Diameter of metric spaces

Question

Let $(X,d)$ be a metric space. The diameter of a set $A\subset X$ is defined to be

diam$(A)=$sup{$d(x,y):x,y\in$ A}

(b)suppose $A_1,...A_n$ is a finite collection of subsets of $X$ each with finite diameter. Prove that $\cup_{i=1}^n A_i$ has finite diameter.

For this question, does the collection of $A_1,...A_n$ need to be disjoint? It seems that the collection of $A_1,...A_n$ shouldn't be disjoint from part (c). I think that two subsets of a metric space cannot be infinitely far away from each other. For any $x,y$ in the space, $d(x,y)$ is always finite but just don't know how to prove this.

(c)Prove that the union of $A_\alpha$ has finite diameter if the intersect of $A_\alpha$ if a non-empty set and there exists a constant $M$ such that diam($A_\alpha$)$\leq M$ for all $\alpha$.

For this part, my idea is to prove the union of $A_\alpha$ has finite diameter, I need to show that diam($A_\alpha$) has a least upper bound $M$ and the collection of $A_i$ isn't disjoint.

Answer 1

Yes, it's the "maximum distance" between any two points in the set, except it's a $\sup$ -- there might be no maximum. To see this, just look at the sets $[0,1]$ and $(0,1) \subset \mathbb{R}$. The diameter of both is $1$, although the latter set has no max distance.

For (a), do not show that the boundary is an empty set. It could be nonempty, like in the example I said. The diameter of $\bar{A}$ could be bigger than that of $A$, since you may have added points. However, the points you added had to be limit points of the old set, by definition of closure. You can use the definition of limit point to show that the diameter didn't change when you added in those points.

Good luck!

Answer 2

we can prove (b) with a mathematical induction. when $n$ is 2, get $a_1$ and $a_2$ from $A_1$ and $A_2$. now when we are finding $sup\{d(x,y): x, y \in A\}$ if both of $x$ and $y$ be from one of the $A_1$ or $A_2$ then the calculated distance is'nt more than maximum of diameter of $A_1$ and $A_2$, and when $x$ and $y$ are'nt in a same set like when $x$ is from $A_1$ and $y$ is from $A_2$ then we have from triangle inequality: $$d(x,y) \leq d(x,a_1)+d(a_1,a_2)+d(a_2,y)\leq diam (A_1)+d(a_1,a_2)+diam(A_2)$$ thus $d(x,y)$ is less than a fixed value and the calculated suprimum is less than this fix value.

for proving (c) it is enoght that we set a fix element in the intersection of all sets and transfer throght it with using triangle inequlity for bounding distants of two arbitrary point from the above.