Union of path connected pairwise not disjoint subsets

Question

Problem

Let $(X,d)$ be a metric space and let $\mathcal A$ be a family of path connected subsets of $X$ such that for every pair of sets $A,B \in \mathcal A$ there are $A_0,...,A_n \in \mathcal A$ with $A_0=A$ and $A_n=B$, and $A_i \cap A_{i+1}\neq \emptyset$ for each $i=0,...,n-1$. Show that $\bigcup_{A \in \mathcal A} A$ is path connected.

What I've tried to do was the following:

Pick $x \neq y$ in the union. We want to show that these two points can be connected by a path (a continuous function $f:[0,1] \to \bigcup A$ with $f(0)=x$ and $f(1)=y$). There are two possible cases:

1)If $x$ and $y$ lie in the same set, then that these two points are path connected follows from the hypothesis of the problem.

2)If $x \in A$ and $y \in B$ with $A \neq B$, there are $A_0,...,A_n$ sets of the family with $A_0=A$ and $A_n=B$, and $A_i \cap A_{i+1} \neq \emptyset$ for $i=0,...,n-1$.

Take $x_i \in A_i \cap A_{i+1}$ for $i=0,...,n-1$. There is a continuous function $f_0:[0,1] \to \ A_0$ with $f(0)=x$ and $f(1)=x_0$, for $0<i<n$, there is a continuous function $f_i:[0,1] \to A_i$ with $f_i(0)=x_{i-1}$ and $f_i(1)=x_i$. For $i=n$, we have $f_n:[0,1] \to A_n$ continuous with $f_n(0)=x_{n-1}$ and $f_n(1)=y$.

I don't know what to do in order to get a path from $x$ to $y$ using these functions. Some linear combination of the constructed functions is not necessarily going to work since the union $\cup_{i=0,...,n} A_i$ is not a vector space.

Any help would be greatly appreciated. Thanks in advance.

Answer 1

The basic construction of "adding" two paths: suppose $f: [0,1] \rightarrow X$ is a path from $a$ to $b$, and $g: [0,1] \rightarrow X$ is a path from $b$ to $c$. So these are continuous maps with $f(0) = a, f(1) = b, g(0) = b, g(1) = c$.

Then define $h: [0,1] \rightarrow X$ by $h(t) = f(2t)$ when $0 \le t \le \frac{1}{2}$ and $h(t) = g(2t-1)$ for $\frac{1}{2} \le t \le 1$.

Then $h$ is well-defined, as $h(\frac{1}{2}) = f(1) = g(0) = b$, so the definitions agree on the overlap. And as $h$ is defined on two closed subsets, and is continuous on the separate parts (as composition of continuous functions), $h$ is continuous, and $h(0) = f(0) = a$ and $h(1) = g(1) = c$, so $h$ is a path from $a$ to $c$, which we can denote by $f \ast g$ (first $f$ then $g$).

Using this idea it's easy to show that the statement holds for two sets, and then we can apply induction (which is easiest here) to do the finite case.

Answer 2

Suppose $x\in A_i$ and $y\in A_j$. We may assume without loss of generality that $i<j$. Choose points $x_k\in A_k\cap A_{k+1}$ for $i\leq k<j$. Concatenate paths from $x$ to $x_i$, $x_i$ to $x_{i+1}$, etc., and finally from $x_{j-1}$ to $y$. Since this is probably for an assignment, you may want to phrase this as a proof by induction on $j-i$.