Is there a better way - space is not path connected

Question

Let $U, V$ be nonempty, open sets of $\Bbb R^n$ such that $U \cap V = \emptyset$.

We are asked to show that $U \cup V$ is not path connected.

I managed to find a proof but it's rather technical and not as simple as we would expect, since intuitively this should be "obvious".

Can you find a better, shorter, more geometric proof? One that doesn't invoke intermediate value theorem?

My proof (if you are interested):

Suppose $U \cup V$ is path connected. Then there is a continuouos function $\gamma :[0, 1] \to U \cup V$ such that $\gamma (0) = u \in U$ and $\gamma (1) = v \in V$.

Let's define another function $f: U \cup V \to \{0, 1\}$, defined by $f(x) = \begin{matrix} 0, x \in U \\ 1 , x\in V\end{matrix}$.

Turns out that this function is continuous. Why?

Since $U$ is open, $\forall x \in U \exists \delta>0$ such that $B(x, \delta) \subset U$. Then $|x-y| < \delta$ implies that $x,y \in U$ which means $|f(x)-f(y)| = |0-0| = 0 < \epsilon$ for all $\epsilon$, so $f$ is continuous on $U$.

Through a similar argument, it is continuous on $V$, and since they are disjointed we don't have to worry about conflicting values of $f$ and can infer $f$ is continuous on $U \cup V$.

To sum up: $\gamma: [0, 1] \to U \cup V$ is continuous, $f: U \cup V \to \{0, 1\}$ is continuous, so their composition $f \circ \gamma:[0, 1] \to \{0, 1\}$ is also continuous. But that can't be because it doesn't agree with intermediate value theorem, so our assumption that $U \cup V$ is connected has to be false.

Surely there is a better way??

Answer 1

The simplest proof of its contrapositive is quite short.

Suppose $U \cup V$ is path connected. Hence it is connected. Since $U$ and $V$ are open, nonempty, and $U \cup V = U \cup V$, by the definition of connectedness, it follows that $U \cap V \neq \emptyset$.

Answer 2

As stated, it is false. You must also assume that $U,V\neq\emptyset$.

Assuming this, you can take $\gamma\colon[0,1]\longrightarrow U\cup V$ continuous such that $\gamma(0)\in U$ and that $\gamma(1)\in V$. Then:

• $\gamma^{-1}(U)\neq\emptyset$;
• $\gamma^{-1}(V)\neq\emptyset$;
• since $\gamma$ is continuous, $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ are open sets;
• since $U\cap V=\emptyset$, $\gamma^{-1}(U)\cap\gamma^{-1}(V)=\emptyset$.

But then $[0,1]$ would be disconnected, which is not true.