The union of a finite sequence of connected sets that meet pairwise is connected Let $X_1,\ldots,X_n$ be a sequence of connected sets s.t. $X_i\cap X_{i+1}\ne\emptyset$. The goal is to show that $\bigcup_{i=1}^n X_i$ is connected.
There is a theorem that says that if the intersection of a family of connected sets is nonempty, then their union is connected. I have to use this result to prove the above by induction.
The case of $n=2$ is trivial. Suppose it holds for $n$. It would be sufficient to show that $\bigcap_{i=1}^{n+1}X_i\ne\emptyset$. But how to deduce this?
For example take $n=3$. Then $X_1\cap X_2\ne\emptyset$ and $X_2\cap X_3\ne\emptyset$. Why would $X_1\cap X_2\cap X_2$ have to be nonempty? One strategy would be to show that empty intersection would force one of the $X_i$ to be disconnected. For this $X_2$ might seem like a good candidate. Obviously $X_1\cap X_2\subset X_2$ and $X_2\cap X_3\subset X_2$. But neither $X_2\cap(X_1\cup X_3)=X_2$ nor are $X_1\cap X_2$, $X_2\cap X_3$ open sets of $X_2$. What to do?
 A: Proof by induction base case is trivial assume its true for $k$ such sets WTS true for $k+1$ sets.
Let $A= X_1 \cup X_2 $ we have that A is a connected set and by above definition we now have that $ A \cap X_3 \neq \emptyset $ so we write out $ A \cup X_3 \cup ... \cup \space X_{k+1} $ is a collection of k pairwise connected sets, hence by induction the case holds for all $n \in \Bbb N $
A: Let $Y_n=\cup_{i=1}^nX_i.$
$Y_1$ is connected.
If $Y_n$ is connected then, because $X_{n+1}$ is connected and because $Y_n\cap X_{n+1}\supseteq X_n\cap X_{n+1}\ne \emptyset,$ the 2-set case implies $Y_n\cup X_{n+1}$ is connected. And of course $Y_n\cup X_{n+1}=Y_{n+1}.$
This is a common method for reducing an induction on $n$ to the case $n=2.$
Remark. As a corollary, if $\{X_n: n\in \Bbb N\}$ is a family of connected sets with $X_n\cap X_{n+1}\ne \emptyset$ for each $n\in \Bbb N$ then $Y=\cup_{n\in \Bbb N}X_n$ is connected:
By contradiction, suppose $\{A,B\}$ is a disconnection of $Y.$ That is, $A,B$ are open sets with $A\cup B\supseteq Y$ and $(A\cap Y)\cap (B\cap Y)=\emptyset$ but $A\cap Y\ne \emptyset \ne B\cap Y.$ Take $a\in A\cap Y$ and $b\in B\cap Y.$ There exists $n\in \Bbb N $ such that $\{a,b\}\subseteq \cup_{i=1}^nX_n,$ but this implies that $\{A,B\}$ is a disconnection of $\cup_{i=1}^nX_n.$
