# Thread: complex analysis

1. ## complex analysis

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

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

Also, I'm clueless about #2 and #3. Some help please.

2. About your question of #1 the answer is yes,

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

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

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

even for those elements in your sequence so

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

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

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

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

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

Both sets are closed but unbounded. And this
$\displaystyle |a_0-b_0|=\inf\{|a-b|:a\in A, b\in B\}$ 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 $\displaystyle A,B$ 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 $\displaystyle A$ or $\displaystyle B$ bounded?

6. Originally Posted by mabruka

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

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

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

even for those elements in your sequence so

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

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

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

I was answering his question about problem #1:

Originally Posted by dori1123
Question about #1, if are closed and bounded, does this mean every sequence in is bounded?
It is enough the set be bounded so that every sequence in there is bounded.

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

Sorry about the confusion.