Prove that if S and T are domains that have at least one point in common, then S union T is also a domain
I wrote: A domain is a set that is open and connected. The union of open sets is easily open.
To prove that S∪T is connected when S∩T is nonemtpy, let A and B be any two distinct points of S∪T and let C be a point of S∩T.
Consider point A. If it is not point C and is in set S, then since A and C are in S and S is a domain, there is a patch entirely in S connecting A and C. Since all points of that path are in S they are also in S U T. Likewise, if A in not C but is an element of T, then there is path in S U T connecting A and C.
So, point A is either the same point as C, or is connected to C by a path entirely in S∪T.
By the same reasoning, point B is either C or connected to C by a path entirely in S∪T.
Since A and B are distinct, at most one of them is C. So either A and B are directly connected by a path in S∪T (because one of them is C) or are connected by the two-part patch from A to C then from C to B, with both parts in S∪T.
Since any two points in S∪T are connected by a path in S∪T, then S∪T is connected.
My professor commented that I need to prove SuT is open, but I don't know what else to include that would show that.
Best Answer
You're right about the union of the open sets being open.
Now, the "non-empty" part is easy too.
For the connectendness, I suggest that you go from the definition: assume that the union is disconnected. Then there are two non-empty open sets $U$ and $V$ that have no intersection and that split your final domain in two: $(S\cup T)\subseteq (U\cup V)$. Now, show that it is impossible: $S$ is completely inside $U$ or $V$. Say, it is in $U$. Then $(S\cap T)\subseteq U$ too. But then, since $T$ is connected, it is entirely inside $U$ too. So, $V$ is empty, which contradicts the definition of non-connectedness.