The continuity of a function defined on $\mathbb R/2\pi \mathbb Z$, and the distance for $\mathbb R/2\pi \mathbb Z.$ I'm handling a function $f$ from $\mathbb R/2\pi \mathbb Z$ to $\mathbb R$, $f : \mathbb R/2\pi \mathbb Z \to \mathbb R$.
And I was told that the meaning of [$f$ is continuous at  $a\in \mathbb R/2\pi \mathbb Z$] is : $$\forall \epsilon >0, \exists \delta>0 \ \mathrm{\ s.t.\ } d'(x, a)<\delta \Rightarrow |f(x)-f(a)|<\epsilon$$, where
$d'(x,y)=\min \{|a-b|\mid a\in x,b\in y \}$ for $x,y\in \mathbb R/2\pi \mathbb Z$.
But I'm wondering whether this $d'$ is metric in this situation, and whether $\{|a-b|\mid a\in x, b\in y\}$ indeed has minimum value.
Accourding to
https://en.wikipedia.org/wiki/Metric_space#Quotient_metric_spaces,

given two equivalence classes $[u]$ and
$[v]$, we define $$
     d'([u],[v]) = \inf\{d(p_1,q_1)+d(p_2,q_2)+\dotsb+d(p_{n},q_{n})\} $$ where the infimum is taken over all finite sequences $(p_1, p_2,
 \dots, p_n)$ and $(q_1, q_2, \dots, q_n)$ with $[p_1]=[u], [q_n]=[v],
 [q_i]=[p_{i+1}], i=1,2,\dots, n-1$.


In general this will only define a
pseudometric, i.e. $d'([u],[v])=0$ does not necessarily imply that
$[u]=[v]$. However for nice equivalence relations , it is a metric.

I wonder the definition $d'(x,y)=\min \{|a-b|\mid a\in x,b\in y \}$ for $x,y\in \mathbb R/2\pi \mathbb Z$ is equivalen to this.
And according to this wikipedia page, $d'$ is not metric in general, but in special case, $d'$ can be metric.
Now, I have considered whether $d'$ defined by $d'(x,y)=\min \{|a-b|\mid a\in x,b\in y \}$ for $x,y\in \mathbb R/2\pi \mathbb Z$ is metric or not.
・$d'(x,y)=0 \iff x=y$
・$d'(x,y)=d'(y,x)$
are almost obvious, but I cannot see $d'$ satisfies triangle inequality and $\min \{|a-b|\mid a\in x, b\in y\}$ really exists.
Could you explain about these things?
 A: In general, the metric $d'$ that you define doesn't necessarily satisfy the triangle inequality as explained here: Why are quotient metric spaces defined this way?
However, in this case, you have a group quotient rather than just a metric space quotient, which means that the metric $d'$ that you define is the same as the usual quotient metric.
In particular, the group structure implies that $[x] = [y]$ is equivalent to $[x - y] = [0]$, and so your metric can be written as
$$
d'([x], [y])
= \min_{k\in \mathbf{Z}} |x - y + 2k\pi|.
$$
Now, this metric does satisfy the triangle inequality.
To see this, let $x, y, z \in \mathbf{R}$ and let $m^*, n^*$ be the minimizers
\begin{align*}
m^* 
&= \operatorname*{argmin}_{m \in \mathbf{Z}} |x - y + 2m\pi|, \\
n^* 
&= \operatorname*{argmin}_{m \in \mathbf{Z}} |y - z + 2n\pi|,
\end{align*}
which yields
\begin{align*}
d'([x], [z])
&\leq |x - z + 2(m^* + n^*)\pi| \\
&\leq |x - y + 2 m^* \pi| + |y - z + 2 n^*\pi| \\
&= d'([x], [y]) + d'([y], [z]).
\end{align*}
