Is $M=[0,1)$ with $d(x,y)=\min\{|x-y|,1-|x-y|\}$ a complete metric space?

Question

Let $M=[0,1)$. For $x,y\in M$, define

$$d(x,y)=\min\{|x-y|,1-|x-y|\}.$$

Then it can be shown that $d$ is a metric on $M$. Positive definiteness and symmetry are easy to show. The frustrating part is to prove the triangle inequality which is done in this post.

Is this metric space $(M, d)$ complete?

I was thinking to find a counter example but I couldn't find any. Also, I cannot prove that it is complete!

There is a comment under this post which says

This is the distance induced by the canonical mapping of the unit interval $[0,1)$ to the unit circle $S^1=\mathbb{R}/\mathbb{Z}$.

Can someone elaborate on this?

Here are some of my observations.

1. For this metric we have,

$$d(x,y)= \begin{cases} |x-y|, & 0\le|x-y|<\frac{1}{2}\\ 1-|x-y|, & \frac{1}{2} \le |x-y| < 1 \end{cases}. $$

2. The following properties are valid.

$$ \begin{array}{l} & d(x,y) \le |x-y|, \\ & d(x,y) \le 1-|x-y|, \\ & d(x,y) \le \frac{1}{2}. \end{array} $$

3. A sequence which is Cauchy/Convergent with respect to $|.|$ is Cauchy/Convergent with respect to $d$. But the converse is not true. See the below example.

4. The sequence $$ a_n= \begin{cases} 1-\frac{1}{n},&\quad n \text{ is even} \\ \frac{1}{n},&\quad n \text{ is odd} \end{cases} $$ is not Cauchy with respect to $|.|$ but it is Cauchy with respect to $d$. It is not convergent with respect to $|.|$ but it is convergent to $0$ in $M$ with respect to $d$!

5. If we try to add $\{1\}$ to $M$ then $d$ loses its definiteness property on $M\cup\{1\}$ and won't be a metric on $M\cup\{1\}$ anymore! The reason is that $d(1,0)=0$ but $1\ne 0$.

Answer 1

The metric space $S^1=\mathbb{R}/\mathbb{Z}$ is formally defined as $\{ \overline{x} \, \colon x\in\mathbb{R}\}$ where $\overline{x} := \{x+n : n\in\mathbb{Z}\}$. For instance, one element of $S^1$ is $\overline{1.5} = \{\dots, -1.5, -0.5, 0.5, 1.5, 2.5, \dots\}$. It is usually referred to as the torus and is endowed with the following distance $\delta$ :

Let $\delta(\bar{x},\bar{y})$ be the smallest distance (in $\mathbb{R}$) between an element of $\bar{x}$ and an element of $\bar{y}$. For instance, one can check that $\delta(\overline{1.5}, \overline{3.4}) = 0.1$.

Now you can show that there is an homeomorphism between $(M,d)$ and $(S^1,\delta)$.

More precisely, show that $\delta(\overline{x},\overline{y}) = d(x,y)$ whenever $x,y \in M$. Then any Cauchy sequence in $M$ can be mapped to a Cauchy sequence in $S^1$, and hence converges. That proves that $(M,d)$ is complete.

Side Note: They are also homeomorphic to $(\mathbb{U}, \lvert \cdot \rvert)$, where $\mathbb{U}$ is the set of complex numbers with modulus $1$, and that's why $S^1$ is called "the unit circle" in the comment.

A Direct Approach: Take a Cauchy sequence $(x_n)$ in $(M,d)$. If $\limsup x_n < 1$, show that $(x_n)$ converges to this limit in $(M,d)$. Else, we have $\limsup x_n = 1$. Hence, there is a subsequence $x_{\phi(n)}$ that converges in the metric space $([0,1],|.|)$ to $1$. Show that $x_{\phi(n)}$ converges to $0$ in $(M,d)$. Deduce that $x_n$ actually converges to $0$ in $(M,d)$.