$\mathbb R^n=U \cup V$ where $n\ge 2$ and $U, V$ open and connected is the intersection $U\cap V$ connected?
My attempt is: to try Mayer Vietoris sequence, since $\mathbb R^n$ is contractible its homology group is $H_k(\mathbb R^n)=\begin{cases}\mathbb Z \quad k=0 \\ 0 \quad else\end{cases}$
And we have reduced short exact sequence as
$$0\to H_0(U\cap V)\to H_0(U)\oplus H_0(V)\to \underbrace{H_0(\mathbb R^n)}_{\mathbb Z}\to 0$$
Since they are all connected but not necessary path connected and we are not given anything about locally compactness about spaces I cannot conclude that $H_0(U\cap V)\cong \mathbb Z$.
Any hint and answer would be appreciated.
According to Wikipedia:
An open subset of $\mathbb R^n$ is locally path-connected, so it is connected if and only if it is path-connected.
The proof of this isn't hard. We'll only prove connected implies path-connected. The other way (path-connected implies connected) is well-known, following from the fact that $[0,1]$ is connected.
Assume $X$ is connected and locally path-connected. let $x\in X$ and let $U$ be the set of points which are path-connected to $x$ and $V$ the set of all points path-disconnected from $x.$ Then $U\cup V=X$ and $U\cap V=\emptyset.$ We will show $U,V$ are open.
If $y\in U$ then pick a neighborhood $W$ of $y$ which is path-connected. Then every element of $W$ is also path-connected to $x,$ so $y\in W\subseteq U.$ So $U$ is open.
If $y\in V,$ then pick a neighborhood $W$ of $y$ path-connected. Then if any point in $W$ is path-connected to $x,$ $y$ would be. So we must have $W\subseteq V.$ So $V$ is open.
Since $x\in U,$ $U$ is non-empty, so connectedness means $V=\emptyset,$ and $U=X,$ so $X$ is path-connected.