Let $X$ be a set. Consider the following properties referring to a function $d : X \times X \to \mathbb R$:
$d(x, y) = 0$ if and only if $x = y$;
$d(x, z) \le d(z, y) + d(y, x)\;$ whatever $\;x, y, z \in X$.
a) Show that any metric $d$ on $X$ satisfies the above two properties.
b) Show that if $d$ satisfies both the first and second properties, then $d$ is a metric over $X$.
c) Let $X = \mathbb R$. Show that:
(i). $d(x, y) = x − y$ satisfies the first property, but not the second;
(ii).$d(x, y) = |x − y| + 1$ satisfies the second property, but not the first.
If it is metric then:
$d(x, x) = 0$ whatever $x \in X$;
$d(x, y) = d(y, x)$ whatever that looks like $x, y \in X$;
$d(x, z) \le d(x, y) + d(y, z)$ whatever that looks like $x, y, z \in X$.
a)(=>) $d(x, y) = d(y, x) = 0 = d(x,x) = d(y,y) \to x = y$ (<=) $x = y \to d(y,y) = d(x,x) = 0 = d(y, x) = d(x,y)$
$d(x, z) \le d(z, y) + d(y, x) \to d(x, z) \le d(y, z) + d(x, y) \to d(x , z) \le d(x, y) + d(y, z)$
b) is not the same thing as item a)?
c) $d(x, y) = x − y = 0 = d(y,x) = y-x \to x = y $
(i).$d(x, z) \le d(z, y) + d(y, x) \to x-z \le z-y + y-x \to x-z \le z-x$
Or that it is true only for $x = z$.
(ii).$d(x, y) = d(y, x) = |x − y| + 1 = 0 $ else if $x=y \to |x − x| +1 \ne $0
$d(x, z) \le d(x, y) + d(y, z) \to |x − z| + 1\le |x − y| + 1 + |y − z| + 1 \to |x − z| \le |x − y| + |y − z| + 1$
Stuck here,
the other attempts are acceptable?
Thanks.