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An Oddtown family of four clubs on four citizens, and why a fifth cannot be added
Example
Take the four singletons on :
Their incidence matrix over is the identity matrix
so .
Facts & Assumptions
Given: the four singleton sets above.
Oddtown families have at most members (Oddtown: distinct with every odd and every () even satisfy ).
The singleton family attains that bound (The singletons form an Oddtown family, so the bound is attained for every ).
Verification
Each singleton has odd size , and any two distinct singletons meet in the empty set of even size , so the family satisfies Oddtown.
The displayed incidence matrix has pairwise orthogonal rows and , exactly as the Oddtown proof predicts.
A fifth set cannot be adjoined: by [L1] an Oddtown family on four points has at most four members, and [L2] says the bound is attained already.
Depends on
- Oddtown: distinct $A_1,\dots,A_m\subseteq[n]$ with every $\lvert A_i\rvert$ odd and every $\lvert A_i\cap A_j\rvert$ ($i\ne j$) even satisfy $m\le n$
- The $n$ singletons form an Oddtown family, so the bound $m\le n$ is attained for every $n$
- A finite family of subsets of $[n]$ and its incidence matrix over $F$
- $\langle v_A,v_B\rangle$ is the image of $\lvert A\cap B\rvert$ in $F$; over $\mathbb{F}_2$ it is $0$ or $1$ according to the parity of $\lvert A\cap B\rvert$
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- L. Babai and P. Frankl, Linear Algebra Methods in Combinatorics, §1.1 (standard reference, not scraped)