6

Prove that $d (A \cup B) \leq d(A)+ d(B)$, given that $ A \cap B \neq \varnothing$.

Here $d$ stands for the diameter of the set. Please note that my knowledge is limited to metric spaces only, with no knowledge of topology at all.

My attempt:

I'm trying to use the relation that $A \subset B \Rightarrow d(A) \leq d(B)$

I prove this result in the following way:

If $a,b \in A$, then $a,b \in B$, so { $d(a,b) : a,b \in A$ }$ \subset$ {$d(a,b) : a,b \in B$}

So, sup { $d(a,b) : a,b \in A$ }$ \subset$ sup {$d(a,b) : a,b \in B$}

Is this part correct?

Now, using this result:

$ A \subset A \cup B$ and $ B \subset A \cup B$

So $d(A) \leq d(A \cup B)$ and $d(B) \leq d(A \cup B)$

Adding, $d(A)+ d(B) \leq 2d(A \cup B)$

This is where I'm stuck.

Diya
  • 2,532
  • 3
  • 29
  • 59

3 Answers3

6

Consider two points $x,y$ in $A\cup B$. If they are both in $A$, then $d(x,y)\leq d(A)$. If they are both in $B$, then $d(x,y)\leq d(B)$. Otherwise, suppose $x\in A$ and $y\in B$. Let $z$ be a point in $A\cap B$. Then by the triangle inequality we have $$d(x,y)\leq d(x,z)+d(z,y)\leq d(A)+d(B)$$ Taking sups we obtain that $d(A\cup B)\leq d(A)+d(B)$.

Matt Samuel
  • 58,164
  • While very concise, this answer glosses over the one key point that I personally really needed. So, I want to supply that point:

    Let $D$ be the set of all distances between points in $A \cup B$, that is $D = {d(a,b) \mid a, b \in A \cup B}$. Then, $\text{diam }A \cup B = \sup D$. In all cases, we have $d(a,b) \leq \text{diam } A + \text{diam } B$. Therefore, $\text{diam }A + \text{diam }B$ is an upper bound for $D$. But $\sup D$ is the least upper bound of $D$, so $\sup D \leq \text{diam } A + \text{diam }B$. And, since $\sup D = \text{diam } A \cup B$, we have the desired result.

     – 1Teaches2Learn Jun 13 '21 at 05:23
0

This is not true. Consider the set $X=[0,1]\cup [2,5]$ then $diamX=5$ but $diam[0,1]$ $=$ $1$ and $diam[2,5]=3$ and their sum is less than 5.

-2

You need $A\cap B\ne \varnothing$. What happens if $A$ and $B$ are just two different points?

The more general inequality (any $A$ and $B$) should have $d(A,B)$ in it. See here for detail. In this case, $A\cap B\ne \varnothing$ implies $d(A,B)=0$.

Community
  • 1
Quang Hoang
  • 15,854