Unions of bounded subsets of a metric space

Question

Let $(X,d)$ be a metric space, $A,B \subset X$. I'm using the definiton of diameter of a set

$\delta(A)=sup_{x,y \in A}d(x,y)$

So, $A$ is bounded if $\delta(A)$ is finite.

My attempt.

Let $A$ and $B$, two bounded subset of $X$, so

$\delta(A)=sup_{x,y \in A}d(x,y)<k_1$ and $\delta(B)=sup_{x,y \in B}d(x,y)<k_2$, $k_1,k_2 \in \mathbb{R}$

Fixing $x_1 \in A$, we have

$sup_{x_1,y \in A}d(x_1,y)<k_1$, so, $A\subset B(x_1,k_1)$

$B(x_1,k_1)$ is the open ball centered in $x_1$ with radius $k_1$. Proceeding the same way, we have

$sup_{x_2,y \in B}d(x_2,y)<k_2$, so, $B\subset B(x_2,k_2)$

Take $k=k_1+k_2$, we have

$A \cup B \subset B(x,k)$, for any $x$

$sup_{x,y \in A \cup B} d(x,y)< k $, so $\delta(A\cup B)<k$.

Is this valid?

Please don't rely on the title of your MSE questions for important information (such as the statement of the problem). Please make the body of your questions self-contained. As for the mathematics, the diameter of $A \cup B$ can be much bigger than the sum of the diameters of $A$ and $B$, so your argument can't be right. Try thinking about an example like two intervals in $\Bbb{R}$. — Rob Arthan, Aug 08 '23 at 19:53
Why this rule? Bounded sets in metric space would be better? — ends7, Aug 08 '23 at 20:01
Are you asking about making the body of the question self-contained? If so, it's because people often read the body without looking at the title, and there have been (and maybe still are) interfaces to MSE that don't display the title properly. My advice is to write the body of the question first and then write the title as a summary of what you have written. (And, yes, something like "Unions of bounded subsets of a metric space" would be a good title for your question.) — Rob Arthan, Aug 08 '23 at 20:12

score 4 · Answer 1 · answered Aug 08 '23 at 20:10

The $\sup$s don't really help. If $A$ is not bounded, $\sup_{x, y \in A} d(x, y)$ is not defined. The usual definition of a bounded subset of a metric space says that $A$ is bounded if there is $r \in \Bbb{R}$ such that for every $x, y \in A$, $d(x, y) < r$.

So if $A$ and $B$ are both bounded, we have $r, s \in \Bbb{R}$ such that $d(x, y) < r$ whenever $x, y \in A$ and $d(x, y) < s$ whenever $x, y \in B$. Now choose $a \in A$ and $b \in B$ and let $t = r + s + d(a, b)$. Then for $x, y \in A \cup B$, we must have $d(x, y) < t$ (trivially if $a, b$ are both in $A$ or both in $B$, and using the triangle inequality twice if one is in $A$ and the other is in $B$).

The above argument makes a tacit assumption that $A$ and $B$ are both non-empty. I'll leave it to you to make the necessary adjustments to deal with the case when either or both of them is empty.

score 3 · Accepted Answer · answered Aug 08 '23 at 19:55

This is wrong consider $(\mathbb{R}, | \cdot |)$ and $(0, 1)$, $(5, 6)$. Then $(0, 1) \subset B\left(\frac{1}{2}, 1\right)$ and $(5, 6) \subset B\left(\frac{11}{2}, 1\right)$ but $\not \exists x \in (0, 1)\cup(5, 6)$ such that $(0, 1)\cup (5, 6) \subset B(x, 2)$.