Download American Mathematical Monthly, volume 117, number 1, january by Daniel J. Velleman PDF

By Daniel J. Velleman

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Example text

For any containment-free society S, there is a left-right alternating containment-free society T such that S T. Proof. Suppose S is not left-right alternating. Then, since there is an equal number of left and right endpoints, there must be a pair of consecutive right endpoints somewhere. Continuing to the right, one must eventually reach a left endpoint. Thus, there must be a sequence of endpoints of the form p x q y r z somewhere in the society. (Note that x = y, since each voter has only one right endpoint.

Note that a S ( p) = a S (q) + 1, because supp S ( p) = supp S (q) ∪ {x} (and x ∈ supp S (q)); continuing this reasoning, we have a S ( p) = a S (q) + 1 = a S (r ) + 2. Let S be the society that results from swapping y and z , so that p x q y r z becomes p x q z r y . ) Since we are expanding A y and A z without altering any other agreement sets, this preserves (k, m)-agreeability. This also does not introduce any containments: when we moved z to the left, it did not pass any other left endpoints; similarly, when we moved y to the right, it did not pass any other right endpoints.

M We prove the sharpness condition (b) in Section 2, and establish the lower bound (a) in Section 4. ) Further Questions. 2 is not just motivated by an interest in approval voting. Rather, approval voting provides a convenient presentation of a broader mathematical problem: given a family of sets, is their intersection nonempty? Treating each set as the approval set of a voter, this amounts to asking whether there is a single platform with the approval of all voters. Or, even if a family of sets has empty intersection, is there a sense in which many of the sets must have nonempty intersection?