Complete Metric Space example

Question

Let $X=(0,1]$ and $d:(0,1]\times(0,1] \to \mathbb{R} $ be a function defined as

$$d(x,y):=\left|\frac 1x - \frac 1y\right| $$

I have managed to show that d is infact the distance by using the axioms of a Metric.

However I am struggling to show that it in fact a complete metric space

Answer 1

The space $[1,+\infty)$ is complete with respect to the usual metric. Now, use the fact that$$\begin{array}{ccc}(0,1]&\longrightarrow&[1,+\infty)\\x&\mapsto&\frac1x\end{array}$$is an isometry, if you consider the metric $d$ in $[0,1]$ and the usual metric in $[1,+\infty)$.

Answer 2

Another proof is that $|\frac{1}{x_n} - \frac{1}{x_m}| = |\frac{x_n-x_m}{x_nx_m}|$. If this sequence is Cauchy, then given some $\epsilon$, there is $N$ such that $n,m>N$ implies this quantity is less than $\epsilon$. Since $x_n, x_m \in (0,1]$, the product is also in $(0,1]$, and so $x_nx_m \epsilon <\epsilon$. So being Cauchy actually implies $|x_n -x_m| < \epsilon$, so being Cauchy in this metric implies being Cauchy in the usual metric, which we know implies convergence.