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 $\displaystyle \color{red}\overline{A_1}\cap A_2 = \overline{A_2}\cap A_1 = \emptyset.$

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.