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11.13 Relation

Basic_concepts__Copy_.pdf

Workshop 5

Exercise 3

State whether the following relations are reflexive, symmetric, transitive:

  • (a) \(E = \mathbb{Z}\) and \(xRy\;\iff\;x=-y\)

no * (b) \(E = \mathbb{R}\) and \(x R y \iff \cos^2(x) + \sin^2(y) = 1\)

no * (c) \(E = \mathbb{N}\) and \(xRy\;\iff\;\exists p,q\geq1,y=px^{q}\) (where \(p\) and \(q\) are integers)

no

Which of the preceding examples are equivalence relations?

Exercise 4

Is the orthogonality (perpendicular) relation between two lines in the plane symmetrical? Reflexive? Transitive?

yes, no, no

Exercise 5

On \(\mathbb{R}^2\), we define the equivalence relation \(R\) by:

\[ (x, y) R (x', y') \iff x = x' \]

Prove that \(R\) is an equivalence relation.

Reflexivity: \((x,y)R(x,y) \iff x=x\)

Symmetry: \((x,y)R(x^{\prime},y^{\prime})\iff x=x^{\prime}~~~~~~~(x^{\prime},y^{\prime})R(x,y) \iff x^{\prime}=x\)

Transitivity: \((x,y)R(x^{\prime},y^{\prime})\;\iff\;x=x^{\prime}\) and \((x^{\prime},y^{\prime})R\left(x'',y''\right)\;\;\iff\;\;x'=x''\)

Then \((x,y)R(x'',y'')\;\iff\;x=x''\)

(*) Then determine the equivalence class of an element \((x_0,y_0)\in\mathbb{R}^2\).

\([(x_0,y_0)]=\{(x_0,y):\forall y\in \R\}\)

Exercise 6

We define on \(\mathbb{R}\) the relation \(x R y\) if and only if:

\[ x^2-y^2=x-y \]
  • Show that \(R\) is an equivalence relation.

Reflexivity: \(xRx\;\iff\;x^2-x^2=x-x\)

Symmetry: \(xRy\;\;\iff\;x^2-y^2=x-y~~~~~~~yRx\;\;\iff\;y^2-x^2=y-x\)

Transitivity: \(xRy\;\;\iff\;x^2-y^2=x-y\) and \(yRz\;\;\iff\;y^2-z^2=y-z\)

Then we have \(x^2-x=z^2-z\), thus we have \(xRz\;\;\iff\;x^2-z^2=x-z\)​ * (*) Calculate the equivalence class of an element \(x\) of \(\mathbb{R}\). How many elements are there in this class?

If \(x-y=0\), then \(y=x\)

If \(x-y\neq 0\), then \(x+y=1\Rightarrow y=1-x\)

Thus we have two elements

Exercise 7

Let \(E\) be a set. We define on \(P(E)\), the set of subsets of \(E\), the following relation:

\[ ARB ~if~A=B\text{ or }A=B^{c}\\\text{ where }B^{c}\text{ is the complement of }B\text{ in }E. \]

Prove that \(R\) is an equivalence relation.

Reflexivity: \(ARA~if~A=A~or~A=A^{c}\)

Symmetry: \(ARB~if~A=B\text{ or }A=B^{c}\) \(BRA~if~B=A\text{ or }B=A^{c}\)

Transitivity: \(ARB~if~A=B\text{ or }A=B^{c}\) and \(BRC~if~B=C\text{ or }B=C^{c}\)

\(ARC~if~A=C\text{ or }A=C^{c}\)

Exercise 8

Let \(E\) be a non-empty set and \(\alpha \subset\mathcal{P}(E)\) non-empty, verifying the following property:

\[ \forall X,Y\in\alpha,\exists Z\in\alpha,Z\subset(X\cap Y). \]

We define on \(\mathcal{P}(E)\) the relation \(\sim\) by:

\[ A\sim B\;\iff\;\exists X\in\alpha,X\cap A=X\cap B. \]

Prove that this defines an equivalence relation on \(\mathcal{P}(E)\).

Transitivity: \(A\sim B\;\;\iff\;\;\exists X_1\in\alpha,X_1\cap A=X_1\cap B.\)

\(B\sim C\;\iff\;\exists X_2\in\alpha,X_2\cap B=X_2\cap C.\)

Since \(\forall X_1,X_2\in\alpha,\exists Z\in\alpha,Z\subset(X_1\cap X_2)\), then \(A\sim B\;\iff\;\exists Z\in\alpha,Z\cap A=Z\cap B\) and \(B\sim C\;\;\iff\;\;\exists Z\in\alpha,Z\cap B=Z\cap C\)

Thus

\(A\sim C\;\;\iff\;\;\exists Z\in\alpha,Z\cap A=Z\cap C\)

(*) What are the equivalence classes of \(\emptyset\) and \(E\)?

Equivalence Class of \(\emptyset\) (\([\emptyset]\))

The equivalence class \([\emptyset]\) consists of all subsets \(A \subseteq E\) such that there exists some \(X \in \alpha\) where:
\(X \cap A = X \cap \emptyset = \emptyset.\)

This means that \(A\) must be disjoint from at least one set in \(\alpha\). In other words:
\([\emptyset] = \{ A \subseteq E \mid \exists X \in \alpha, \; A \cap X = \emptyset \}.\)

Interpretation: \([\emptyset]\) includes all subsets of \(E\) that do not share any elements with some member of \(\alpha\).


Equivalence Class of \(E\) (\([E]\))

The equivalence class \([E]\) consists of all subsets \(A \subseteq E\) such that there exists some \(X \in \alpha\) where:
\(X \cap A = X \cap E = X.\)

This implies that \(X \subseteq A\), meaning \(A\) contains at least one set from \(\alpha\). Therefore:
\([E] = \{ A \subseteq E \mid \exists X \in \alpha, \; X \subseteq A \}.\)

Interpretation: \([E]\) includes all subsets of \(E\) that contain some member of \(\alpha\).