Math Problem Statement

Verify that the inclusion relation \subset Is a partial ordering on the power set s where s={a,b,c}

Solution

To verify that the inclusion relation \subset is a partial ordering on the power set P(S)\mathcal{P}(S), where S={a,b,c}S = \{a, b, c\}, we need to check if the relation satisfies the three properties of a partial order:

  1. Reflexivity: For every subset AP(S)A \in \mathcal{P}(S), AAA \subseteq A (every set is a subset of itself).
  2. Antisymmetry: If ABA \subseteq B and BAB \subseteq A, then A=BA = B.
  3. Transitivity: If ABA \subseteq B and BCB \subseteq C, then ACA \subseteq C.

Let's verify these properties one by one.

1. Reflexivity

The power set P(S)\mathcal{P}(S) contains all the subsets of SS: P(S)={,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}\mathcal{P}(S) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\} For each of these subsets AP(S)A \in \mathcal{P}(S), it's clear that AAA \subseteq A holds (every set is a subset of itself). Hence, reflexivity is satisfied.

2. Antisymmetry

The antisymmetry property requires that if ABA \subseteq B and BAB \subseteq A, then A=BA = B. In other words, if two sets are mutually subsets of each other, they must be equal.

For example:

  • If A={a}A = \{a\} and B={a}B = \{a\}, then clearly A=BA = B, since they are mutually subsets of each other.

In general, if ABA \subseteq B and BAB \subseteq A, the only way this can happen is if the sets contain exactly the same elements. Thus, antisymmetry holds.

3. Transitivity

The transitivity property requires that if ABA \subseteq B and BCB \subseteq C, then ACA \subseteq C. This means that if AA is a subset of BB and BB is a subset of CC, then AA must also be a subset of CC.

For example:

  • If A={a},B={a,b},C={a,b,c}A = \{a\}, B = \{a, b\}, C = \{a, b, c\}, we can see that ABA \subseteq B and BCB \subseteq C, and therefore ACA \subseteq C.

This holds for all subsets in P(S)\mathcal{P}(S), so transitivity is satisfied.

Conclusion

Since the inclusion relation \subset satisfies reflexivity, antisymmetry, and transitivity, it is indeed a partial ordering on the power set P(S)\mathcal{P}(S) of S={a,b,c}S = \{a, b, c\}.


Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

  1. What is the difference between a partial order and a total order?
  2. Can you give an example where a relation is reflexive and transitive but not antisymmetric?
  3. How can we represent a partial order using a Hasse diagram?
  4. If SS had more elements, how would the power set and partial order relation change?
  5. Can the inclusion relation \subset be a total order on some set?

Tip: A total order is a partial order where every pair of elements is comparable. In contrast, in a partial order, some elements might be incomparable.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Relations
Partial Order
Power Set

Formulas

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Theorems

Reflexivity
Antisymmetry
Transitivity

Suitable Grade Level

Grades 11-12