$X = (0, +\infty)$. Function $d(x, y) = \left|\frac{1}{x} - \frac{1}{y}\right|$. Is $(X, d)$ complete?

Question

I'm having trouble understanding and solving this problem. For $(X,d)$ to be complete, every Cauchy sequence of points in $X$ has to have a limit in $X$. Let's take a sequence $(x_n)$ in $X$. For every ε > 0, there exists an $N$ such that for all $m, n ≥ N$, $$|1/x_m - 1/x_n| < ε$$ And as $m, n$ go to infinity, $1/x_m$ and $1/x_n$ go to $0$, making $|1/x_m - 1/x_n|$ converge to $0$. Is this even partially right? Thank you in advance.

Answer 1

You have to show that for any Cauchy sequence $(x_n)\in X$, there exists a point $x_0$, such that $x_n\to x_0$ as $n\to\infty$.

Let then $(x_n)$ be a Cauchy sequence. By definition, for all $\varepsilon>0$, there exists an $N\in\mathbb{N}$, such that for all $n,m\geq N$:

$$d(x_n,x_m)\leq \varepsilon$$

Using the definition of $d$, this is equivalent to

$$\bigg|\frac{1}{x_n}-\frac{1}{x_m}\bigg|\leq \varepsilon$$

Recall that we want to show that $x_n\to x_0$ for some $x_0\in X$. This means showing that for all $\delta>0$, there exists an $N_\delta\in\mathbb{N}$ such that for all $n>N$, we have $d(x_n,x_0)\leq \delta$. Which using the definition of $d$ yields

$$\bigg|\frac{1}{x_n}-\frac{1}{x_0}\bigg|\leq\delta.$$

Can you continue from here?

Answer 2

Hint: Consider the sequence $x_n = n \in X = (0,\infty)$. Is $(x_n)$ a Cauchy sequence in $(X, d)$ ? Can $(x_n)$ converge to some $x_0 \in X$ ?

Answer 3

If $X=(0,+\infty)$ you can derive few conclusions. First of all, you need a Cauchy sequence $x_n$. It is familiar that if $\lim_{n \to \infty}x_n=x_0 \neq 0$ then $\lim_{n \to \infty}\frac{1}{x_n}=\frac{1}{x_0}$. This means that if you take a Cauchy sequence that is convergent in $\mathbb{R}^+$ with standard topology (and does not convergent to 0), it will also be convergent in your set. If you take some sequence that does converge to 0 you will have a problem with the Cauchy property of that sequence in accordance to your metrics. Since we have commented out all of the convergent sequences in $\mathbb{R}^+$, we should see how divergent sequences behave. This is the point where you take some infinitely large sequence ($\lim_{n\to \infty}x_n=+\infty$). For example $x_n=n$. In this case, you will find that this sequence is Cauchy sequence, but it is not convergent since $n\to+\infty$. This way you remove piece-by-piece sequences that do not suit your problem. Also, this does not mean that you will always find a proper counterexample this way, but it can help.

Answer 4

Let $f:(0,\infty)\to (0,\infty)$ be defined by $f(x)=x^{-1}.$ Then $d(f(x),f(y))=|x-y|.$ Thus $f$ is an isometry from $(0,\infty)$ with the standard metric to $X=((0,\infty),d).$ Clearly $(0,\infty)$ with the standard metric is not complete (as it is not closed in $\mathbb{R}$), hence $X$ is not complete.