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By Bjorner A., Stanley R.P.

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Example text

Suppose that |A ∪ B| > 2. Since 1 (and not 0) is written into position 62 {i, j} that means that there is some partition C ∪ D = {1, 2, . . , n} into nonempty disjoint subsets C and D such that i ∈ C , j ∈ D and all possible edges {c, d} with c ∈ C , d ∈ D and {c, d} = {i, j} are already marked with 0. Clearly, we must have A ⊆ C and B ⊆ D, so in particular all edges between a node in A and a node in B have already been inspected. Also, all edges between two nodes both in A have by the induction assumption been inspected, and similarly for B.

Draw (curvilinear) triangles on the torus so that each edge of a triangle is also the edge of some other 54 triangle, and the 2 endpoints of each edge are not the pair of endpoints of any other edge. The triangles should cover the torus so that each point on the torus is in exactly one of the triangles, or possibly in an edge where two triangles meet or at a corner where several triangles meet. We can think of this as cutting the rubber surface of an inner tube into small triangular pieces. Figure 8 shows one way to do this using 14 triangles.

Borsuk’s Theorem. If the k-dimensional sphere is covered by k + 1 closed sets, then one of these sets must contain a pair of antipodal points. Borsuk’s Problem. Is it true that every set of diameter one in k-dimensional real space Rk can be partitioned into at most k + 1 sets of smaller diameter? This work of Borsuk has interacted with combinatorics in a remarkable way. In 1978 L´aszl´o Lov´asz (b. 1948) solved a difficult combinatorial problem — the “Kneser Conjecture” from 1955 — by using Borsuk’s theorem.

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Combinatorial miscellany by Bjorner A., Stanley R.P.

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