Complete quotient metric space

Question

Let $(X,d)$ a metric space and $\sim$ a equivalence relation such that :

1. $\forall x\in X$ : $[x]=\{y\in X \vert y \sim x \}$ is closed.
2. If $[x] \neq [y]$ : $d([x],[y])=d(a,[y]), \forall a\in[x]$

Define in $\dfrac{X}{\sim}$ : $D([x],[y])=d([x],[y])$ what is a metric (I've proved it). Prove that if $(X,d)$ is complete then $(\dfrac{X}{\sim},D)$ is complete.

Any idea how to proceed? the truth I tried to find some succession of cauchy in $ X $ from one in $\dfrac{X} {\sim}$ but I could not.

Answer 1

Let $X = \mathbb{R} \times (0,\infty)$ and $\sim$ the equivalence relation for $X$ with the equivalence classes

• $E = \{0\} \times (0, \infty)$,
• $R(a) = \{(x,\frac{a}{x}) : 0 < x \}$ for all $a > 0$, and
• $L(a) = \{(x,\frac{a}{x}) : x < 0 \}$ for all $a < 0$.

Each of those equivalent classes is closed within $X$.

Though $E \neq R(1)$, $D([(0,1)], [(1,1)]) = D(E,R(1)) = 0$.

Thusly $D([a],[b]) = \min \{ d(x,y) : x \in [a], y \in [b] \}$ is not a metric for $X/\!\sim$.

Perhaps if $X$ were compact $D$ could be a metric.

However upon close inspection of the definition of $D$, a salient condition appears.  Namely, for all $x,y \in [a]$, $d(x,[b]) = d(y,[b])$.

Let $([x_j])$ be a Cauchy sequence. Define $a_1 = x_1$, $a_{j+1}$ a point in $[x_{j+1}]$ with $d(a_j,a_{j+1}) = d(a_j,[x_{j+1}])$. Show $(a_j)$ is a Cauchy sequence, and it converges to a point $a$. Show ([$x_j$]) converges to [a].