Metric space proof I'm learning calculus and we have the following problem in our book:

$(X,d)$ is a metric space. Let $ \emptyset  \neq A_1 \subset A_2
 \subset X$. Prove that if $A_2$ is a bounded set, then $A_1$ is also a
  bounded set and the diamater of $A_1 \leq $ diameter of $A_2$

We would usually prove it on our lectures, but since there are not lectures anymore, I don't have proof. 
My idea was to use the suprema, but not really sure how to form the proof.
Could you please help me?
Thanks
 A: If $A_2$ is bounded this means that it has a bound; that is, there is a positive real number $b$ such that $d(x,y)\leq b$ for all $x,y\in A_2$. Now $A_1$ is a subset of $A_2$ so in particular $d(x,y)\leq b$ for all $x,y\in A_1$. But this is saying exactly that $A_1$ is bounded (and that $b$ is a bound).
Now the diameter is the infimum of all bounds, so if we let $d_i$ be the diameter of $A_i$ then 

$d_i\leq b$ for all bounds $b$ of $A_i$ ($i=1,2$). 

Since $d_2$ is itself a bound of $A_2$ it is automatically a bound of $A_1$ (as noted in the first paragraph). Therefore, taking $b=d_2$ and $i=1$ we get $d_1\leq d_2$.
It's worth thinking about why it was assumed that $A_1$ is not empty, and convincing yourself that the diameter is itself a bound.
A: As a first step you should consult the definition of diameter in order to actually look at what it is you need to prove. The statement about the diameters is just the pretty packaging. To prove anything you have to look inside. 
So, following the definition of diameter: $\Delta(A)=\bigvee \{d(x,y)\mid x,y\in A\}$ is the diameter of any subset $A\subseteq X$. This includes the case $A=\emptyset $ as well as any set whose diameter is infinite. There is really no need to exclude any of these options (though some books do). So, what you need to show is that if $A_1\subseteq A_2\subseteq X$, then $\Delta(A)=\bigvee \{d(x,y)\mid x,y\in A_1\}\le \bigvee \{d(x,y)\mid x,y\in A_2\}$. Now if you look at it long enough you should note that the join (another word for supremum) on the left is computed over a subset of values the join on the right is computed over. In other words, if you write $S_1=\{d(x,y)\mid x,y\in A_1\}$ and similarly for $A_2$, then $\Delta(A_1)=\bigvee S_1$ and $\Delta(A_2)=\bigvee S_2$. The key is noting that $S_1 \subseteq S_2$ (which follows since $A_1\subseteq A_2$). So, it all boils down to a general property of computing joins: Whenever $S_1\subseteq S_2$ (and these are just subsets of the set $[0,\infty ]$ of the (extended) non-negative reals) it follows that $\bigvee S_1 \le \bigvee S_2$. Stated differently, taking joins is a monotone operation. The whole exercise boils down to that fact.  
A: Its a matter of using the definitions correctly and setting them up right.

$A_2$ is bounded.

By definition that means that there exist a $r_2 \in \mathbb R^+$ so that for any two $x,y\in A_2$ that $d(x,y) \le r_2$.

$A_1 \subset A_2$ means any $w\in A_1$ will have $x\in A_2$.

So proof that $A_1$ is bound using $A_1 \subset A_2$ and $A_2$ is bounded.
That is

Prove there is an $r_1$ so that for every $w,u\in A_1$ we have $d(w,u)<d_1$ given that for every $z \in A_1$ then $z_\in A_2$ and given that for any $x,y\in A_2$ that $d(x,y)< r_2$.

Well, that should be obvious if we put it that way:
For every $w,u\in A_1$ then $w,u\in A_2$ and so $d(w,u) < r_2$.
So $A_1$ is bounded.
Now for a bounded set, $A$ there is an $r \in \mathbb^+$ so that for any $x,y\in A$ thatn $d(x,y)\le r$. So in other words,  the set of $D_A= \{d(x,y)|x,y\in A\}\subset R$ is bounded above and so

Definition: If $A$ is bounded then the diameter, $d_A$ of $A$ is $\sup D_A=\sum\{d(x,y)|x,y\in A\}$.  Every (non-empty) bounded set has a diameter.

So proving $d_{A_1} \le d_{A_2}$....
Consider the sets $D_{A_1} = \{d(x,y)|x,y\in A_1\}$ and $D_{A_2} = \{d(x,y)|x,y\in A_2}$.
If $x,y \in A_1 \subset A_2$ then $x,y\in A_2$ and if $m=d(x,y) \in D_{A_1}$ then $m=d(x,y)\in D_{A_2}$ so $D_{A_1} \subset D_{A_2}$.
Have you proven that if $M \subset N$ and $N$ is bounded above then $M$ is also bounded above and $\sup M \le \sup N$?
From that it follows.  $d_{A_1} = \sup D_{A_1}\le \sup D_{A_2} = d_{A_2}$.  And we are done.
... addendum....

Prop:  If $C\subset \mathbb R$ is bounded above and $A \subset C$ then $A$ is bounded above and $\sup A \le \sup C$.
Pf:  $C$ bounded above and a real subset means $\sup C$ exist and for all $c \in C$ then $c\le \sup C$.  If $a\in A\subset C$ then $a\in C$ so $a\le \sup C$ so $A$ is bounded above by $\sup C$ and $\sup A$ exist.
Now if $\sup A > \sup C$ then $\sup C$ wouldn't be an upper bound of $A$ which we argued it must be.  So $\sup A \le \sup C$.

A: Hint: It is crucial to note that for a function $f: \Omega \rightarrow \mathbb{R}$ and $A \subseteq B \subseteq \Omega$ we have $\sup_{x \in A} f(x) \leq \sup_{x \in B} f(x)$ (assuming both suprema exist).
