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Say we have a metric space $(\mathbb{R}^+, d)$

where the distance function is

$d(x,y) = |x - y| + | 1/x - 1/y |$

Then I argue that this metric space is not complete: If we look at the Cauchy sequence $1/x$, which is contained in the metric space, we see that the limit of the sequence $\lim_{x\to \infty} \frac{1}{x} = 0$ is not in the metric space. Hence, the metric space is not complete.

Am I doing something wrong or is this a valid argument/proof?

L.F. Cavenaghi
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Fredrik
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2 Answers2

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Your proof has a serious flaw: The sequence $\frac{1}x$ is not Cauchy in this metric space. Note that (for $x,y\in \mathbb N$) $$d(1/x,1/y)=|x-y| + |1/x - 1/y| > |x-y| \geq 1$$ This means that there can be no $N$ such that $d(1/x,1/y)<1$ for all $x,y>N$, contradicting that it is a Cauchy sequence.

Milo Brandt
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  • Thanks Milo, I guess I just have a very poor idea of how the distance function should be interpreted in relation to the metric space itself. This helped me to get a better picture of it :) – Fredrik Apr 19 '16 at 00:22
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Milo Brandt has shown what is wrong with your argument. In fact $\langle\Bbb R^+,d\rangle$ is complete. Here’s a hint outlining how you might prove that.

HINT: Suppose that $\sigma=\langle x_n:n\in\Bbb N\rangle$ in $\langle\Bbb R^+,d\rangle$ is $d$-Cauchy.

  • Show that $\sigma$ is Cauchy in the Euclidean metric, and conclude that $\sigma$ converges to some $x\in\Bbb R$ in the Euclidean metric.
  • Show that $x\ge 0$.
  • Show that if $x$ were $0$, $\sigma$ wouldn’t be Cauchy after all.
  • Show that if $x>0$, $\sigma$ converges to $x$ in the metric $d$ as well as in the Euclidean metric.
  • Put the pieces together to conclude that $\langle\Bbb R^+,d\rangle$ is complete.
Daniel Fischer
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Brian M. Scott
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  • Thanks! A lot more work than i imagined though. Can you tell me why the Euclidean metric is of any interest in relation to this metric? – Fredrik Apr 19 '16 at 00:18
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    @Fredrik: You’re welcome. Basically the metric $d$ is a way to adjust the Euclidean metric just enough to get one that is complete on $\Bbb R^+$. Notice that $d(x,y)$ and $|x-y|$ are very nearly the same when $x$ and $y$ are large; it’s only when you get near $0$ that $d(x,y)$ and $|x-y|$ differ greatly, and that difference is carefully arranged so that sequences that converge to $0$ in the usual metric are no longer Cauchy. But $d$ is still enough like the Euclidean metric to generate the same topology. – Brian M. Scott Apr 19 '16 at 00:24
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    @Fredrik You can actually get that this metric is just the Euclidean metric in disguise - consider the map $f(x) = x - \frac{1}x$. You can prove that $d(x,y)=|f(x)-f(y)|$. This observation lets you lift basically every property (for instance, completeness) of $\mathbb R$ (with the usual metric) to the new metric space. – Milo Brandt Apr 19 '16 at 00:34
  • @Brian: That is a very good explanation and I think it is a smart strategy :) With some work, I am now confident that I will be able to make a solid proof of it being complete. Thanks a lot Brian! – Fredrik Apr 19 '16 at 00:37