$H = [1,\infty)$ with $d(x,y) = |\frac{1}{x} - \frac{1}{y}|$ is not Compact by covers

Question

Let (M,d) be a metric space where M = (0, $\infty$) and $d(x,y) = |\frac{1}{x} - \frac{1}{y}|$

Prove that $H = [1,\infty)$ is not compact by covers.

H is clearly bounded and closed, so I am unsure as to how the proof would work. I would start by establishing {${G_\alpha}$}$_{\alpha\in\mathbb I}$ a cover of H. But I do not know how to proceed.

Answer 1

Let $D(x,y)=|x-y|$ be the usual metric of $\mathbb{R}$. I'll try to make everything as clear as possible:

The Heine-Borel theorem is the following:

Theorem: Let $A$ be a nonempty subset of $\mathbb{R}$. Then $A$ is compact, as a subset of the metric space $(\mathbb{R},D)$, if, and only if,

1. There exists $a,b\in\mathbb{R}$ such that $A\subseteq[a,b]$; and
2. $A$ is a closed subset of $(\mathbb{R},D)$.

Here's the problem

• The metric $d$ is not $D$. The set $H$ is $d$-bounded, but not $D$-bounded.

• By the Heine-Borel theorem, $(H,D_{H\times H})$ is not compact, because it is an unbounded subset of $(\mathbb{R},D)$.

• The map $(H,d)\to((0,1],D_{(0,1]\times(0,1]})$, $x\mapsto 1/x$, is an isometry. By the Heine-Borel theorem, $((0,1],D_{(0,1]\times(0,1]})$ is not compact, because it is not closed, so $(H,d)$ is also not compact.

Answer 2

Hint: $(0, 1]\to H:x\mapsto \frac1x$ is an isometry and therefore a homeomorphism.

Here $(0,1]$ is endowed with the usual metric. You can build a suitable open cover of it and transport it.

Answer 3

If you need a direct way: take the open cover $\{G_n\}_{n \in \mathbb{N}}$, where $G_n = [1, n+1)$.