No proper dense open subsets of $\mathbb{R}^n$?

Question

Let $U$ be a dense open subset of $\mathbb{R}^n$, a positive integer, and suppose that $X:=\mathbb{R}^n-U$ is $0$-dimensional; in the sense of the Lebesgue covering dimension. Then is $U$ homeomorphic to $\mathbb{R}^n$ if and only if $U=\mathbb{R}^n$?

Answer 1

What about $$\mathbb{R}^2 \setminus \lbrace (x,0), x \geq 0 \rbrace \quad \quad ?$$

Answer 2

You are right. Clearly, if $U=\mathbb{R}^n$, $U$ is a dense open subset of itself.

For the other implication, let us assume $\exists U\subsetneq \mathbb{R}^n$, a dense open subset of $\mathbb{R}^n$, with $V:=\mathbb{R}-U$ $0$-dimensional. Since $V$ is $0$-dimensional, for any $\mathbf{y}\in V$, $B^n_{\epsilon}(\mathbf{y})\cap U\neq \emptyset,\;\forall \epsilon>0$ (Where $B^n_\epsilon(\mathbf{y})$ is $n$-dimensional ball, centered in $\mathbf{y}$ and radius $\epsilon$,).

Let, $$x_0\in \bigcap_{\epsilon >0} B^n_{\epsilon}(\mathbf{y})\cap U,$$ Then, since $U$ is an open set, the $\exists\delta >0\;|\;B_{\delta}^n(\mathbf{x}_0)$ is contained in $U$. But, by construction, $\mathbf{y}\in B_{\delta}^n(\mathbf{x}_0)$, for any $\delta >0$. This is a contradiction, since we assumed $\mathbf{y}\notin U$.