Union of an increasing sequence on connected sets

Question

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.

Answer 1

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.

Answer 2

@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 $\forall i, A_i \subset A_{i+1}$.