Let $A$ and $B$ be $2$ closed subsets of a metric space $E$. Suppose $A \cap B$ and $A \cup B$ are path connected. Show that $A$ and $B$ are path connected.

Proof: it suffices to show that A is path connected.

Suppose $a,b \in A \Rightarrow a,b \in A \cup B$. As $A \cup B$ is path connected, there exists a path $\phi: [0,1] \to A \cup B$ such that $\phi(0) = a$ and $\phi(1) = b$.

If $\phi([0,1]) \cap B = \emptyset$, then we are done.

If not, there exist: $$t_o = \inf\{t \in [0,1]: \phi(t) \in B \}$$ $$t_1 = \sup\{t \in [0,1]: \phi(t) \in B \}$$ As $\phi$ is continuous, $A$ and $B$ are closed, we have $\phi(t_0), \phi(t_1) \in A \cap B$.

As $A \cap B$ is path connected, there exists a path $f:[t_0,t_1] \to A \cap B$ such that: $f(t_0) = \phi(t_0)$ and $f(t_1) = \phi(t_1)$.

Now, let $g : [0,1] \to A$, defined as: $$g(t) = \begin{cases} f(t), & t \in [t_0,t_1] \\ \phi(t), & \text{otherwise} \end{cases}$$

Then $g$ is continuous and $g(0) = \phi(0) = a$, $g(1) = \phi(1) = b$. Done.

There is one point that I'm not clear. Let $E = [0,1] \cap \phi^{-1}(B)$. Then the proof implicitly says that $E = [t_0,t_1]$, that would mean $\phi^{-1}(B)$ is connected a priori.

$A \cap B$ and $A \cup B$ are path connected, thus connected, thus $A$ and $B$ are connected. But why $\phi^{-1}(B)$ is connected?

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    $\begingroup$ No $E$ need not be $[t_0,t_1]$ but just contained in! the path can go in and out of $B$ any number of time it pleases! We are just interested to know when it first entered and when it left for good! $\endgroup$ – baharampuri Jan 1 '16 at 21:59
  • $\begingroup$ I don't understand. If $A\cup B$ is path-connected, isn't it trivial that $A$ is path-connected? $\endgroup$ – vhspdfg Jan 1 '16 at 22:40
  • $\begingroup$ No, $\phi([0,1])$ is included in $A \cup B$ but may be not in $A$ $\endgroup$ – SiXUlm Jan 1 '16 at 22:58
  • $\begingroup$ @vhspdfg Take $A=[0,1]\cup [2,3]$ and $B=[1,2]$ - two closed subsets of the real line whose union is path connected (the intersection isn't of course). $\endgroup$ – Mark Bennet Jan 1 '16 at 23:17
  • $\begingroup$ Oh, how daft of me. Can't think today. $\endgroup$ – vhspdfg Jan 1 '16 at 23:18

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