The diameter in the metric space

Question

The diameter of the nonempty set $A$ in a metric space $(X,d)$ is given by : $$ \delta(A)=\sup_{x,y\in A}d(x,y). $$

$1)$ Show that if : $A\subset B$

Then: $\delta(A)\leq\delta(B)$

$2)$ Then show that the sufficient and necessary condition for $ \mathit{\delta}{\mathrm{(}}{A}{\mathrm{)}}\mathrm{{=}}{0} $

is that $A$ contain only one element.

I have just try to prove that according to the definition for the part one For the part two the same but stuck a bit on the first part cause I don't know how I can express it by $ \mathrm{\leq} $.

Answer 1

For part 1, argue two things.

1. Take two sets of real numbers $S$ and $T$, where $S$ is the set of all distances $d(a_1, a_2)$ between two points of $A$, and $T$ is the set of all distances $d(b_1, b_2)$ between two points of $B$. Then $S \subseteq T$.
2. For any two sets of real numbers $S$ and $T$, if $S \subseteq T$, then $\sup S \leq \sup T$.

For part 2, note that if $A$ has two distinct points $a \neq b$, then $d(a, b) > 0$ (because $d(a, b) = 0$ iff $a = b$). Then you know $\delta(A) = \sup_{x, y \in A} d(x,y) \geq d(a, b) > 0$