M=13 case of FUNC: Difference between revisions
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[[Lemma 7]]: If <math>\mathcal{A}</math> contains a 5 element set, then <math>\mathcal{A}</math> is Frankl's. | [[Lemma 7]]: If <math>\mathcal{A}</math> contains a 5 element set, then <math>\mathcal{A}</math> is Frankl's. | ||
[[Lemma 8]]: If <math>\mathcal{A}</math> contains two 6 element sets intersecting at five elements, then <math>\mathcal{A}</math> is Frankl's. | [[Lemma 8]]: If <math>\mathcal{A}</math> contains two 6 element sets intersecting at five elements, then <math>\mathcal{A}</math> is Frankl's. | ||
[[Lemma 9]]: <math>\mathcal{A}</math> is Frankl's. Note that this relies on the prior lemmas, which so far haven't all been proved. | [[Lemma 9]]: <math>\mathcal{A}</math> is Frankl's. Note that this relies on the prior lemmas, which so far haven't all been proved. | ||
Revision as of 21:07, 11 November 2016
On this page, we attempt a proof of the m=13 case of Frankl's conjecture. This will probably end up being a long case analysis, with the cases being tackled out of order, so bear with me while the page is in progress. In all results below, we will try to prove that the set [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's (satisfies the conjecture), but will assume the opposite. We will use the terminology defined in the paper "The 11 element case of Frankl's Conjecture" by Ivica Bosnjak and Petar Markovic.
Lemmas
Note that the lemmas will get moved around in the process of proof. Right now this is an outline of the different things to prove.
Lemma 1: If [math]\displaystyle{ \mathcal{A} }[/math] contains 2 three element sets with a two element intersection, then [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's. (Not completed)
Lemma 2: If [math]\displaystyle{ \mathcal{A} }[/math] contains 2 intersecting 3 element sets, then [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's.
Lemma 3: If [math]\displaystyle{ \mathcal{A} }[/math] contains 2 three element sets, then [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's.
Lemma 4: If [math]\displaystyle{ \mathcal{A} }[/math] contains a four element set and a three element subset, then [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's.
Lemma 5: If [math]\displaystyle{ \mathcal{A} }[/math] contains a 3 element set, then [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's.
Lemma 6: If [math]\displaystyle{ \mathcal{A} }[/math] contains a 4 element set, then [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's.
Lemma 7: If [math]\displaystyle{ \mathcal{A} }[/math] contains a 5 element set, then [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's.
Lemma 8: If [math]\displaystyle{ \mathcal{A} }[/math] contains two 6 element sets intersecting at five elements, then [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's.
Lemma 9: [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's. Note that this relies on the prior lemmas, which so far haven't all been proved.
