Construct a partition of the subsets of {1,2,3,4,5} into symmetric chains.

Name: Problem 33, Chapter 5: Introductory Combinatorics
Uploaded: 2020-05-08T02:43:14-08:00
Duration: 4 min 59 s
Channel: Heather Zimmers
Description: Construct a partition of the subsets of {1,2,3,4,5} into symmetric chains.

Construct a partition of the subsets of {1,2,3,4,5} into...

Key Concepts

Combinatorial Partitioning

Combinatorial partitioning involves dividing a set into disjoint subsets that together cover the entire original set. In the case of the Boolean lattice, the goal is to partition the set of all subsets into symmetric chains. This method not only organizes the structure in a meaningful way but also reveals properties about the distribution of subsets across different ranks, thus playing a significant role in combinatorial optimization and analysis.

Symmetric Chain Decomposition

A symmetric chain decomposition is a partitioning of a graded poset, like the Boolean lattice, into chains that are symmetric relative to the middle rank. Each chain starts at a lower level and reaches a peak at a level equidistant from the maximum level as its minimum, ensuring that the chain is balanced around the center. This concept is crucial in combinatorics for proofs related to extremal set theory and results such as Sperner's theorem.

Boolean Lattice

A Boolean lattice is a structured collection of all subsets of a finite set, organized by the subset relation. This structure is a classic example of a graded poset, where the rank (or level) of each subset is determined by its cardinality. The Boolean lattice is fundamental in combinatorics, especially in problems involving counting and ordering subsets by size.

Posets and Chains

A partially ordered set (poset) is a set equipped with a relation that orders some of its elements in a way that is reflexive, antisymmetric, and transitive. Within a poset, a chain is a subset of elements such that every pair of elements is comparable. In the context of the Boolean lattice, chains represent sequences of subsets where each is contained in the next, providing a linear structure through the hierarchy of subset sizes.

Transcript

00:02 Okay, in this problem we're going to write down all the subsets of the set with elements a, b, c, d, and e.

00:09 And remember in the reading it said, if you have n elements, you'll have two to the n subsets.

00:15 So here we have five elements, so we're going to have two to the fifth subsets, and two to the fifth is 32.

00:22 So we're expecting to come up with 32 subsets.

00:26 And it helps to figure that out ahead of time so we can keep track and make sure we didn't get too many or we didn't get too few.

00:31 So let's do this in a systematic way, starting with the sets that have zero elements.

00:38 Well, that's pretty easy.

00:39 There's only one.

00:40 That would be the empty set.

00:42 Now, the other extreme would be the sets that have five elements.

00:46 There's only one.

00:48 It's the set with elements a, b, c, d, and e.

00:53 So let's keep track.

00:54 So far, we have one, and we have one.

00:57 So we have 30 more to go.

00:59 Not bad.

01:00 Now let's come up with the sets that have one element.

01:06 Okay.

01:06 I think we can figure this out pretty easily.

01:09 We could have a, or we could have b, or we could have c, or we could have d, or we could have e.

01:17 So there's five of those.

01:20 Now the other extreme, if it has four elements, that means we're leaving out one.

01:25 So finding the sets with four elements is very similar to finding the sets with one element.

01:30 So let's do that next.

01:34 So which one are we going to leave out? we could leave out a, and we would have b, c, d, and e.

01:41 We could leave out b and we would have a, c, d, and e.

01:46 We could leave out c and we would have a, b, d, d, and e.

01:51 We could leave out d and we would have a, b, c, and e.

01:55 Or we could leave out e and we would have a, b, c, and d.

02:01 How many did we get? we got five.

02:05 Makes sense that we got five both times, one element, or four elements.

02:09 So now we're up to 12.

02:11 Okay, so we have 20 more.

02:13 And the ones that we're missing are the sets that have two elements and the sets that have three elements.

02:19 Chances are there will be 10 of each.

02:21 It seems like we have a bit of a symmetry thing going on here.

02:25 So let's shoot for 10 of each.

02:26 So let's see what we can do...

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