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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?

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up vote 1 down vote accepted

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.

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That's a clearer way of putting it, thanks. –  sonicboom Nov 1 '12 at 19:58
    
@sonicboom: My pleasure. –  Brian M. Scott Nov 1 '12 at 20:04
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