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.

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

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

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