# Prove that if $X_{n}\cap X_{n+1}\neq\emptyset$, $X=\bigcup X_{n}$ is connected

Let $$X_{1},\dots X_{m}$$ a finite collection of subsets of a metric space $$M$$ such that $$X_{n}\cap X_{n+1}\neq \emptyset$$ for all $$n$$. Show that $$X=\cup X_{n}$$ is connected.

How can I prove that?

I tried to do the contrapositive: If $$X=\cup X_{n}$$ is not connected, exists $$1\leq n < m$$ such that $$X_{n}\cap X_{n+1}=\emptyset$$, but I didn't work. Can someone give me any some tips?

• Each $X_n$ should be connected. – eloiprime Oct 13 '18 at 2:37
• Maybe induction... – Eduardo Longa Oct 13 '18 at 2:51

As mentioned, the result is false unless each space $$X_n$$ is connected so let's assume so.

Here is an alternative approach: a metric space $$(Y,d)$$ is connected if and only if any continuous function $$Y \to \{0,1\}$$ is constant. If you had a non-constant function, preimages of $$0$$ and $$1$$ would disconnect $$Y$$. Reciprocally, if $$Y$$ can be disconnected, sending one open set to $$0$$ and the other one to $$1$$ gives a continuous function.

Let $$z_n \in X_n \cap X_{n+1}$$ for each $$n$$. To show that $$\bigcup_n X_n$$ is connected, let $$\varphi: \bigcup_n X_n \to \{0,1\}$$ continuous and let's see that $$\varphi$$ is constant. Since each $$X_n$$ is connected, the restriction $$\varphi_n = \varphi|_{X_n} : X_n \to \{0,1\}$$ ought to be constant, i.e. $$\varphi_n \equiv c_n \in \{0,1\}$$. Now, it suffices to see that $$c_n \equiv c_1$$ for all $$n$$. We proceed by induction: the base case is trivial. As for the inductive step,

$$c_{n+1} = \varphi_{n+1}(z_n) = \varphi_n(z_n) = c_n \stackrel{I.H.}{=} c_1,$$

and so we are done.

• Nice criative answer! Thank you – Mateus Rocha Oct 13 '18 at 14:26

As stated in the comments, each $$X_n$$ must be connected. For example, if $$M=X_1$$ is the discrete space of two points, then $$X=X_1$$ is not connected.

Suppose that $$X$$ is not connected. Then there exist disjoint nonempty open subsets $$U$$ and $$V$$ of $$X$$ which cover $$X$$. Since each $$X_n$$ is connected, we have, for each $$n$$, $$\mbox{(X_n\subseteq U and X_n\cap V=\emptyset) or (X_n\subseteq V and X_n\cap U=\emptyset)}.$$ Furthermore, since $$U$$ and $$V$$ are both nonempty, we have $$m>1$$. Now use the assumption $$\mbox{X_n\cap X_{n+1}\ne\emptyset for all n}$$ to contradict our choice of $$U$$ and $$V$$.