What is order

Review of sets, relations functions(1)
Exercise 1-16(2)

Suppose \(A\) is a partitioned set and \(P,Q\) are two partitions such that for each \(p \in P\) there exists a \(q \in Q\) with \(A_p=A_q\)

  1. Show that for each \(p \in P\) there is at most one \(q \in Q\) such that \(A_p = A_q\)

  2. Show that for each \(q \in Q\) there is a \(p \in P\) such that \(A_p = A_q\)

Solution(1)
  1. Suppose \(q' \ne q\). If they are both equal to \(A_p\) then they are equal to each other, but a partition rule is that \(q' \ne q\) must have an empty intersection (and \(A_p\) cannot be empty by the other rule).

  2. By part 1, the mapping between part labels is a bijection, so there is an inverse map as well.

Exercise 1-20(2)

Finish proof Proposition 1.19. Suppose that \(\sim\) is an equivalence relation on a set A, and let P be the set of \(\sim\)-closed and \(\sim\)-connected subsets.

  1. Show that each part \(A_p\) is nonempty

  2. Show that \(p \ne q \implies A_p \cap A_q = \varnothing\)

  3. Show that \(A = \bigcup_{p \in P} A_p\)

Proof(1)
  1. Part of the definition of \(\sim\)-connected is being nonempty

  2. Suppose \(a \in A\) is in the intersection. Then \(a \sim p\) and \(a \sim q\) for some elements \(p \not\sim q\) arbitrarily selected from \(A_p, A_q\). But this is impossible because \(\sim\) is transitive, so this must be impossible.

    • Every \(a \in A\) is part of some equivalence class which is a \(\sim\)-closed and \(\sim\)-connected set, so \(A \subseteq \bigcup_{p \in P} A_p\)

      • The equivalence class is \(\sim\)-closed because two elements being \(\sim\)-related implies they are in the same equivalence class.

      • The equivalence class is \(\sim\)-connected because equivalence classes are nonempty and the equivalence relation is transitive.

    • The constituents of \(A_p\) were defined to be subsets of \(A\), so unioning these will also be a subset of \(A\), i.e. \(\bigcup_{p \in P} A_p \subseteq A\).

    • Therefore \(A = \bigcup_{p \in P} A_p\).

Function(1)

A function from \(S\) to \(T\)

A relation \(F \subseteq S \times T\) such that \(\forall s \in S:\ \exists! t \in T:\ (s,t) \in F\)

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Partition(1)

A partition of set \(A\)

A set \(P\) (‘part labels’) and, for each \(p \in P\), a nonempty subset (‘pth part’) \(A_p \subseteq A\) such that:

  1. \(A = \bigcup_{p \in P}A_p\)

  2. \(p \ne q \implies A_p \cap A_q = \varnothing\)

Two partitions are considered the same if the partitioned groups are the same, the labels don’t matter.

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Partitions are equivalences(2)

There is a bijection between ways to partition a set \(A\) and the equivalence relations on \(A\)

Proof(1)
  • Every partition gives rise to a distinct equivalence relation

    • Define \(a \sim b\) to mean \(a,b\) are in the same part. This is a reflective, symmetric, and transitive relation given the definition of a partition.

  • Every equivalence relation gives rise to a distinct partition.

    • Define a subset \(X \subseteq A\) as \(\sim\)-closed if, for every \(x \in X\) and \(x' \sim x\), we have \(x' \in X\).

    • Define a subset \(X \subseteq A\) as \(\sim\)-connected if it is nonempty and \(\forall x,y \in X:\ x \sim y\)

    • The parts corresponding to \(\sim\) are precisely the \(\sim\)-closed and \(\sim\)-connected subsets.

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Quotient(1)

A quotient of a set under an equivalence relation.

This is equivalent to the parts of the partition associated with the equivalence relation.

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Relation(1)

A relation between sets \(X\) and \(Y\)

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