Is the metric well-defined?

Question

Let $(X,d)$ be a pseudo-metric space and fix $z\in X$. Define $x\sim y\;$ iff. $\;d(z,x)=d(z,y)$.

Then define $\hat{X}=X/\sim$ and $\hat{d}:\hat{X}\times\hat{X}\rightarrow\mathbb{R}\;$ by $\;\hat{d}([x],[y])=d(x,y)$.

I am trying to show that $\hat{d}$ is well-defined.

To begin the proof, I fix $x\in X$ and let $\,y_0\in [y].\;$ Then, $\;d(z,y)=d(z,y_0)$.

I need to show that $d(x,y)=d(x,y_0)$.

Answer 1

It's not… take $X = \Bbb R, d(x,y) = |x-y|$ and $z = 0$

Then we have: $x\sim y \iff |x| = |y|$

And get $[1] = \{-1,1\} = [-1], [2] = \{-2,2\} = [-2]$

But $$\hat{d}([1],[2]) = |1-2| = 1 \not= 3 = |-1-2| = \hat{d}([-1],[2])$$

Answer 2

It’s false in general. Let $X=\Bbb R^2$, and define $d$ by

$$d\big(\langle x_0,y_0\rangle,\langle x_1,y_1\rangle\big)=|x_0-x_1|\,;$$

$d$ is a pseudometric on $X$. Take $z=\langle 0,0\rangle$, so that $\langle x_0,y_0\rangle\sim\langle x_1,y_1\rangle$ iff $|x_0|=|x_1|$, so $\langle -1,0\rangle\sim\langle 1,0\rangle$. Now let $x=y=\langle 1,0\rangle$ and $y_0=\langle -1,0\rangle$; then

$$d(x,y)=d\big(\langle 1,0\rangle,\langle 1,0\rangle\big)=0\,,$$

but

$$d(x,y_0)=d\big(\langle 1,0\rangle,\langle -1,0\rangle\big)=|1-(-1)|=2\,.$$

Answer 3

Usually, when $d$ is a pseudometric we define

$$x \sim y \iff d(x,y)=0\tag{1}$$

This is the same as the equivalence relation

$$x \sim y \iff \forall z: d(x,z)=d(y,z)\tag{2}$$ which might be what your text was going for, really.

Using the more usual $(1)$ we have a metric on $\hat{X}$, defined by $\hat{d}([x], [y]) = d(x,y)$ and a tedious and easy check shows that this is well-defined (i.e. does not depend on the representative of a class $[x]$) and indeed a real metric.

Well-definedness: if $[x']=[x]$ and $[y']= [y]$ implies $x \sim x'$ and $y \sim y'$ (i.e. $d(x,x')=0, d(y,y')=0$) and so: $$d(x',y') \le d(x',x) + d(x,y) + d(y',y) = 0 + d(x,y)+0= d(x,y)$$

and a symmetrical argument also gives $d(x,y) \le d(x',y')$ hence $d(x,y)=d(x',y')$. QED.