# Strengthening the intermediate value theorem to an "intermediate component theorem"

Let $f$ be a continuous function on a closed Euclidean ball (in dimension $\ge 2$) that is negative in the center of the ball and positive on its boundary.

On any path from the center of the ball to its boundary, there must be some point where $f=0$, by the intermediate value theorem.

I would like to strengthen this, and say there exists an "intermediate zero component" between the center and the boundary: Some connected component of $\left\{f=0\right\}$ which any such path must traverse.

Is it possible to prove this without using heavy tools such as the Jordan-Brouwer separation theorem?

• I'd be particularly interested in the case where $f \ne 0$ has exactly two components. Then of course it suffices to show that $f = 0$ is connected. This is a sort of converse to Jordan-Brouwer, which I don't know if it is hard. Commented Nov 2, 2014 at 1:31
• Is there something wrong with saying "find the point given by IVT and take its connected component"? Commented Nov 10, 2014 at 5:01
• @EricStucky Why do you believe that the connected component must encircle the origin? In fact, you could easily choose a point (with the IVT) that isn't part of an encircling connected component. Commented Nov 10, 2014 at 5:07
• No reason, I just didn't read carefully enough. Commented Nov 10, 2014 at 5:08
• In the case when $f\ne 0$ has exactly two components, you are asking for a proof of unicoherence of $\mathbb R^n$. I think unicoherence is not as hard as Jordan-Brouwer, but I don't have any reference with its proof.
– user147263
Commented Nov 10, 2014 at 6:42

I'm not sure this is simpler than what you had in mind, but one thing you can do is approximate $f$ and try to pass through the smooth category.
By uniform approximation there is a smooth map $g$ such that $\|f-g\|_\infty\leq \epsilon$, and then by Sard's theorem there is a regular value $y$ of $g$ in the range $[-\epsilon,\epsilon]$, so that $\{g=y\}$ is a codimension-$1$ embedded submanifold. In particular $\{g=y\}$ has finitely many components, and at least one of these must separate $0$ from $\partial B$ (some simple argument is needed here). Thus for every $n$ there is some connected closed set $C_n$ separating $0$ from $\partial B$ such that $|f|\leq1/n$ on $C_\epsilon$.
Pass to a subsequence of $(C_n)$ such that $C_n$ converges in Hausdorff distance to some set $C$. Since $C_n\subset \{|f|\leq 1/n\}$ for all $n$ we have $C\subset\{f=0\}$, and since each $C_n$ is connected so is $C$. It remains only to show that $C$ separates $0$ from $\partial B$. Let $\gamma$ be any continuous path from $0$ to $\partial B$. Since each $C_n$ separates $0$ from $\partial B$ there is some point $x_n$ of $\gamma$ in $C_n$. For every $\epsilon>0$ eventually $x_n$ is in the $\epsilon$-neighbourhood of $C$, so any limit point of $(x_n)$ is a point of $\gamma$ in $C$. Thus $\gamma$ crosses $C$.