# Could there exist two open, disjoint sets in $R^n$ s.t. their union is path connected?

Could there exist two open, disjoint sets in $R^n$ s.t. their union is path connected?

I don't really know where to start with this, but for right now I think that trying to prove that for a set $U$, the set $U \cup U^c$ cannot be path connected. For any disjoint set $V$ , $V \subset U^c$, therefore $V \cup U$ cannot be path connected.

Is this correct? Thanks for your time.

• What do you mean $U \cup U^c$ is not path-connected? $U \cup U^c$ is $\mathbb R^n$. Also, how do you deduce that $V \cup U$ is not path-connected? It's certainly possible for a path-connected set to be a subset of a non-path-connected set. – Erick Wong Mar 1 '14 at 16:09
• You can't use $U\cup U^c$, because that is the entire space, and that is here path-connected. The baisc idea isn't bad, however. – Daniel Fischer Mar 1 '14 at 16:09
• So what I should've said was that $U \cup int(U^c)$ can't be path connected. Is proving this as simple as using norms and the mean value theorem? – ILoveLev Mar 1 '14 at 16:16

## 1 Answer

If both sets $U$ and $V$ are non-empty, then the very existence of $U$ and $V$ shows that $U \cup V$ is disconnected, as these sets form a disconnection of the union. And disconnected implies not path-connected (as path connected implies connected)