Consider $\mathcal{F} = {\{ f:\mathbb{S}^3 \to \mathbb{R}^3; f\mbox{ is smooth} \}}$ and the metric $d: \mathcal{F}\times \mathcal{F} \to \mathbb{R} $ $$d(f,g) = |f-g| = \sup_{p\in \mathbb{S}^3} \|f(p) - g(p)\|, $$ it is easy to see that $(\mathcal{F}, d)$ is a metric space. Let $T^2 \subset \mathbb{S}^3$ be the set $$T^2 = \frac{1}{\sqrt{2}}\cdot \mathbb{T}^2 = \left\{(x_1,x_2,x_3,x_4)\in \mathbb{S}^3 ; \ x_1^2 + x_2^2 = x_3^2 + x_4^2 = \frac{1}{2} \right\}. $$
Question: Is the set $\mathcal{A} = \{f\in \mathcal{F};\ f(p)\neq 0 \ \ \forall\ p\in T^2 \}$ dense in $\mathcal{F}$?
This seems true but I'm a little confused about how I supposed to approach this problem. Can anyone help me?