1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Let $f_i$ adjoin $f_{i-1}$. $\endgroup$ – Qiyu Wen Jun 21 '16 at 1:49
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $j-i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.