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sav08:sets_and_relations [2010/02/26 21:56]
vkuncak
sav08:sets_and_relations [2015/04/21 17:30] (current)
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To denote large or infinite sets we can use set comprehensions:​ $\{ x.\ P(x) \}$ is set of all objects with property $P$. To denote large or infinite sets we can use set comprehensions:​ $\{ x.\ P(x) \}$ is set of all objects with property $P$.
-$+\begin{equation*} y \in \{ x. P(x) \} \ \leftrightarrow\ P(y) y \in \{ x. P(x) \} \ \leftrightarrow\ P(y) -$+\end{equation*}

Notation for set comprehension:​ $\{ f(x)|x. P(x) \} = \{ y. (\exists x. y=f(x) \land P(x)) \}$ Notation for set comprehension:​ $\{ f(x)|x. P(x) \} = \{ y. (\exists x. y=f(x) \land P(x)) \}$

Sometimes the binder $x$ can be inferred from context so we write simply $\{ f(x) | P(x) \}$.  In general there is ambiguity in which variables are bound. (Example: what does the $a$ in $f(a,b)$ refer to in the expression: Sometimes the binder $x$ can be inferred from context so we write simply $\{ f(x) | P(x) \}$.  In general there is ambiguity in which variables are bound. (Example: what does the $a$ in $f(a,b)$ refer to in the expression:
-$+\begin{equation*} \{a \} \cup \{ f(a,b) \mid P(a,b) \} \{a \} \cup \{ f(a,b) \mid P(a,b) \} -$+\end{equation*}
does it refer to the outerone $a$ as in $\{a\}$ or is it a newly bound variable? The notation with dot and bar resolves this ambiguity. does it refer to the outerone $a$ as in $\{a\}$ or is it a newly bound variable? The notation with dot and bar resolves this ambiguity.

Subset: $A \subseteq B$ means $\forall x. x \in A \rightarrow x \in B$ Subset: $A \subseteq B$ means $\forall x. x \in A \rightarrow x \in B$

-$+\begin{equation*} A \cup B = \{ x. x \in A \lor x \in B \} A \cup B = \{ x. x \in A \lor x \in B \} -$ +\end{equation*}
-$+\begin{equation*} A \cap B = \{ x. x \in A \land x \in B \} A \cap B = \{ x. x \in A \land x \in B \} -$ +\end{equation*}
-$+\begin{equation*} A \setminus B = \{ x. x \in A \land x \notin B \} A \setminus B = \{ x. x \in A \land x \notin B \} -$+\end{equation*}

Boolean algebra of subsets of some set $U$ (we define $A^c = U \setminus A$): Boolean algebra of subsets of some set $U$ (we define $A^c = U \setminus A$):
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Note that sets can be nested. Consider, for example, the following set $S$ Note that sets can be nested. Consider, for example, the following set $S$
-$+\begin{equation*} S = \{ \{ p, \{q, r\} \}, r \} S = \{ \{ p, \{q, r\} \}, r \} -$+\end{equation*}
This set has two elements. The first element is another set. We have $\{ p, \{q, r\} \} \in S$. Note that it is not the case that  This set has two elements. The first element is another set. We have $\{ p, \{q, r\} \} \in S$. Note that it is not the case that

Suppose that we have a set $B$ that contains other sets.  We define union of the sets contained in $B$ as follows: Suppose that we have a set $B$ that contains other sets.  We define union of the sets contained in $B$ as follows:
-$+\begin{equation*} ​\bigcup B = \{ x.\ \exists a. a \in B \land x \in a \} ​\bigcup B = \{ x.\ \exists a. a \in B \land x \in a \} -$+\end{equation*}
As a special case, we have As a special case, we have
-$+\begin{equation*} ​\bigcup \{ a_1, a_2, a_3 \} = a_1 \cup a_2 \cup a_3 ​\bigcup \{ a_1, a_2, a_3 \} = a_1 \cup a_2 \cup a_3 -$+\end{equation*}
Often the elements of the set $B$ are computed by a set comprehension of the form $B = \{ f(i).\ i \in J \}$. Often the elements of the set $B$ are computed by a set comprehension of the form $B = \{ f(i).\ i \in J \}$.
We then write We then write
-$+\begin{equation*} ​\bigcup_{i \in J} f(i) ​\bigcup_{i \in J} f(i) -$+\end{equation*}
and the meaning is and the meaning is
-$+\begin{equation*} ​\bigcup \{ f(i).\ i \in J \} ​\bigcup \{ f(i).\ i \in J \} -$+\end{equation*}
Therefore, $x \in \bigcup \{ f(i).\ i \in J \}$ is equivalent to $\exists i.\ i \in J \land x \in f(i)$. Therefore, $x \in \bigcup \{ f(i).\ i \in J \}$ is equivalent to $\exists i.\ i \in J \land x \in f(i)$.

