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I am trying to determine a general criteria in order to conclude whether a given metric $d$ induces a complete metric space $(ℝ,d)$. There seems to be the following result:

Let $d(x,y)=|f(x)−f(y)|$, where $f$ is an injective function defined on $ℝ$, with values in $ℝ$, and $|\cdot |$ is the usual absolute value function on $ℝ$. Then the image of $f$ is a closed set if and only if $(ℝ,d)$ is complete.

To prove this, one must show that a Cauchy sequence converges, i.e. that $f(x_n)$ converges if $|f(x_n)−f(x_m)|<ε$ for any $ε>0$, if $m$ and $n$ are large enough. Intuitively, if the image of $f$ is closed, it must contain all its limit points, including the limit of $f(x_n)$. But why does $f(x_n)$ converge in the first place? Why does $f$ have to be injective, and how is the converse true? Thank you for any help!

Example: consider the metric defined by $d(x,y)=|x^3-y^3|$. Let $f: ℝ \longrightarrow ℝ$ be the injective function defined by $f(x)=x^3$. The image of $f$ is $ℝ$ which is closed, so $(ℝ,d)$ is complete. On the other hand if $d(x,y)=|\arctan{x}-\arctan{y}|$, $Im(f)=]-\pi/2;\pi/2[$ is not closed, so $(ℝ,d)$ is not complete.

Note: this questions is in the continuity of this one: $(\mathbb{R},d)$ is not always a complete space?

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    $\begingroup$ Note that if $f$ is not injective, then $d$ is not a metric, since you will have $d(x,y) = 0$ for some $x \neq y$. $\endgroup$ – Bungo Sep 24 '15 at 0:59
  • $\begingroup$ @Bungo: yes that makes a lot of sense. Thanks. Yes I meant $f(x_n)$, my mistake, I have made the changes in the question. $\endgroup$ – Kuifje Sep 24 '15 at 1:05
  • $\begingroup$ What is that double-bar thingy, $\|\cdot\|$ $\endgroup$ – GEdgar Sep 24 '15 at 1:06
  • $\begingroup$ @GEdgar: $|| \cdot ||$ would be the absolute value in $\mathbb{R}$. Indeed this should be clarified, thanks! $\endgroup$ – Kuifje Sep 24 '15 at 1:08
  • $\begingroup$ You also didn't say that $f$ has real values. $\endgroup$ – GEdgar Sep 24 '15 at 1:09
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Your guess is correct. Note that now $(\mathbb R, d)$ is isometric to $(f(\mathbb R), d_0)$, where $d_0$ is the standard metric $d_0(x, y) = |x-y|$. Then try to show that $(Y, d|_Y)$ in a complete metric space $(X, d)$ is complete if and only if it is closed.

Remark: This is really a general construction: Let $(Y, d)$ be a metric space and $ f:X \to Y$ be an injective map. Then one can define a metric $f^*d$ on $X$ by

$$(*)\ \ f^*d(x_1, x_2) = d(f(x_1), f(x_2)).$$

This metric on $X$ has the property that $f: (X, f^*d) \to (f(X), d)$ is an isometry (To be an isometry you need only to satisfy $(*)$)

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  • $\begingroup$ Why would an injective mapping need be an isometry? $\endgroup$ – Anthony Peter Sep 24 '15 at 1:17
  • $\begingroup$ @AnthonyPeter : Note in this case the isometry is given by $f:\mathbb R\to f(\mathbb R)$ (there are reasons why you need an injective $f$: see the comment below the question) $\endgroup$ – user99914 Sep 24 '15 at 1:25
  • $\begingroup$ @AnthonyPeter: the answer is here: math.stackexchange.com/questions/568415/… $\endgroup$ – Kuifje Sep 24 '15 at 1:26
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    $\begingroup$ @JohnMa: Yes, I can see why $Y$ is complete if and only if it is closed in $X$. So the proof would have the two following steps? : 1. Show $(\mathbb{R},d)$ is isometric to $(f(\mathbb{R}),d_0)$. 2. $f(\mathbb{R})$ is a subspace of $\mathbb{R}$, show that it is complete if and only if it closed. $\endgroup$ – Kuifje Sep 24 '15 at 2:17
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    $\begingroup$ @JohnMa: Thanks for the guidelines. $\endgroup$ – Kuifje Sep 24 '15 at 2:27

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