# Union of an increasing sequence on connected sets

Let $$(A_i)$$ be a family of strictly increasing connected sets, then I have to prove that $$A=\bigcup_{i\in I}A_i$$ is connected.

By contradiction, I assume that $$A$$ is not connected. Then there exist two open, non empty and disjoint sets $$U$$ and $$V$$ such that $$A= U\cup V,$$

as $$A_i$$ is connected, we have $$\forall i\in I, [A_i\subset U~\text{or}~A_i\subset V]$$

How to prove that $$\forall i\in I, A_i\subset U~\text{or}~\forall i\in I, A_i\subset V$$

to get a contradiction?

Thank you.

• If the statement you want to prove is not true, then the sets $A_i$ will not be an increasing sequence of sets, that is easy to show - again by contradiction. – Kaind Sep 20 '18 at 19:39
• Is $I$ a directed set? Or in what sense are the $A_i$ increasing? – Guido A. Sep 20 '18 at 19:44

## 2 Answers

Recall that a topological space $$D$$ is disconnected if and only if there exists a non-constant continuous function $$\phi: D \to \{0,1\}$$.

Now we can give a proof by the contrapositive as you want: if $$A$$ is disconnected, some $$A_i$$ must be disconnected.

Concretely, let $$\phi : A \to \{0,1\}$$ be continuous and non-constant, i.e. suppose that $$A$$ is disconnected. Then there exist $$a,b \in A$$ such that $$\phi(a) = 0$$ and $$\phi(b) =1$$. Since $$A = \bigcup_{i \in I}A_i$$, there exist $$i,j \in I$$ such that $$a \in A_i$$ and $$b \in A_j$$. Suppose without loss of generality that $$i \leq j$$, since the other case is symmetrical. Now, since $$(A_i)_{i \in I}$$ is an increasing family, $$A_i \subseteq A_j$$ and so $$a,b \in A_j$$. Thus, the restriction

$$\phi |_{A_j} : A_j \to \{0,1\}$$

is continuous and non-constant, which proves that $$A_j$$ is disconnected.

@Ashish's comment I think makes it easier to prove. If you have $$\forall i \in I, [A_i \subset U \text{ or } A_i \subset V]$$ and we know that $$\exists i , A_i \subset U$$ and $$\exists j, A_j \subset V$$ (because otherwise $$U$$ or $$V$$ would be empty), then your sets can't be increasing $$A_i \not\subset A_j$$ and $$A_j \not\subset A_i$$.

P.S. I assume that by increasing you mean that $$\forall i, A_i \subset A_{i+1}$$.