On the vanishing of derivative of surjective differentiable function $\mathbb R \to \mathbb R^2$ Let $f,g : \mathbb R \to \mathbb R  $ be differentiable function such that the function $H:\mathbb R \to \mathbb R^2$ given as $H(t)=(f(t),g(t)), \forall t \in \mathbb R$ is surjective i.e. for every $(a,b) \in \mathbb R^2, \exists t \in \mathbb R$ such that $a=f(t), b=g(t)$. Then how to prove that $\exists t_o \in \mathbb R$ such that $f'(t_o)=g'(t_o)=0$ ?  
 A: Assume no such $t_0$ exists. Let us consider some vertical line $v\ni H(t)$ for some $t$. Since $H$ is differentiable, if $f'(t)\neq 0$, then by the definition of a derivative there is some open neighborhood $U_t$ of $t$ such that $(U_t\sim t) \cap v = 0$. However, this means that if there is no point on $v$ with derivative 0, then for each $t$ such that $H(t)\in v$ there is an open neighborhood $U_t$ of $t$ such that the elements in the set $\{U_t\}$ are pairwise disjoint, since we can take the open set such that it does not contain any other $t$, then shrink it in half, implying there are only countably many such sets, contradicting the uncountability of $v$.
Thus for every vertical line $v$ there must be some point $t_v$ such that $f'(t_v)=0$. Now consider two such points $t_{v_1}, t_{v_2}$ for $v_1\neq v_2$. In between these points, there must be some open set, since there must be a point between them with derivative not equal to zero by the Mean Value Theorem, thus by the definition of derivative there is an open neighborhood of that point such that no point in that set has derivative 0. Thus for every point $t_v$, there is an open set $U_{t_v}$ such that the elements of $\{U_{t_v}\}$ are disjoint, so the set is countable, contradiction, proving the claim.
It is a bit late for me, so I'm logging off, but let me know if there are any confusing or ambiguous bits in the proof and I'll come back to them tomorrow!
