I'm reading about the isometric embedding and the definition goes like this. Let $(X, d)$ and $(Y, d_1)$ be metric spaces and $f$ a mapping of $X$ into $Y$. Let $Z=f(X)$, and $d_2$ be the metric induced on Z by $d_1$. If $f: (X, d) → (Z, d_2)$ is an isometry, then $f$ is said to be an isometric embedding of $(X, d)$ in $(Y, d_1)$. I'm not able to understand the metric induced by another metric part. Can someone give an example so that I can understand this?
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The definition is $d_2(f(x),f(y))=d_1(x,y)$. For this to be a metric we have to assume that $f$ is injective.
For example let $f:\mathbb R \to \mathbb R$ be defined by $f(x)=e^{x}$. the range of this function is $(0,\infty)$ and the induced metric $d_2$ on the range is defined by $d_2(x,y)=|\log \,x -\log \, y |$ (assuming that the domain of $f$ is given the usual metric).
Kavi Rama Murthy
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So, $(\mathbb R, d)$, $(\mathbb R, d2)$ where $d$ is usual euclidean metric and $d_2(x,y)=|e^{x}-e^{y}|$ is isometric with given mapping $f$? – Vamshi Kumar Kurva Jul 23 '19 at 12:36
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@VamshiKumarKurva The second space is not $(\mathbb R, d_2)$. It is $((0,\infty),d_2)$ and it is isometric with the first space. – Kavi Rama Murthy Jul 23 '19 at 12:42
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If I select $x=2, y=1$, then $d(x,y)=1$, $f(y) = e, f(x) = e^2$. $d2(f(x), f(y)) = |e^{f(1)} - e^{f(2)}|$. Is that right? – Vamshi Kumar Kurva Jul 23 '19 at 12:46
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Sorry, there was a mistake. I have edited my answer. – Kavi Rama Murthy Jul 23 '19 at 13:10
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Thank you. That makes sense. – Vamshi Kumar Kurva Jul 23 '19 at 14:55
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But, it seems like $d_2$ is actually induced by $d$ and $f$, rather than $d_1$. – Vamshi Kumar Kurva Jul 23 '19 at 15:08