now on the one hand, if $A \subset \mathbb{R}^n$ is open in the $p$ norm then for $a\in A$ there exists $\varepsilon >0$ such that the $p-$Ball is a subset of $A$.

$$B_\varepsilon^{(p)} (a) =\{x\in X : \lVert x-a \rVert_p < \varepsilon \} \subset A$$

It clearly follows that $B_\varepsilon^{(q)}(a) = \{x\in X : \lVert x-a \rVert_q <\lVert x-a \rVert_p < \varepsilon \}\subset B_\varepsilon^{(p)} (a) \subset A$

On the other hand, if for all $a\in A$ there exists $\epsilon >0$ such that $B_\epsilon^{(q)} (a) \subset A$ then just set $\epsilon'= \sup_{x\in A } \lVert x-a\rVert_q n^{\frac{1}{p}}$. Clearly $B_{\epsilon'}^{(p)} =\{x\in X : \lVert x-a \rVert_p < \epsilon' \} \subset A$

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The norm inequality and its proof is given here: math.stackexchange.com/questions/76739/… –  Squirtle May 26 '14 at 19:48