Question

Let A and B be connected sets, and A intersection B closure is non empty set ,prove that A union B is connected.

Answer 1

Answer:

Suppose A∪B is not connected. Thus, there exists a separation (U,V) of A∪B (i.e there are non-empty disjoint open sets U,V of A∪B such that U∪V=A∪B).

Since A⊆A∪B is an open set of the topological space A and V is open in A∪B, then A∩V is an open set of the space A (with the subspace topology). By the same token, A∩U is also an open set of A.

Clearly, (A∩U)∪(A∩V)=A. Hence, if A∩U and A∩V are both nonempty, then we've arrived at a pair of disjoint nonempty open sets of A whose union is A → contradiction because A is connected.

Hence, either A∩U=∅ or A∩V=∅ (but not both, since the points of A must be in one of them). Let's suppose, without loss of generality, that A∩V=∅ (and so A∩U≠∅).

Now, just like for A, we have B∩U=∅ or B∩V=∅ (but not both). If B∩V=∅, then (A∪B)∩V=(A∩V)∪(A∩V)=∅, and hence V=∅ (since V⊆A∪B)→ absurd. Thus B∩U=∅ and B∩V≠∅.

From this, A=U (if x∈A, then x∈U or x∈V, and thus x∈U; reciprocally, if x∈U then x∈A or x∈B, and so x∈A). Similarly, B=V.

Therefore, a and B are a separation of A∪B (in fact, by what we've just seen this is the only possible separation of A∪B). Now, let x∈A∩B¯¯¯¯. Since x∈A, x∉B. but B is closed and so B=B¯¯¯¯. Hence, x∈A, x∉B¯¯¯¯ → absurd.

Therefore, A∪B is connected. □