Why the graph of a smooth function $f$: $\mathbb{R}\rightarrow\mathbb{R^n}$ is a curve in $\mathbb{R^n}$? This must be a silly question, but I'm really confused about the title, why it is a curve (one dimension in $\mathbb{R^n}$ not, for example, a plane (two dimension in $\mathbb{R^n}$?)
 A: It is not a silly question at all, and in an answer given by Robert Israel to a similar question, a non-trivial argument involving the Hausdorff dimension proves that the graph of a differentiable function $f:\mathbb{R}\to\mathbb{R}^n$ in $\mathbb{R}^{n+1}$ cannot have positive Lebesgue measure of dimension higher than $1$. In particular, it cannot be a plane.
A: There are continuous functions $f:\>I\to{\mathbb R}^2$ mapping the interval $I:=[0,1]$ onto the full unit square in ${\mathbb R}^2$, so called Peano curves. But there are no smooth such curves. Assume that $f$ is $C^1$ on $I$, and that $f'(t)\ne0\in{\mathbb R}^n$ for all $t\in I$. Consider a point $\tau\in I$, and assume that $f_1'(\tau)\ne0$. Then $\tau$ is the midpoint of an interval $J\subset I$ with $f'(t)>0$ (say) for all $t\in J$. This implies that $f_1$ is invertible on $J$, so that $t=f^{-1}_1(x_1)$ for all $x_1\in J'$, whereby $J'$ is a neighborhood of $\xi:=f_1(\tau)$. This allows to produce the set $f(J)$, a part of the full set $f(I)$, in the form
$$x_1\mapsto\bigl(x_1,\phi_2(x_1),\ldots,\phi_n(x_1)\bigr)\qquad(x_1\in J')\tag{1}$$
whereby $$\phi_k(x_1):=f_k\bigl(f_1^{-1}(x_1)\bigr)\qquad(2\leq k\leq n)\ .$$
Now you should accept that $(1)$ is the parametric representation of a curve, namely the graph of the function $$x_1\mapsto(\phi_2(x_1),\ldots, \phi_n(x_1)\bigr)\qquad(x_1\in J')\ .$$
