Let $d$ and $p$ be two metrics on a set $X$ and let $m$ and $n$ be positive constants such that $md(x,y) \leq p(x,y) \leq nd(x,y)$ for every $x,y \in X$. Show that every open ball for one metric contains an open ball with the same center for the other metric.
Well to show that every open ball for $p$ contains and open ball for $d$, I have the following -
We have a $p$ open ball $B_{\epsilon}^p(x)$ and we want to find a $\delta > 0$ such that $B_{\delta}^d(x) \subseteq B_{\epsilon}^p(x)$
We know that $d(x,y) \geq \frac{p(x,y)}{n}$
So if we take $\delta = \frac{\epsilon}{n}$ we have
$B_{\epsilon}^p(x) = \{ y \in X | p(x,y) < \epsilon\}$ $B_{\delta}^d(x) = \{ y \in X | d(x,y) < \frac{\epsilon}{n}\}$
and the $d$ open ball will be the same size or smaller than the $p$ open ball.
It's hard to explain it properly with just notation, it seems alot clearer when I draw out diagrams on paper and sub in actual numbers for $n$ and $\epsilon$. Have I got the general idea correct?