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I understand the definition of an isometric embedding, (an injective, distance preserving map) but I don't understand what it means for a metric space to be isometrically embedded in another space. Does it mean that all maps from the first space to the other need to be an isometric embedding? Does just one? I tried searching online but I couldn't find any examples.

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    $\begingroup$ Just one. Like how $\mathbb{Q}$ is embedded in $\mathbb{R}$. It doesn't make sense for the condition to be on all maps, because you can define maps arbitrarily. $\endgroup$ – Bruce Zheng Oct 11 '14 at 17:14
  • $\begingroup$ Distance preserving map is always injective for metric spaces. There is a need for mentioning a distance preserving map to be injective in cases of pseudo-metric spaces. $\endgroup$ – Jave Dec 26 '18 at 15:45
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Let $(X,d)$ & $(Y,d_1)$ be a metric spaces. Then $X$ is isometrically embedded to another space means $\exists f:X \to Y$ s.t $d(x,y)=d_1(f(x),f(y)) \forall x,y\in X$

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Let $(S,d)$ and $(T,e)$ be two metric spaces. A function $f$ from $S$ to $T$ is called and isometry iff $e(fx,fy)=d(x,y)$ for all $x,y\in S$. So any isometry $f$ from $S$ onto a subset of $T$ is what is called and embedded into another space. Since $f$ preserves the only given structure on $S$, the metric, we can consider $S$ as a subset of $T$

NB: For any metric space $(S,d)$ there is a complete metric space $(T,e)$ and an isometry $f$ from $S$ onto a subset of $T$, as in the case of the rational numbers, and $T$ is called the completion of $S$. So in this case, $S$ is isometrically embedding in $T$.

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