Proof that function $F: \mathbb{R}^n \to \mathbb{R}$ has uncountable infinite roots

Question

Let $n \geq 2$ and $F: \mathbb{R}^n \to \mathbb{R}$ continous. Let $a,b \in \mathbb{R}^n $ with $ F(a) > 0 $ and $F(b) < 0$.

I have already shown that $F$ has at least one root.

In the next step I want to prove that $F$ has uncountable infinite roots.

Given an explicite function, for instance $g(x,y)=x + y$ it's easy to see, that $g$ has uncountable infinite roots $(x,-x) \in \mathbb{R}^n$ due to the fact that $\mathbb{R}$ is a uncountable infinite set and $x \in \mathbb{R}$.

Knowing that an example isn't a proof I would appreciate some help to prove that theorem for a arbitrary function $F$ such as given above.

Answer 1

Since $F(\vec b)<0$ and $F$ is continuous, there exists $\varepsilon>0$, such that for all $k\in(1-\varepsilon,1+\varepsilon), F(k\vec b)<0$. Now define $$g_k(t)=F(\vec a+t(k\vec b-\vec a))$$ $g_k$ is a continuous function from $[0,1]$ to $\Bbb R$ such that $g_k(0)>0$ and $g_k(1)<0$. So $g_k$ has a root between $0$ and $1$ by intermediate value theorem. Also, solution vectors corresponding to $g_k$ for distinct $k$ are distinct because they lie on distinct lines intersecting at $\vec a$.

Further, because the set $(1-\varepsilon,1+\varepsilon)$ is uncountably infinity, we are done.

P.S. I am assuming that $\vec a$ and $\vec b$ are linearly independent, otherwise find another vector $\vec a'$ in a small neighbourhood of $\vec a$ which is linearly independent to $\vec b$ and $F(\vec a')>0$.

Answer 2

Just construct a family of curves connecting $a$ and $b$.

For example, let $H$ be the hyperplane $H = \{c \in \Bbb R^n: |ac| = |bc|\}$.

For every point $h \in H$, we have either $F(h) \leq 0$, in which case there is a zero on the line segment $ah$, or $F(h) \geq 0$, in which case there is a zero on the line segment $bh$.

The unions $ah \cup bh$ are disjoint for different $h$'s, so there are uncountably infinitely many zeros of $F$, one corresponding to each $h \in H$.