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Let $f: \mathbb{R}^d \to \mathbb{R}$ be a continuous function. Let $t \in (\inf(f), \sup(f))$ and define $C = f^{-1} (t)$. Is it true that the Hausdorff dimension of C is $\geq d -1$? If no how does one construct a counter example?

I believe the following argument works for $d = 2$:

$A = f^{-1}((-\infty, t))$ and $B= f^{-1}((t,\infty))$ are two open sets whose complement is contained in $C$. If the Hausdorff dimension of $C$ was $< 1$, then $C$ would be totally disconnected. Hence, $\mathbb{R}^2 \setminus C$ would be disconnected, which is implossible.

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I retagged with dimension-theory to help bring out the fact that this is about Hausdorff dimension to those who use tags as filters – David White Jul 28 '11 at 20:22
up vote 6 down vote accepted

The boundary of $A = f^{-1}((-\infty, t))$ and $B= f^{-1}((t,\infty))$ is $C = f^{-1} (t)$. Therefore $C$ has Hausdorff dimension at least $d-1$, using this MO entry. I recommend Sergei Ivanov's response for a simple proof. Strictly speaking, the quoted MO entry would require that $A$ or $B$ is bounded, but read below.

Sergei Ivanov's argument can be adapted for a direct proof as follows. One can find balls $A'\subset A$ and $B'\subset B$ of equal radius. Consider the line $L$ connecting the centers of these balls, and the planes orthogonal to $L$ passing through the centers. These planes intersect $A'$ and $B'$ in two parallel disks of dimension $d-1$ and equal radius. Applying the intermediate value theorem to $f$ restricted to the lines parallel to $L$, one sees that the orthogonal projection of $C$ to either disk is surjective, hence $C$ has Hausdorff dimension at least $d-1$.

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Indeed, more is true: (1) The topological dimension is $\ge d-1$. And (2) the Hausdorff dimension is $\ge$ the topological dimension.

For (1) note that $C$ is a closed set that separates $\mathbb R^d$.

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