Continuous $f:\mathbb{R}\rightarrow\mathbb{R}^n$ with dense image is surjective? Let $f:\mathbb{R}\rightarrow\mathbb{R}^n$ be continuous with dense image. Must $f$ be surjective?
Intuitively it seems true, I can't picture a curve in $\mathbb{R}^2$ satisfying this without covering everything. In the case $n=1$, I can prove it quite easily using intermediate value theorem.
 A: No, $f$ need not be surjective. It is not too hard to imagine a space-filling-like curve that misses isolated points.
Indeed, take any plane-filling curve and map it to a new curve in the plane by way of the complex exponential map $z\mapsto e^z$. This is space-filling except it never hits the origin.
Even more, if you take a space-filling curve and map its $y$-component through the $\arctan$ function you get a curve that fills the strip $\Bbb R\times(-\pi/2,\pi/2)$. If you then map this through $z\mapsto e^{2z}$, you get a continuous map with dense image that misses the entire negative real number line.
A: No, a continuous function with dense image is not necessarily surjective. Let us consider a sequence of points $(x_n)_{n\in {\bf N}}$ that is dense in ${\bf R}^n$. Consider the function that sends the interval $[n,n+1]$ to the segment $[x_n, x_{n+1}]$. This gives a function from $[0, \infty)$ to ${\bf R}^n$ whose image contains the dense set of points ${x_n}$ and is contained in a countable union of straight lines.
A countable union of straight lines has a dense complement by the Baire category theorem. Note also that a countable union of straight lines is of zero two dimensional Lebesgue measure. It has zero area.
