The question is as follows:
$A \subset B$, and $B$ is bounded in $n$-space
Show: the diameters, $\operatorname{diam}(A) \leq \operatorname{diam}(B)$
I came up with the following 2 attempts, but I can't proceed further.
Attempt 1: Proof by contradiction
1/ By definition of diameter:
$$\operatorname{diam}(A) = \sup\{\operatorname{dist}(a,b) \mid a,b \in A\}$$
$$\operatorname{diam}(B) = \sup\{\operatorname{dist}(c,d) \mid c,d \in B\}$$
2/ Suppose $\operatorname{diam}(A) > \operatorname{diam}(B)$,
then I claim there is a $(x,y)$ [where $x,y \in A$] s.t. $\operatorname{dist}(x,y) > \sup\{\operatorname{dist}(c,d) \mid c,d \in B\}$.
Thus, I claim we get a contradiction to the fact that $A$ is strictly contained in $B$.
But… is this idea OK?
Attempt 2: Direct proof
Since $A \subset B$ and $B$ is bounded
$$\implies \sup(A) \leq \sup(B) \quad\text{and}\quad \inf(A) \leq inf(B)$$
Thus, if I take the distance, the "biggest" distance of $A$ should be the distance between $\sup(A)$ and $\inf(A)$, which I claim is less than or equal to distance between $\sup(B)$ and $\inf(B)$. Thus $\operatorname{dist}(A) \leq \operatorname{dist}(B)$.
But I feel really shaky about this direct proof
Would someone please help me on this question?
Thanks in advance ^_^
Best Answer
Unfortunately the idea of $\sup A$ doesn't make sense if we're working in general $\mathbb R^n$. We don't have a total ordering in that space.
However, from its definition: $$ \text{diam}\,A=\sup\{d(x,y):x,y\in A\}\leq\sup\{d(x,y):x,y\in B\supset A\}=\text{diam}\,B. $$
The reason for this $\leq$ is because we're taking the supremum over a larger set, thus it must attain at least the same supremum.