We analogously define intersection of elements in the set: We analogously define intersection of elements in the set:
-$+\begin{equation*} ​\bigcap B = \{ x. \forall a. a \in B \rightarrow x \in a \} ​\bigcap B = \{ x. \forall a. a \in B \rightarrow x \in a \} -$+\end{equation*}
As a special case, we have As a special case, we have
-$+\begin{equation*} ​\bigcap \{ a_1, a_2, a_3 \} = a_1 \cap a_2 \cap a_3 ​\bigcap \{ a_1, a_2, a_3 \} = a_1 \cap a_2 \cap a_3 -$+\end{equation*}
We similarly define intersection of an infinite family We similarly define intersection of an infinite family
-$+\begin{equation*} ​\bigcap_{i \in J} f(i) ​\bigcap_{i \in J} f(i) -$+\end{equation*}
and the meaning is and the meaning is
-$+\begin{equation*} ​\bigcap \{ f(i).\ i \in J \} ​\bigcap \{ f(i).\ i \in J \} -$+\end{equation*}
Therefore, $x \in \bigcap \{ f(i).\ i \in J \}$ is equivalent to $\forall i.\ i \in J \rightarrow x \in f(i)$. Therefore, $x \in \bigcap \{ f(i).\ i \in J \}$ is equivalent to $\forall i.\ i \in J \rightarrow x \in f(i)$.

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Pairs: Pairs:
-$+\begin{equation*} (a,b) = (u,v) \iff (a = u \land b = v) (a,b) = (u,v) \iff (a = u \land b = v) -$+\end{equation*}
Cartesian product: Cartesian product:
-$+\begin{equation*} A \times B = \{ (x,y) \mid x \in A \land y \in B \} A \times B = \{ (x,y) \mid x \in A \land y \in B \} -$+\end{equation*}

Relations $r$ is simply a subset of $A \times B$, that is $r \subseteq A \times B$. Relations $r$ is simply a subset of $A \times B$, that is $r \subseteq A \times B$.

Note: Note:
-$+\begin{equation*} A \times (B \cap C) = (A \times B) \cap (A \times C) A \times (B \cap C) = (A \times B) \cap (A \times C) -$ +\end{equation*}
-$+\begin{equation*} A \times (B \cup C) = (A \times B) \cup (A \times C) A \times (B \cup C) = (A \times B) \cup (A \times C) -$+\end{equation*}

=== Diagonal relation === === Diagonal relation ===

$\Delta_A \subseteq A \times A$, is given by $\Delta_A \subseteq A \times A$, is given by
-$+\begin{equation*} ​\Delta_A = \{(x,x) \mid x \in A\} ​\Delta_A = \{(x,x) \mid x \in A\} -$+\end{equation*}

==== Set operations ==== ==== Set operations ====
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==== Relation Inverse ==== ==== Relation Inverse ====

-$+\begin{equation*} r^{-1} = \{(y,x) \mid (x,y) \in r \} r^{-1} = \{(y,x) \mid (x,y) \in r \} -$+\end{equation*}

==== Relation Composition ==== ==== Relation Composition ====

-$+\begin{equation*} r_1 \circ r_2 = \{ (x,z) \mid \exists y. (x,y) \in r_1 \land (y,z) \in r_2\} r_1 \circ r_2 = \{ (x,z) \mid \exists y. (x,y) \in r_1 \land (y,z) \in r_2\} -$+\end{equation*}

Note: relations on a set $A$ together with relation composition and $\Delta_A$ form a //monoid// structure: Note: relations on a set $A$ together with relation composition and $\Delta_A$ form a //monoid// structure:
-$+\begin{equation*} \begin{array}{l} \begin{array}{l} r_1 \circ (r_2 \circ r_3) = (r_1 \circ r_2) \circ r_3 \\ r_1 \circ (r_2 \circ r_3) = (r_1 \circ r_2) \circ r_3 \\ r \circ \Delta_A = r = \Delta_A \circ r r \circ \Delta_A = r = \Delta_A \circ r \end{array} \end{array} -$+\end{equation*}

