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How to prove that all maximal independent sets of a matroid have the same cardinality.

Provided a matroid is a 2-tuple (M,J ) where M is a finite set and J is a family of some of the subsets of M satisfying the following properties:

  1. If A is subset of B and B belongs to J , then A belongs to J ,
  2. If A, B belongs to J , |A| <= |B|, and x belongs to A - B, then there exists y belongs to B - A such that (B U {x})- {y} belongs to J.

The members of J are called independent sets.

Ami Tavory
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user3622539
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    Try our sistersite http://math.stackexchange.com/ . This question is off topic here, as it has nothing to do with programming. – Matsmath Aug 18 '16 at 19:26
  • I'm voting to close this question as off-topic because it is not a practical programming question. Math or Computer Science may be a better place to ask. – m69's been on strike for years Aug 19 '16 at 01:43
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    @m69 I disagree - this is a topic standardly taught in undergrad algorithm courses with very practical application (such as minimum spanning tree), and the site is full of graph-theoretic questions not really directly more related to programming than this. In any case, if you disagree, FYI, when you close as OffTopic, there's already an option recommending migrating it to a different stackexchange. – Ami Tavory Aug 19 '16 at 06:36
  • @AmiTavory I still think it's iffy, but it takes 5 people to close the question, so if you're right and no one agrees, then nothing will happen. (Unfortunately, the off-topic->migration option only lists a few sites, and neither cs nor math are among them.) – m69's been on strike for years Aug 19 '16 at 09:08

2 Answers2

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Assume to the contrary that |A| < |B|, and A is not maximally independent.

Consider the following Venn diagram

enter image description here

Clearly B \ A (the only-blue part) is nonempty, as the cardinality of B is larger than that of A. Also, clearly A \ B (the only-orange part) is nonempty, as otherwise A ⊂ B, and, by definition, A is not maximally independent.

Hence, by the exchange property, there is some x &in; A \ B, y &in; B \ A, such that B ∪ {x} \ {y} &in; J as well. Let's call this set C. Note that if we would draw the Venn diagram for A and C (now the blue circle is C):

  1. |B| = |C| (the blue circle has the same size)

  2. |(A \ {x}) \ C| < |A \ B| (the only-orange part is smaller than before)

Now we can repeat the argument about A and C, and so on. Note, however, that we can't repeat it indefinitely, as A is assumed to be finite. Hence, at some point we will reach the contradiction that the orange set is completely contained in the blue set, which we already saw before is impossible (that would mean, by definition, that it is not maximally independent).

Ami Tavory
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We will do this using proof by Contradiction. Let's assume that all maximal independent set of a matroid does not have the same cardinality. Thus there must some set A and set B so that both are maximal independent set. Without the loss of generality let us take j A j < j B j i.e cardinality of A is less than cardinality of B. Let j A j = P and j B j = Q , P < Q . Now Let X 2 A-B and Y 2 B-A. X and Y will always exist since A is maximal and dierent from B. Using the second property of Matroid we can make B1 = f B [ X g - f y g which is also independent set and j B1 j = Q . We can continue picking an element from X' 2 A-Bi and an element from Y' 2 Bi-A and insert X' and remove Y' to make a new independent set which has the cardinality Q untill there is no element in A-Bi. Since A-Bi = thus A Bi. But Bi is also and independent set with cardinality Q. Now we can say that A is not maximal which is a contradiction and thus our assumption was wrong.Thus j A j = j B j which implies there can be no two maximal independent set with dierent cardinalities. So all maximal independent set of a matroid have the same cardinality.

anonymous
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