Proposition
Let {An} be a sequence of connected subspaces of X, such that An ∩ An+1 ≠ ∅ for all n.
∪An is connected.
I will show two different proofs.
Proof.
A1 ∩ A2 ≠ ∅ means that A1 and A2 has a point in common, implying A1 ∪ A2 is connected.
A2 ∩ A3 ≠ ∅, ( A1 ∪ A2 ) ∩ A3 ≠ ∅ implies A1 ∪ A2 ∪ A3 is connected.
By iterating this infinitely, we find that ∪An is connected.
....
Is this a correct proof?
In topology, we need to be careful when we so something infinitely.
For example, when we intersect an open set infinitely many times, the result might not be open.
The fact that A1 ∪ A2 ∪ ... ∪ An is connected for all n does not mean that ∪An is connected.
(As an analogy, think of a limit. f(x) < A for all x doesn't mean that lim f(x) < A beacuse lim f(x) = A can be true)
So, I will try a different method.
In the method above, we can still conclude that (A1 ∪ A2 ∪ .... ∪ An) is connected, because it is the result of finite iterations.
Now, If we set Bn = (A1 ∪ A2 ∪ .... ∪ An) , then ∪An=∪Bn.
Also, all of Bn has a point in common p, such that p∈A1.
I already showed Bn is connected.
Therefore, the arbitrary union of Bn, which is ∪Bn is connected.
Now, the proof is done.
Different proof.
This proof is a proof by reduction to absurdity.
Let's say ∪An is NOT connected.
Then there exists a seperation U, V.
A1 ⊂ ∪An and A1 is connected, so A1 ⊂ U or A1 ⊂ V.
Without loss of generality, let's say A1 ⊂ U.
Now, with An, let's say An ⊂ U.
An+1 ⊂ U or An+1 ⊂ V by the same logic with A1.
If An+1 ⊂ V, An ∩ An+1 = ∅ because U ∩ V = ∅, a contradiction.
So, An+1 ⊂ U if An ⊂ U.
By mathematical induction, An ⊂ U for all n.
This means V is empty, resulting in a contradiction.
We can conclude ∪An is connected.
증명의 오류는 댓글로 남겨주시면 감사합니다.
If there are any errors, let me know in the comments.