Let $X=(0,1]$ and $d(x,y)=\|\frac{1}{x}-\frac{1}{y}\|$ for $x,y \in X$. Show that $(X,d)$ is complete. Let $X=(0,1]$ and $d(x,y)=\|\frac{1}{x}-\frac{1}{y}\|$ for $x,y \in X$. Show that $(X,d)$ is complete.
I want to show that every Cauchy sequence has a limit point in X.
Let $(x_1,x_2,...)$ be a Cauchy sequence. If $\epsilon \in R_{>0}$, then there exists $N \in Z_{\ge 0}$ such then if $m,n \ge N$, then $|\frac{1}{x_n}-\frac{1}{x_m}| \lt \epsilon$.  Since we don't know that does this sequence converge to, I am confused about how to prove this sequence is convergent in X.
 A: Suppose that $x_n$ is  Cauchy sequence in $(0,1]$. Then $|\frac{1}{x_n} - \frac{1}{x_m}|\to 0$ as $n,m \to \infty$. However, this means that $\{\frac 1{x_n}\}$ is a Cauchy sequence in the standard absolute value norm. Therefore, $\{\frac 1{x_n}\}$ is a convergent sequence(by completeness of $\mathbb R$ under the standard metric), which converges to some real number $x \in \mathbb R$.
All you have to show now is that $x \neq 0$, $x_n \to \frac 1x$ in your metric, and that $\frac 1x \in (0,1]$. Then, you will be done.
Since $x_n \in (0,1]$, each $\frac{1}{x_n}$ is greater than or equal to $1$, so $x$ being the limit has to be greater than or equal to $1$. This also shows that $\frac 1x  \in (0,1]$. 
I leave you to see that  $x_n \to \frac 1x$ in your metric. It is a straightforward rearrangement.
A: As an alternative approach (which particularly I find more intuitive in this case) consider reasoning with an isometry instead of Cauchy sequences.
Call $X$ metric $ d_X(x,y) = d(x,y)$ and let $ Y = [1, \infty],\ d_Y(x,y) = \|x - y\| $ and $ f:X \to Y:x \mapsto \frac{1}{x} $.
Notice that $ d_X(x,y) = d_Y(f(x),f(y)) $ and that $f$ is a bijection between $X$ and $Y$. Therefore $f$ is an isometry between $(X, d_X)$ and $(Y, d_Y)$. Since $(Y, d_Y)$ is a closed subspace of the complete space $(\mathbb{R}, d_Y)$, hence complete, it follows that $(X, d_X)$ is complete as well.
