Showing that $(S,d)$ is a metric space for $S = (0,1]$ and $d(x,y) = |1/x - 1/y|$ Let $S$ be a half-open interval $(0,1]$. If we define $d$ on $S$ by
$$d(x,y) := \left|{1\over x} - {1\over y}\right|\;,$$
then show that $d$ is a metric on $S$. Also, prove that $(S,d)$ is a complete metric space.
I think that this problem would begin by showing the $3$ properties of a metric space. And then show that it is Cauchy?
 A: Just as two topological spaces are identical in terms of topological structure precisely when there is a homeomorphism between them, two metric spaces are identical in terms of metric structure precisely when there is an isometry between them. In both cases you could say that they’re the same space under a different name. My suggestion in the comments was that $S$ with the metric $d$ is structurally identical to the ray $[0,\to)$ with the usual metric. The idea is that a point $x\in S$ corresponds to $\frac1x\in[1,\to)$. This map $x\mapsto\frac1x$ really is an isometry: if $x,y\in S$, the $d$-distance between $x$ and $y$ is exactly the same as the ordinary distance between $\frac1x$ and $\frac1y$ in $[1,\to)$. And the space $[1,\to)$ with the usual metric is complete, so we can use this correspondence to show that $\langle S,d\rangle$ is complete as well.
Suppose that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence in $\langle S,d\rangle$. This means that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $d(x_k,x_n)<\epsilon$ whenever $k,n\ge m$, i.e., that
$$\left|\frac1{x_k}-\frac1{x_n}\right|<\epsilon\qquad\text{whenever}\qquad k,n\ge m_\epsilon\;.$$
We want to show that the sequence converges to some $x\in S$. Look at the corresponding sequence $$\left\langle\frac1x_n:n\in\Bbb N\right\rangle\tag{1}$$ in $[1,\to)$. For convenience of notation let $y_n=\frac1{x_n}$, so that $(1)$ is just $\langle y_n:n\in\Bbb N\rangle$. Is this a Cauchy sequence in $[1,\to)$ with the usual metric? Absolutely: if $k,n\ge m_\epsilon$ (that’s the same $m_\epsilon$ as before), then
$$|y_k-y_n|=\left|\frac1{x_k}-\frac1{x_n}\right|=d(x_k,x_n)<\epsilon\;.$$
And $[1,\to)$ is complete, so $\langle y_n:n\in\Bbb N\rangle$ converges to some limit $y$.
At this point I’ve done most of it; can you use $y$ to find the limit of the original sequence $\langle x_n:n\in\Bbb N\rangle$ in $S$ and show that what you find really is the limit?
