Showing this metric space is complete

Let $X=(0,1]$ and $d(x,y)=\left|\frac{1}{x}-\frac{1}{y}\right|$. I've proven $(X,d)$ is a metric space but I don't know how to show its completeness. How can I do that?

• Start with a Cauchy Sequence in $(0,1]$ and show that it has a limit in $(0,1]$
– C.S.
Commented Oct 12, 2014 at 8:59
• {1/x_n} is a Cauchy sequence but its limit 0 isn't in (0,1] Commented Oct 12, 2014 at 9:01
• what is $1/x_n$??? Commented Oct 12, 2014 at 9:03
• Assuming that your $\frac1{x_n}$ is simply $\frac1n$, it’s not a Cauchy sequence with respect to the metric $d$: $$d\left(\frac1n,\frac1m\right)=|n-m|\;.$$ Commented Oct 12, 2014 at 9:03

3 Answers

HINT:

The map $x \mapsto \frac{1}{x}$ from your space to $[1, \infty)$ with the usual metric is an isometry.

Expanding on orangeskid's hint-answer, "isometry" means that if we call that map $f$, the distance on $X$ $d$ and the usual metric $d_e$, then $d(x,y)=d_e(f(x),f(y))$. This can be used to turn $x_n$ into a Cauchy sequence in $[1,\infty)$, thus finding a limit there and getting back to $X$.

$\frac{1}{n}$ is a Cauchy sequence in $[0,1]$ or any subset of $\mathbb{R}$ containing all its terms with the Euclid metric, not with your metric, where, as Brian M. Scott pointed out, the distances are never less than one.

Finally, a comment on format: it is better to provide a generic title like "Proving a metric space is complete" or "Proving a certain metric space is complete" and then putting the definition into the question rather than placing the definition of the space and distance in the title and then simply stating you can't find a proof in the question. I'll edit the question accordingly, but keep this in mind for the future :).

Let $x_n$ be a Cauchy sequence in $(X,d)$. This is equivalent to $$\forall_{\epsilon>0}\exists_{N \in \mathbb{N}}\forall_{n,m>N} \left|\frac{1}{x_n}-\frac{1}{x_m}\right| < \epsilon$$ But then $\frac{1}{x_n} =:y_n$ is a Cauchy sequence in $(\mathbb{R},p)$ with $p(x,y) = |x-y|$. Because $(\mathbb{R},p)$ is complete there exists a number $x$ such that $\lim_{n \to \infty}y_n = x$. Note that $x \geq 1$, because $0<x_n\leq 1$. $\lim_{n \to \infty}y_n = x$ is again equivalent by the definition of convergence to $$\forall_{\epsilon>0}\exists_{N \in \mathbb{N}}\forall_{n,m>N} \left|\frac{1}{x_n}-x\right| < \epsilon$$ But this is the same as $$\forall_{\epsilon>0}\exists_{N \in \mathbb{N}}\forall_{n,m>N} d(x_n,\frac{1}{x}) < \epsilon$$ which means that $\lim_{n \to \infty}x_n = \frac{1}{x} \in X$.