complex analysis

**dori1123** · February 6th 2010, 07:57 PM

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?

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

**mabruka** · February 6th 2010, 10:30 PM

About your question of #1 the answer is yes,

think about it:

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.

**mabruka** · February 6th 2010, 10:41 PM

About #2 think of this pair of sets:

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

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

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

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

**mabruka** · February 6th 2010, 10:45 PM

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

**dori1123** · February 8th 2010, 03:54 PM

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?

**Bruno J.** · February 8th 2010, 04:13 PM

Originally Posted by mabruka

About your question of #1 the answer is yes,

think about it:

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 $|z-2|<1$ ?

**mabruka** · February 8th 2010, 04:22 PM

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!

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 $a_0, b_0$ .

Sorry about the confusion.

Thread: complex analysis

LinkBack

Thread Tools

Display

complex analysis

Similar Math Help Forum Discussions

Complex Analysis - Question on differentiability of a complex function

Complex Analysis: Proof that a straight line in Complex Plane is a connected set

Complex Analysis: find one complex number w such that w is not in the range f(z)= e^z

[SOLVED] Complex analysis, determine the set in the complex plane that satisfies...

Complex analysis: Show limit of complex function does not exist.

Search Tags