Hausdorff dimension of inverse images. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:24:49Z http://mathoverflow.net/feeds/question/71514 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71514/hausdorff-dimension-of-inverse-images Hausdorff dimension of inverse images. Helge 2011-07-28T19:59:16Z 2011-07-28T21:24:15Z <p>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?</p> <hr> <p>I believe the following argument works for $d = 2$: </p> <p>$A = f^{-1}((-\infty, t))$ and $B= f^{-1}((t,\infty))$ are two open sets whose complement is contained in $C$. If the Hausdorff dimension of $C$ was $&lt; 1$, then $C$ would be totally disconnected. Hence, $\mathbb{R}^2 \setminus C$ would be disconnected, which is implossible.</p> http://mathoverflow.net/questions/71514/hausdorff-dimension-of-inverse-images/71516#71516 Answer by GH for Hausdorff dimension of inverse images. GH 2011-07-28T20:37:12Z 2011-07-28T20:52:45Z <p>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 <a href="http://mathoverflow.net/questions/40593/hausdorff-dimension-of-the-boundary-of-an-open-set-in-the-euclidean-space-lower" rel="nofollow">this MO entry</a>. 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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Intermediate_value_theorem" rel="nofollow">intermediate value theorem</a> 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 $d-1$.</p> http://mathoverflow.net/questions/71514/hausdorff-dimension-of-inverse-images/71520#71520 Answer by Gerald Edgar for Hausdorff dimension of inverse images. Gerald Edgar 2011-07-28T21:24:15Z 2011-07-28T21:24:15Z <p>Indeed, more is true: (1) The topological dimension is $\ge d-1$. And (2) the Hausdorff dimension is $\ge$ the topological dimension. </p> <p>For (1) note that $C$ is a closed set that separates $\mathbb R^d$.</p>