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So I need to show that an isometry is always continuous, (from $M$ to $N$ in this case) and my first thought was to show that for some $p,q\in M$, $\exists$ $\varepsilon\gt0$ such that $d_M(p,q)\lt \varepsilon$, and for some $f(p),f(q)\in N$, $d_N(f(p),f(q))$ must also be less than epsilon. But then I realize that's not enough, because if an isometry preserves distance, then if $d_M(p,q)=\alpha$ for some $\alpha\in M \implies \alpha\in N$ and therefore $d_N(f(p),f(q))$ must also be equal to $\alpha$. Is this reasoning right? Is this even in the right direction of a proof?

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Couldn't you just say "take $\delta$ equal to $\varepsilon$"? –  Giuseppe Negro Feb 12 '13 at 19:14
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up vote 2 down vote accepted

Note: I think you meant $\alpha\in\mathbb{R}$ and not $\alpha\in M$, and anyway that statement is unclear.

Your general reasoning is correct, but the proof is vague. To show continuity do this: Let $y\in Im(f)$, and let $x \in M$ with $f(x)=y$. Now let $\epsilon>0$. We need to show that there's a $\delta > 0$ such that for every $z\in B(x,\delta)$ (the open ball of size $\delta$ around $x$), it holds that $f(z)\in B(y,\epsilon)$.

Can you see what $\delta$ should be?

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Well I guess my point was that if $d_M(p,q)=\alpha$ and $\alpha$ would have to be in $M$. Since the function $f$ is isometric from $M$ to $N$ then $\alpha$ would have to also be in $N$. Is that right? Since isometry preserves distance? –  TheHopefulActuary Feb 12 '13 at 21:41
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I think you might be confused about metrics. $d_{M}(p,q) \in \mathbb{R}$ by definition. $M$ is some metric space, and not necessarily $\mathbb{R}$, so you can't say that $\alpha \in M$. –  Mark Feb 12 '13 at 21:58
    
Oh I see what you're saying now. I did a little rereading, thanks! –  TheHopefulActuary Feb 12 '13 at 22:14
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