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 a lot clearer when I draw out diagrams on paper and sub in actual numbers for $n$ and $\epsilon$. Have I got the general idea correct?
Best Answer
Your choice of $\delta$ does indeed work. Here’s a way of explaining it that may be a bit clearer.
You want to choose your $\delta$ so that if $d(x,y)<\delta$, then $p(x,y)<\epsilon$; that will ensure that if $y\in B_\delta^d(x)$, then $y\in B_\epsilon^p(x)$ and hence that $B_\delta^d(x)\subseteq B_\epsilon^p(x)$. You know that $p(x,y)\le nd(x,y)$, so if $d(x,y)<\delta$, then $p(x,y)<nd(x,y)<n\delta$. As long as you choose $\delta\le\frac{\epsilon}n$, you’ll then have
$$p(x,y)<nd(x,y)<n\delta\le\epsilon\;,$$
which is exactly what you need.