The (n-1)-dimensional Hausdorff measure of any k-antichain in [0,1]^n is at most kn, and for n=2 there exist k-antichains attaining exactly 2k.
A continuous analogue of Erd\H{o}s' $k$-Sperner theorem
1 Pith paper cite this work. Polarity classification is still indexing.
abstract
A \emph{chain} in the unit $n$-cube is a set $C\subset [0,1]^n$ such that for every $\mathbf{x}=(x_1,\ldots,x_n)$ and $\mathbf{y}=(y_1,\ldots,y_n)$ in $C$ we either have $x_i\le y_i$ for all $i\in [n]$, or $x_i\ge y_i$ for all $i\in [n]$. We show that the $1$-dimensional Hausdorff measure of a chain in the unit $n$-cube is at most $n$, and that the bound is sharp. Given this result, we consider the problem of maximising the $n$-dimensional Lebesgue measure of a measurable set $A\subset [0,1]^n$ subject to the constraint that it satisfies $\mathcal{H}^1(A\cap C) \le \kappa$ for all chains $C\subset [0,1]^n$, where $\kappa$ is a fixed real number from the interval $(0,n]$. We show that the measure of $A$ is not larger than the measure of the following optimal set: \[ A^{\ast}_{\kappa} = \left\{ (x_1,\ldots,x_n)\in [0,1]^n : \frac{n-\kappa}{2}\le \sum_{i=1}^{n}x_i \le \frac{n+ \kappa}{2} \right\} \, . \] Our result may be seen as a continuous counterpart to a theorem of Erd\H{o}s, regarding $k$-Sperner families of finite sets.
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math.CA 1years
2019 1verdicts
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On $k$-antichains in the unit $n$-cube
The (n-1)-dimensional Hausdorff measure of any k-antichain in [0,1]^n is at most kn, and for n=2 there exist k-antichains attaining exactly 2k.