TITLE: Existence of open map from $\mathbb{R}$ to $\mathbb{R}^2$
QUESTION [5 upvotes]: Is there an open map from $\mathbb{R}$ to $\mathbb{R}^2$? I am guessing that non-existence can be derived from the fact that $\mathbb{R}$ and $\mathbb{R}^2$ are not locally homeomorphic, but cannot write down a rigorous argument. Any help is appreciated, thanks!

REPLY [4 votes]: Chris Culter's answer shows that there does exist such a map.  However, it is impossible for such a map to be continuous.  Indeed, suppose $f:\mathbb{R}\to\mathbb{R}^2$ is continuous and open.  Then $U=f((0,1))$ is open in $\mathbb{R}^2$, and $f([0,1])$ is compact and hence closed and bounded in $\mathbb{R}^2$.  It follows that $U$ is bounded and $f([0,1])$ is its closure, so the boundary of $U$ must be contained in $f(\{0,1\})$.  But any nonempty bounded open subset of $\mathbb{R}^2$ must contain more than two boundary points (in fact, uncountably many).  For instance, fix any point $p\in U$; then by boundedness of $U$, every line through $p$ must contain at least two boundary points of $U$ (one on each side of $p$).