If $U\subset \mathbb{R^n}$ is open then $\dim U=n?$

Question

If $U\subset \mathbb{R^n}$ is open then $\dim U=n?$ I know this is true from open submanifolds but not sure if it works for any open set.

Answer 1

Any open set of $\mathbb{R}^n$ is an $n$-dimensional submanifold of $\mathbb{R}^n$, for around any point there is a small ball contained in the set, and such a ball is homeomorphic to an open set of $\mathbb{R}^n$, namely, the ball itself!

Answer 2

Let $A = \{(U_i ,\phi_i): i \in I\}$ be an atlas for $\mathbb{R}^n$. Then $A' = \left\{\left(U \cap U_i, \phi_i \bigr|_{U \cap U_i}\right): i \in I\right\}$ is an atlas for $U$. Here $U$ is given the subspace topology of $\mathbb{R}^n$.