Prove that $X_1\cup X_2$ is path connected if and only if both $X_1$ and $X_2$ are path connected

Suppose $X_1\subset [0,\infty)\times\mathbb{R}$ and $X_2\subset (-\infty,0]\times\mathbb{R}$ such that $X_1\cap X_2=\{0\}\times[0,1]$. Prove that $X_1\cup X_2$ is path connected if and only if both $X_1$ and $X_2$ are path connected and non-disjoint.

This was a question that was posed on a previous exam, and I know that it is FALSE after talking to several people. (There is a counter example in the comments.) I know that the statement SHOULD say something about being compact, but I am not sure where it is needed. Given the above statement I know how to do if $X_1$ and $X_2$ are path connected then $X_1\cup X_2$ is path connected, but not the other direction.

I am studying for the next exam and any help about what the statement SHOULD say, and how to prove it would be helpful.

Solution 1:

The problematic part is to show that if $X_1 \cup X_2$ is path-connected then $X_1$ and $X_2$ are also path-connected. As examples given in the comments show, this is not true in general, but it is true, if we assume that $X_1$ and $X_2$ are both closed. I'll sketch a proof leaving you to fill in some of the details. Let me know if it's unclear.

Assume that $X_1$ and $X_2$ are closed subsets of any topological space, that $X_1 \cup X_2$ is pathi-connected and that $X_1 \cap X_2$ is non-empty and path-connected. We want to show that $X_1$ and $X_2$ are both path-connected. By symmetry, it is enough to show that $X_1$ is path-connected.

Claim: any path $\alpha : [0, 1] \to X_1 \cup X_2$ from a point $a \in X_1$ to a point $b \in X_2$ has an initial segment contained in $X_1$ that ends in $X_1 \cap X_2$.

To see this, note that if $a$ or $b$ is in $X_1 \cap X_2$, the claim is clear, so I may assume that $a \not\in X_2 $ and $b \not\in X_1$. Now consider the set $T \subseteq [0, 1]$ comprising all $t$ such that for all $s < t$, $\alpha(s) \in X_1 \setminus X_2$ . Let $u = \sup\,T$. Then $\alpha(u) \in X_1$, because $X_1$ is closed. Moreover, every neighbourhood of $\alpha(u)$ contains a point in $X_2$. So as $X_2$ is closed $\alpha(u) \in X_2$. So the initial segment $\alpha[T]$ of our path $\alpha$ is contained in $X_1$ and ends in $X_1\cap X_2$, as claimed.

Now given $a, b \in X_1$, we need to find a path from $a$ to $b$. To do this, pick $c \in X_2$ and choose paths $\alpha$ from $a$ to $c$ and $\beta$ from $c$ to $b$ (as we may since $X_1 \cup X_2$ is path-connected). Using the claim, we get paths contained in $X_1$ from $a$ and $b$ to points in $X_1 \cap X_2$, and, as the latter is path-connected, that gives us a path contained in $X_1$ from $a$ to $b$.