Proof that union of two connected non disjoint sets is connected

**csuMath&Compsci** · September 26th 2009, 02:34 PM

The following is an attempt at a proof which I wrote up for a homework problem for Advanced Calc. I got some help from my professor briefly the morning it was due, but I'm not quite sure if I grasped exactly what he was saying. I think he said that somehow A_(1) instersects both S_(1) and A_(2) intersects S_(2)... Sorry about the formatting. ( I have written papers in LaTeX but I dont know exactly how to format properly with this GUI editor.) Please show me what I did wrong, because I know I was definitely on the right track with this proof.

Suppose S_(1) and S_(2) are connected non disjoint sets.
Show that S_(1) U S_(2) is connected.

Proof by contradiction:
Assume S_(1) U S_(2) is disconnected.

Then there exists A_(1) and A_(2) such that

1. A_(1) U A_(2) = S_(1) U S_(2)
2. closure {A_(1)} \int A_(2) = closure {A_(2)} \int A_(1)

Case 1: A_(1) and A(2) cannot be S_(1) and S_(2) or S_(2) and S_(1)
respectively.

If either of these cases is true then 2. is violated.

We can then construct S_(1) or S_(2) from a piece of A_(1) and A_(2).
These two pieces then do not intersect by 1.
Then S_(1) or S_(2) is the union of non empty disjoint sets and hence it is disconnected.

----><----- contradiction.

So S_(1) U S_(2) is connected.

**Opalg** · September 27th 2009, 08:22 AM

Originally Posted by csuMath&Compsci

The following is an attempt at a proof which I wrote up for a homework problem for Advanced Calc. I got some help from my professor briefly the morning it was due, but I'm not quite sure if I grasped exactly what he was saying. I think he said that somehow A_(1) instersects both S_(1) and A_(2) intersects S_(2)... Sorry about the formatting. ( I have written papers in LaTeX but I dont know exactly how to format properly with this GUI editor.) Please show me what I did wrong, because I know I was definitely on the right track with this proof.

Suppose S_(1) and S_(2) are connected non disjoint sets.
Show that S_(1) U S_(2) is connected.

Proof by contradiction:
Assume S_(1) U S_(2) is disconnected.

Then there exists A_(1) and A_(2) such that

1. A_(1) U A_(2) = S_(1) U S_(2)
2. closure {A_(1)} \int A_(2) = closure {A_(2)} \int A_(1) I think you mean $\color{red}\overline{A_1}\cap A_2 = \overline{A_2}\cap A_1 = \emptyset.$

Case 1: A_(1) and A(2) cannot be S_(1) and S_(2) or S_(2) and S_(1)
respectively.

If either of these cases is true then 2. is violated.

We can then construct S_(1) or S_(2) from a piece of A_(1) and A_(2).
These two pieces then do not intersect by 1.
Then S_(1) or S_(2) is the union of non empty disjoint sets and hence it is disconnected.

----><----- contradiction.

So S_(1) U S_(2) is connected.

Your proof is along the right lines, but you haven't quite tied it together. You're assuming that $S_1\cup S_2$ is disconnected. That means that the sets $A_1$ and $A_2$ satisfy not only 1. and 2., but also

$3.\quad (S_1\cup S_2)\cap A_1\ne\emptyset \text{ and } (S_1\cup S_2)\cap A_2\ne\emptyset.$

Now you can argue as follows. It's not possible for $S_1\cap A_1$ and $S_1\cap A_2$ both to be nonempty, because $S_1$ is connected. So the whole of $S_1$ must lie in one or other of $A_1$ and $A_2$ , say $S_1\subseteq A_1$ .

Similarly, we must have $S_2\subseteq A_1$ or $S_2\subseteq A_2$ . But if $S_2\subseteq A_1$ then $(S_1\cup S_2)\subseteq A_1$ , which means that $(S_1\cup S_2)\cap A_2$ is empty, contradicting 3.

Therefore $S_2\subseteq A_2$ . But we are told that $S_1\cap S_2\ne\emptyset$ . So choose $x\in S_1\cap S_2$ . Then $x\in S_1\subseteq A_1$ , and $x\in S_2\subseteq A_2$ . Therefore $A_1\cap A_2\ne\emptyset$ . But that contradicts 2. So either way we have a contradiction.

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