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I have such a question:

$${A = \{1, 2, 3\}}$$ $${R \text{ on } P(A)}$$ $${\{3\} \subset a \cap b \leftrightarrow aRb}$$

Question is - if such relation is symmetric, antisymmetric, transitive or reflexive

So, as I far as I understand first of all we have a set 1, 2, 3, so P(A) is

$${\{\{1\} \ \{2\} \ \{3\} \ \{1, 2\} \ \{1, 3\} \ \{2, 3\} \ \{1, 2, 3\} \ \varnothing \}}$$

ok, so let's make a relation that fit to this condition

$${\{3\} \subset a \cap b \leftrightarrow aRb}$$

so,

$${R = \{\{1, 3\} \ \{2, 3\} \ \{1, 2, 3\} \}}$$

and for know, when I know that R is I can say if this set is symmetric, antisymmetric, transitive or reflexive

  1. Symmetric - no
  2. antisymmetric - yes
  3. transitive - no
  4. reflexive - no

What is my question - I have just antisymmetric - yes, but as far as I know the right answer that just symmetric - yes, thus my logic isn't right. What is not right in this logic?

Sirop4ik
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  • Since ${3} \subset a \cap b$, then ${1,2,3}R{2,3}$ and ${1,2,3}R{1,3}$ but ${1,3} \not \sim {1,2}$. – healynr May 08 '20 at 16:24
  • What you're written down is not $R$, since the elements of $R$ are pairs of subsets of $A$; for example $({1,3},{1,2,3})\in R$. – Michael Hoppe May 08 '20 at 16:26
  • @healynr But ${2,3}R{1,3}$. – Michael Hoppe May 08 '20 at 16:28
  • @MichaelHoppe Why? Isn't ${2,3} \cap {1,3} = {3}$ and $\subset$ indicates proper subset? – healynr May 08 '20 at 16:29
  • @healynr no, he means that in your first comment you wrote but {1,3}≁{1,2} , but should be written But {2,3}R{1,3} – Sirop4ik May 08 '20 at 16:32
  • @healynr: Unfortunately, the symbol $\subset$ is ambiguous, and far too many people use it to mean $\subseteqq$, so we can’t be sure which is meant here. However, the OP’s incorrect listing of $R$ does not include ${3}$, so it appears that the OP at least is interpreting it as proper subset. – Brian M. Scott May 08 '20 at 16:46
  • @BrianM.Scott Oh I see thank you. – healynr May 08 '20 at 16:58
  • Did you intend $p\subset q$ to mean $p\subseteq q$ or $p\subsetneq q \text{ ?} \qquad$ – Michael Hardy May 08 '20 at 17:46

1 Answers1

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A relation on the power set of $A$ should be represented by a set of ordered pairs of subsets of $A,$ not by a set of subsets of $A.$

$$ R = \left\{ \begin{array}{cccccccccc} \big(\{3\},\{3\}\big), \, \big(\{3\}, \{1,3\}\big), \, \big(\{1,3\},\{3\}\big),\\ \big(\{2,3\},\{3\}\big),\, \big(\{3\},\{2,3\}\big), \\ \big(\{3\},\{1,2,3,\}\big) , \big(\{1,2,3\},\{3\}\big), \\ \big(\{1,3\},\{1,3\}\big),\, \big(\{1,3\},\{2,3\}\big), \big(\{2,3\},\{1,3\}\big), \\ \big(\{1,3\},\{1,2,3\}\big),\,\big(\{1,2,3\},\{1,3\}\big), \\ \big( \{2,3\},\{2,3\}\big), \big(\{2,3\}, \{1,2,3\}\big),\, \big( \{1,2,3\}, \{2,3\}\big), \\ \big( \{1,2,3\},\{1,2,3\}\big) \end{array} \right\} $$ (At this point I'm not completely sure I didn't miss any.)

This relation is symmetric and transitive. It is not reflexive on the power set of $A=\{1,2,3,\}$ because, for example, it is not true that $\{1,2\} \mathbin{R} \{1,2\}.$

Michael Hardy
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  • Can you explain to me why it is transitive? I thought ${2,3}$ did not relate to ${1,3}$ even though ${2,3}R{1,2,3}$ and ${1,2,3}R{1,3}$. – healynr May 08 '20 at 16:42
  • @healynr : Why would you say it is not true that ${2,3} \mathbin{R} {1,3} \text{ ?}$ The set ${3}$ is a subset of both of them, and that was the definition. – Michael Hardy May 08 '20 at 16:52
  • I assumed that $\subset$ meant proper subset (as opposed to $\subseteq$) but apparently as @Brian M. Scott pointed out that is not always the case. – healynr May 08 '20 at 16:57
  • @healynr : If that means proper subset then it is still the case that ${2,3} \mathbin{R} {1,3}$ since ${3}$ is a proper subset of both. – Michael Hardy May 08 '20 at 17:45
  • but it isn't a proper subset of their intersection, is it? Isn't their intersection ${3}$? – healynr May 08 '20 at 17:45
  • @healynr : I see. In that case the answer would be different. However, the relation should still be represented by a set of ordered pairs of subsets of ${1,2,3}. \qquad$ – Michael Hardy May 08 '20 at 17:48
  • I am afraid that it could be kind of misunderstanding, bacause in my case P(A) is - {{1} {2} {3} {1,2} {1,3} {2,3} {1,2,3} ∅}, this set has an 8 items, because that set A has 3 items, thus P(A) = 2^3 = 8. Thus it is not pair of pairs... – Sirop4ik May 08 '20 at 18:19
  • @AlekseyTimoshchenko : I didn't say it was a pair of pairs, but rather that it is a set of pairs. – Michael Hardy May 08 '20 at 19:14