PSEUDOMETRIC Any explanation i will be so thankful .
a finite pseudometric on a set X is a function satisfying (M1) ,(M2),(M3) and (M2*) d(x,x)=0 What is the difference between a metric and pseudometric ? show that $ {d}{(}{x}{,}{y}{)}{=}\left|{{\mathit{\xi}}_{1}{-}{\mathit{\eta}}_{1}}\right| $ defines a pseudometric on the set of all ordered pairs of real numbers,where $ {x}\mathrm{{=}}{\mathrm{(}}{\mathit{\xi}}_{1}{\mathrm{,}}{\mathit{\xi}}_{2}{\mathrm{),}}\hspace{0.33em}{y}\mathrm{{=}}{\mathrm{(}}{\mathit{\eta}}_{1}{\mathrm{,}}{\mathit{\eta}}_{2}{\mathrm{)}} $
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$\begingroup$The idea behind the metric $d$ is that one wants to designate a "distance" between points on in $\mathbb{R}^2$ by only considering their distance in the first component.
For example if $(3,5)$ and $(4,5)$ are two points, their "distance" will be $$d((3,5), (4,5)) = |3 - 4| = 1.$$
Two show that the term "distance" is applicable we want to show that $d$ is in fact a pseudometric. Note that a (pseudo)metric tries to grasp the intuitive notion of distance and formulates it by mathematical axioms, just as (M1), (M2), (M3) in your case.
Now let's try to check the axioms of a (pseudo)metric. Let therefore $x := (x_1, x_2), y:= (y_1, y_2), z:= (z_1, z_2) \in \mathbb{R}^2$.
- $d(x,x) = |x_1 - x_1| = 0\quad \checkmark$
- $d(x,y) = |x_1 - y_1| = |y_1 - x_1| = d(y,x)\quad \checkmark$
- $d(x,y) = |x_1 - y_1| = |x_1 - z_1 + z_1 - y_1| \leq |x_1 - z_1| + |z_1 - y_1| = d(x,z) + d(z, y) \quad\checkmark$
So, $d$ is a pseudometric. Observe that everything I used were properties of the euclidean metric on $\mathbb{R}$. You should try to prove that $d$ is in fact a pseudometric and not a metric, i.e. find a pair $x,y \in \mathbb{R}^2$ with $x \neq y$ but $d(x,y) = 0$.
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