How can I show diameter of subset is less then diameter of superset?

Question

I know it should be a basic proof but I have stuck. I cannot obtain inequality exactly.

Let $(X,d)$ be a metric space. If $A,B \subseteq X$ with $A \subseteq B$ then prove that $\delta A \le \delta B$

$\delta A = sup\{ d(a_1,a_2) : a_1,a_2 \in A\}$ and $\delta B = sup\{ d(b_1,b_2) : b_1,b_2 \in B\}$

From supremum axioms

i) $d(a_1,a_2) \le \delta A$ and $d(b_1,b_2) \le \delta B$

ii) $\forall \epsilon \gt 0$ $\exists a_{\epsilon_1}, a_{\epsilon_2} \in A$ such that $d(a_{\epsilon_1}, a_{\epsilon_2}) + \epsilon \ge \delta A $

and $\exists b_{\epsilon_1}, b_{\epsilon_2} \in B$ such that $d(b_{\epsilon_1}, b_{\epsilon_2}) + \epsilon \ge \delta B $

Since $a_1, a_2 \in B$ at the same time $d(a_1,a_2)$ must be less then $\delta B$. How should I do for proof in the easiest way? I think there are some incorrect or unnecessary thing in my writtens. Could someone help me please?

Answer 1

Let $a_1,a_2\in A$. Then since $A\subseteq B$, $\delta B \ge d(a_1,a_2)$ by definition. Hence $\delta B$ is an upper bound for the set $$\{d(a_1,a_2) : a_1,a_2\in A\}.$$ Since $\delta A$ is the supremum of this set, we have $\delta A \le \delta B$ by the definition of the supremum as the least upper bound.

Answer 2

Hint: If $A\subseteq B\subseteq {\mathbb R}$, then $\sup A\leq \sup B$.

Answer 3

Generally,if you have $C \subset D \subset \mathbb{R}$ then $\sup C \le \sup D$.

Let $C= \{ d(x,y) | x,y \in A\}$, $D= \{ d(x,y) | x,y \in B\}$.

We see that $C \subset D$.