Criterion for connectedness

**Sheld** · March 24th 2011, 06:02 PM

In my book they go through this proof of the following:

An open set $G \subset \mathbb{C}$ is connected iff for any two points $a,b$ in $G, \; \exists$ a polygon from $a$ to $b$ lying entirely in $G$ .

We suppose $G$ satisfies this condition and $G$ is not connected to obtain a contradiction.

Pf:
Know $G = A \cup B$ where $A,B$ are both open and closed, $A \cap B = \emptyset$ and $A,B$ are nonempty.

If we let $a \in A$ and $b \in B$

$\exists$ a polygon from $a$ to $b$ such that $P \subset G$

We can assume $P = [a,b]$

Define:

$S = {s \in [0,1] : sb + (1-s)a \in A }$
$T = {t \in [0,1] : tb +(1-t)a \in B}$

$S \cap T = \emptyset, S \cup T = [0,1] , 0 \in S, 1 \in T$

It can be shown that $S$ and $T$ are both open, contradicting the connectedness of $[0,1]$ .

-----------------------------------------------------

Now, I do not see how S and T are both open. Can anyone give me a hint please? Thank you.

-Sheld

**Drexel28** · March 24th 2011, 07:19 PM

Originally Posted by Sheld

In my book they go through this proof of the following:

An open set $G \subset \mathbb{C}$ is connected iff for any two points $a,b$ in $G, \; \exists$ a polygon from $a$ to $b$ lying entirely in $G$ .

We suppose $G$ satisfies this condition and $G$ is not connected to obtain a contradiction.

Pf:
Know $G = A \cup B$ where $A,B$ are both open and closed, $A \cap B = \emptyset$ and $A,B$ are nonempty.

If we let $a \in A$ and $b \in B$

$\exists$ a polygon from $a$ to $b$ such that $P \subset G$

We can assume $P = [a,b]$

Define:

$S = {s \in [0,1] : sb + (1-s)a \in A }$
$T = {t \in [0,1] : tb +(1-t)a \in B}$

$S \cap T = \emptyset, S \cup T = [0,1] , 0 \in S, 1 \in T$

It can be shown that $S$ and $T$ are both open, contradicting the connectedness of $[0,1]$ .

-----------------------------------------------------

Now, I do not see how S and T are both open. Can anyone give me a hint please? Thank you.

-Sheld

Haha, that is the hard part of the problem if I'm interpreting your question correctly. Here's a discussion of the problem.

**Sheld** · March 25th 2011, 09:24 AM

Thanks for the link. I think that post thought is talking about the other direction of this iff . I was just wondering how do I show that the sets S and T, as described above, are open?

**Sheld** · March 25th 2011, 11:22 AM

I think I have made progress!

$S \subset A$

Since $A$ is open,

$\forall s \in S, \exists B(s,\epsilon) \subset A$

So if we make epsilon sufficiently small, $B(s,\epsilon) \subset S$

however I'm not sure how small $\epsilon$ needs to be? Do I need to know? If we keep making the ball smaller and smaller, we end up with points only in $S$ , right?

**Opalg** · March 25th 2011, 01:09 PM

Originally Posted by Sheld

I think I have made progress!

$S \subset A$

Since $A$ is open,

$\forall s \in S, \exists B(s,\epsilon) \subset A$

So if we make epsilon sufficiently small, $B(s,\epsilon) \subset S$

however I'm not sure how small $\epsilon$ needs to be? Do I need to know? If we keep making the ball smaller and smaller, we end up with points only in $S$ , right?

I think you probably have the right idea there, although the statement $B(s,\epsilon) \subset A$ does not actually make sense. (In fact, $s\in[0,1]$ , whereas A is a subset of G.)

The point is that the map $f:[0,1]\to G$ defined by $f(s) = sb+(1-s)a$ is continuous, and the set A is open. Therefore the set $S = f^{-1}(A)$ is also open.

**Sheld** · March 25th 2011, 02:20 PM

You are right, my proof does make zero sense.

I can see how that function is continuous. The metric of G is absolute value? G is a subset of the complex numbers. And that theroem on continuity is very handy.

However since this is several chapters before continuity, I wonder if this can be proved without such heavy machinery.

Thank you very much for helping.

**Opalg** · March 26th 2011, 07:29 AM

Originally Posted by Sheld

I can see how that function is continuous. The metric of G is absolute value? G is a subset of the complex numbers. And that theroem on continuity is very handy.

However since this is several chapters before continuity, I wonder if this can be proved without such heavy machinery.

If you don't want to use the heavy machinery, you can get the same result just using epsilons.

Suppose you have a point $x_0 = s_0b+(1-s_0)a$ in the segment [a,b], with $s_0\in S$ so that $x_0\in A$ . We know that A is open, so there exists $\varepsilon>0$ such that if $|x-x_0| < \varepsilon$ then $x\in A$ .

If $|s-s_0| < \varepsilon/|b-a|$ then $\bigl|(sb+(1-s)a) - (s_0b+(1-s_0)a)\bigr| = |(s-s_0)(b-a)| < \varepsilon$ . Therefore $sb+(1-s)a \in A$ and so $s\in S$ . Hence S is open.

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