Question What is the least integer such that every set of at least points in the plane contains collinear points or a subset of points in general position (no three collinear)?
Problem Let be natural numbers with . It follows from the pigeon-hole principle that there exist distinct subsets with . Is it possible to find such a pair in polynomial time?
Conjecture Let be an integer. For every integer , there exists an integer such that for every digraph , either has a pairwise-disjoint directed cycles of length at least , or there exists a set of at most vertices such that has no directed cycles of length at least .
Setup Fix a tree and for every vertex a non-negative integer which we think of as the amount of gold at .
2-Player game Players alternate turns. On each turn, a player chooses a leaf vertex of the tree, takes the gold at this vertex, and then deletes . The game ends when the tree is empty, and the winner is the player who has accumulated the most gold.
Conjecture Let be a Cantor set embedded in . Is there a self-homeomorphism of for every greater than so that moves every point by less than and does not intersect ? Such an embedded Cantor set for which no such exists (for some ) is called "sticky". For what dimensions do sticky Cantor sets exist?
Conjecture Let be a field of characteristic zero. A collection of polynomials in variables defines an automorphism of if and only if the Jacobian matrix is a nonzero constant.
Conjecture Let be the complete funcoid corresponding to the usual topology on extended real line . Let be the order on this set. Then is a complete funcoid.
Proposition It is easy to prove that is the infinitely small right neighborhood filter of point .
If proved true, the conjecture then can be generalized to a wider class of posets.
Problem Determine a computable set of invariants that allow one to determine, given a compact boundaryless 3-manifold, whether or not it embeds smoothly in the 4-sphere. This should include a constructive procedure to find an embedding if the manifold is embeddable.