Let $f\colon U \subseteq \mathbb{R}^n \to \mathbb{R}$ be a non-constant continuous function from an open $n$-dimensional interval to $\mathbb{R}$. Suppose that $f$ hits zero. Show that there is a open subinterval $U' \subseteq U$ such as that:
- The restriction of $f$ to $U'$ is non-constant;
- $f(U')$ does not contain zero.
I've found how to prove this in two ways, one for the case $n = 1$ using suprema and infima, and one using this more general statement from general topology:
Let $U$ be a connected open subset of a connected space $X$ and let $f\colon U\to Y$ be a continuous function from $U$ to a Hausdorff space $Y$. If $f$ is non-constant and $y$ is in the image of $f$, then there is a connected open subset $U' \subseteq U$ such as that $f\colon U' \to Y$ is non-constant and does not hit $y$.
This statement has a relatively small proof. Although, I'm interested in a proof for arbitrary $n \in \mathbb{N}$ that does not resort to concepts of abstract topology, more like the proof for $n = 1$. I've tried to generalize it, but I don't think I can simply take arbitrary products of subintervals of $U$ like it's done in $n = 1$.
The proof for $n = 1$
Let $U = (a_0, b_0)$ and let $$a_1 = \inf\{x \in U | f(x) \neq 0\}$$ $$b_1 = \inf\{x \in (a_1, b_0) | f(x) = 0\}$$
We have $a_1 < b_1$ because if they were equal this would mean that we would have two sequences $\lim x_i = \lim y_i = a_1$ with $\lim f(x_i) = 0$ and $\lim f(y_i) \neq 0$. Clearly $(a_1, b_1) \subseteq (a_0, b_0)$. Now, take any sequence $\{x_i\}$ in $(a_1, b_1)$ converging to $b_1$. Then, the sequence $\{f(x_i)\}$ is a sequence of nonzero terms converging to zero, so $f$ is not constant on $(a_1, b_1) = U'$.
The general topology proof
Take $U'$ to be any connected component of the open $f^{-1}(Y - \{y\})$. Clearly $U'$ is a connected open subset of $X$ (it is open in the open subset $U$) whose image does not contain $y$. Now, $U'$ is not closed in $U$ (because it is a proper open subset of a connected $U$). This means that $U'$ has a cluster point $x \in U - U' = f^{-1}(y)$. Let $\{x_i\}_{i \in I}$ be any sequence converging to $x$. Then $f(x_i)$ converges to $y$ by values different than $y$. Given that $Y$ is Hausdorff, $f(x_i)$ cannot be constant (for the limit is unique and every constant is the limit of itself).