Complete quotient metric space

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.

• $d([x],[y])$ is just the usual infimum definition, so $d([x],[y])=\inf \{d(a,b): a \in [x], b \in [y]\}$? – Henno Brandsma May 12 at 6:25
• Yes, it´s correct. – Juan Daniel Valdivia Fuentes May 12 at 6:47

Let X = Rx(0,oo) and ~ the equivalence relation for X
with the equivalence classes
E = {0}x(0,oo),
R(a) = { (x,a/x) : 0 < x } for all a > 0 and
L(a) = { (x,a/x) : x < 0  } for all a < 0

Each of those equivalent classes are closed within X.
Though E /= 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/~.

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.
So it converges to a point a.
Show ([$$x_j$$]) converges to [a].#

• And how prove that $\dfrac{X}{\sim}$ is complete? – Juan Daniel Valdivia Fuentes May 15 at 15:57
• @JuanDanielValdiviaFuentes. See exit for a sketch of a proof. – William Elliot May 18 at 8:59