Show that this is the set of bases of a matroid
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Let $mathcal{B}$ be the set of bases of a matroid with set of elements $mathcal{E}$. Define: $mathcal{B}^∗ :={mathcal{E}backslash B | B∈mathcal{B}}$.
Show that this is the set of bases of a matroid.
How can I proof this?
matroids
asked Nov 14 at 14:49
plsneedhelp
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Let $mathcal{B}$ be the set of bases of a matroid with set of elements $mathcal{E}$. Define: $mathcal{B}^∗ :={mathcal{E}backslash B | B∈mathcal{B}}$.
Show that this is the set of bases of a matroid.
How can I proof this?
matroids
asked Nov 14 at 14:49
plsneedhelp
434
Show that the basis axioms of a matroid hold, that is, show that the set of complements of bases of a matroid is a nonempty collection of subsets such that for any pair of subsets $B^*_1, B^*_2 in mathcal{B}^*$ and any $e$ in $B^*_1$ not in $B^*_2$ there is an $f in B^*_2 setminus B^*_1$ such that $B^*_1 - {e} cup {f}$ is also in $mathcal{B}^*$. Of course, you'll have to use that the complements of the $B^*_i$ are bases of the original matroid.
– Aaron Dall
Nov 15 at 21:28
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $mathcal{B}$ be the set of bases of a matroid with set of elements $mathcal{E}$. Define: $mathcal{B}^∗ :={mathcal{E}backslash B | B∈mathcal{B}}$.
Show that this is the set of bases of a matroid.
How can I proof this?
matroids
asked Nov 14 at 14:49
plsneedhelp
434
Let $mathcal{B}$ be the set of bases of a matroid with set of elements $mathcal{E}$. Define: $mathcal{B}^∗ :={mathcal{E}backslash B | B∈mathcal{B}}$.
Show that this is the set of bases of a matroid.
How can I proof this?
matroids
matroids
asked Nov 14 at 14:49
plsneedhelp
434
asked Nov 14 at 14:49
plsneedhelp
434
asked Nov 14 at 14:49
plsneedhelp
434
asked Nov 14 at 14:49
plsneedhelp
434
asked Nov 14 at 14:49
plsneedhelp
434
434
Show that the basis axioms of a matroid hold, that is, show that the set of complements of bases of a matroid is a nonempty collection of subsets such that for any pair of subsets $B^*_1, B^*_2 in mathcal{B}^*$ and any $e$ in $B^*_1$ not in $B^*_2$ there is an $f in B^*_2 setminus B^*_1$ such that $B^*_1 - {e} cup {f}$ is also in $mathcal{B}^*$. Of course, you'll have to use that the complements of the $B^*_i$ are bases of the original matroid.
– Aaron Dall
Nov 15 at 21:28
add a comment |
Show that the basis axioms of a matroid hold, that is, show that the set of complements of bases of a matroid is a nonempty collection of subsets such that for any pair of subsets $B^*_1, B^*_2 in mathcal{B}^*$ and any $e$ in $B^*_1$ not in $B^*_2$ there is an $f in B^*_2 setminus B^*_1$ such that $B^*_1 - {e} cup {f}$ is also in $mathcal{B}^*$. Of course, you'll have to use that the complements of the $B^*_i$ are bases of the original matroid.
– Aaron Dall
Nov 15 at 21:28
Show that the basis axioms of a matroid hold, that is, show that the set of complements of bases of a matroid is a nonempty collection of subsets such that for any pair of subsets $B^*_1, B^*_2 in mathcal{B}^*$ and any $e$ in $B^*_1$ not in $B^*_2$ there is an $f in B^*_2 setminus B^*_1$ such that $B^*_1 - {e} cup {f}$ is also in $mathcal{B}^*$. Of course, you'll have to use that the complements of the $B^*_i$ are bases of the original matroid.
– Aaron Dall
Nov 15 at 21:28
Show that the basis axioms of a matroid hold, that is, show that the set of complements of bases of a matroid is a nonempty collection of subsets such that for any pair of subsets $B^*_1, B^*_2 in mathcal{B}^*$ and any $e$ in $B^*_1$ not in $B^*_2$ there is an $f in B^*_2 setminus B^*_1$ such that $B^*_1 - {e} cup {f}$ is also in $mathcal{B}^*$. Of course, you'll have to use that the complements of the $B^*_i$ are bases of the original matroid.
– Aaron Dall
Nov 15 at 21:28
add a comment |
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Show that the basis axioms of a matroid hold, that is, show that the set of complements of bases of a matroid is a nonempty collection of subsets such that for any pair of subsets $B^*_1, B^*_2 in mathcal{B}^*$ and any $e$ in $B^*_1$ not in $B^*_2$ there is an $f in B^*_2 setminus B^*_1$ such that $B^*_1 - {e} cup {f}$ is also in $mathcal{B}^*$. Of course, you'll have to use that the complements of the $B^*_i$ are bases of the original matroid.
– Aaron Dall
Nov 15 at 21:28