Answer to Question #225729 in Discrete Mathematics for Prince

Question #225729
1) Draw the Hasse diagram for inclusion on the set P(S), where S = {a, b, c, d}

2) Let S = {1,2,3,4} with lexicographic order "<=" relation
a. Find all pairs in S x S less than (2, 3)
b. Find all pairs in S x S greater than (3, 1)
c. Draw the Hasse diagram of the poset (S x S, <)
1
Expert's answer
2021-09-01T16:01:50-0400

Part 1 and 2



Part a

(2, 2), (2, 1),(1, 4), (1, 3), (1, 2), (1, 1)

Part b

(3, 2), (3, 3),(3, 4), (4, 1), (4, 2), (4, 3), (4, 4)

Part c

P(A)={,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}}.P(A) = \big \{ \varnothing, \{a\}, \{b\}, \{c\}, \{a, b\}, \{b, c\}, \{a,c\}, \{a, b,c\} \big \}.

Let's show S=(P(A),)S=(P(A), \subseteq) is a poset. Let's show that SS satisfies the following three properties.

  1. Reflexivity. For every xP(A) xxx \in P(A) \ x \subseteq x is trivially true.
  2. Antisymmetry. Let x,yP(A)x, y \in P(A) and xyyzx \subseteq y \wedge y \subseteq z.Then x=yx=y from the definition.
  3. Transitivity. Let x,y,zP(A)x, y,z \in P(A) and xyyxx \subseteq y \wedge y \subseteq x. Then for allaxa \in x we have aya \in y, and therefore aza \in z. Thus xzx \subseteq z.

The Hasse diagram for (P(A),)(P(A), \subseteq) is shown below.


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