1. ## complex analysis

Suppose A and B are nonempty closed bounded subsets of $\mathbb{C}$.
1. Show that there exist $a_0 \in A, b_0 \in B$ such that $|a_0-b_0|=\inf\{|a-b|:a\in A, b\in B\}$.
2. Show that this is not true if the boundedness condition on $A$ and $B$ is dropped.
3. What if only one of $A$ or $B$ is bounded but both are still closed?

Question about #1, if $A, B$ are closed and bounded, does this mean every sequence in $A, B$ is bounded?

You have a bounded set A, u take a sequence there, say $\{ x_n \}_n \subset A$.

A is bounded means that there is a bound M such that:

$|y| \leq M$ $\forall y \in A$

even for those elements in your sequence so

$|x_n| \leq M$ $\forall n \in \mathbb N$

You dont even need the sets to be closed for this.

3. About #2 think of this pair of sets:

$
A = \{x+iy \in \mathbb C : y\geq \frac{1}{x}$
for $x>0 \}$

and take $B=\{x+iy : y\leq 0 \}$

Both sets are closed but unbounded. And this
$|a_0-b_0|=\inf\{|a-b|:a\in A, b\in B\}
$
does not hold.

If by any chance why it does not hold remains unclear, draw both sets and you will see.

4. #3 Must be true. I think #1 proof should work with a slight modification.

5. I am stuck in #3. For #1, I used the fact that both $A,B$ are bounded so that I have two bounded sequences, and then apply the Bolzano-Weierstrass Theorem.
How can I prove that it is true with only $A$ or $B$ bounded?

6. Originally Posted by mabruka

You have a bounded set A, u take a sequence there, say $\{ x_n \}_n \subset A$.

A is bounded means that there is a bound M such that:

$|y| \leq M$ $\forall y \in A$

even for those elements in your sequence so

$|x_n| \leq M$ $\forall n \in \mathbb N$

You dont even need the sets to be closed for this.
You do! What would $a_0, b_0$ be if the two sets were the open disc $|z|<1$ and the open disc $|z-2|<1$?

7. Originally Posted by Bruno J.
You do! What would $a_0, b_0$ be if the two sets were the open disc $|z|<1$ and the open disc $|z-2|<1$?
Every sequence in any of those sets is bounded of course!

Of course you need them to be closed for the existence of $a_0, b_0$.