Moreover, Moreover,
-$+\begin{equation*} ​\emptyset \circ r = \emptyset = r \circ \emptyset ​\emptyset \circ r = \emptyset = r \circ \emptyset -$ +\end{equation*}
-$+\begin{equation*} r_1 \subseteq r_2 \rightarrow r_1 \circ s \subseteq r_2 \circ s r_1 \subseteq r_2 \rightarrow r_1 \circ s \subseteq r_2 \circ s -$ +\end{equation*}
-$+\begin{equation*} r_1 \subseteq r_2 \rightarrow s \circ r_1 \subseteq s \circ r_2 r_1 \subseteq r_2 \rightarrow s \circ r_1 \subseteq s \circ r_2 -$+\end{equation*}

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When $S \subseteq A$ and $r \subseteq A \times A$ we define **image** of a set $S$ under relation $A$ as When $S \subseteq A$ and $r \subseteq A \times A$ we define **image** of a set $S$ under relation $A$ as
-$+\begin{equation*} ​S\bullet r = \{ y.\ \exists x. x \in S \land (x,y) \in r \} ​S\bullet r = \{ y.\ \exists x. x \in S \land (x,y) \in r \} -$+\end{equation*}

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Iterated composition let $r \subseteq A \times A$. Iterated composition let $r \subseteq A \times A$.
-$+\begin{equation*} \begin{array}{l} \begin{array}{l} r^0 = \Delta_A \\ r^0 = \Delta_A \\ r^{n+1} = r \circ r^n r^{n+1} = r \circ r^n \end{array} \end{array} -$+\end{equation*}
So, $r^n$ is n-fold composition of relation with itself. So, $r^n$ is n-fold composition of relation with itself.

Transitive closure: Transitive closure:
-$+\begin{equation*} r^* = \bigcup_{n \geq 0} r^n r^* = \bigcup_{n \geq 0} r^n -$+\end{equation*}

Equivalent statement: $r^*$ is equal to the least relation $s$ (with respect to $\subseteq$) that satisfies Equivalent statement: $r^*$ is equal to the least relation $s$ (with respect to $\subseteq$) that satisfies
-$+\begin{equation*} \Delta_A\ \cup\ (s \circ r)\ \subseteq\ s \Delta_A\ \cup\ (s \circ r)\ \subseteq\ s -$+\end{equation*}
or, equivalently,​ the least relation $s$ (with respect to $\subseteq$) that satisfies or, equivalently,​ the least relation $s$ (with respect to $\subseteq$) that satisfies
-$+\begin{equation*} \Delta_A\ \cup\ (r \circ s)\ \subseteq\ s \Delta_A\ \cup\ (r \circ s)\ \subseteq\ s -$+\end{equation*}
or, equivalently,​ the least relation $s$ (with respect to $\subseteq$) that satisfies or, equivalently,​ the least relation $s$ (with respect to $\subseteq$) that satisfies
-$+\begin{equation*} \Delta_A\ \cup\ r \cup (s \circ s)\ \subseteq\ s \Delta_A\ \cup\ r \cup (s \circ s)\ \subseteq\ s -$+\end{equation*}

==== Some Laws in Algebra of Relations ==== ==== Some Laws in Algebra of Relations ====

-$+\begin{equation*} (r_1 \circ r_2)^{-1} = r_2^{-1} \circ r_1^{-1} (r_1 \circ r_2)^{-1} = r_2^{-1} \circ r_1^{-1} -\ +\end{equation*} ​ -\[+\begin{equation*} r_1 \circ (r_2 \cup r_3) = (r_1 \circ r_2) \cup (r_1 \circ r_3) r_1 \circ (r_2 \cup r_3) = (r_1 \circ r_2) \cup (r_1 \circ r_3) -$ +\end{equation*}
-$+\begin{equation*} (r^{-1})^{*} = (r^{*})^{-1} (r^{-1})^{*} = (r^{*})^{-1} -$+\end{equation*}

Binary relation $r \subseteq A\times A$ can be represented as a directed graph $(A,r)$ with nodes $A$ and edges $r$ Binary relation $r \subseteq A\times A$ can be represented as a directed graph $(A,r)$ with nodes $A$ and edges $r$
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Equivalence classes are defined by Equivalence classes are defined by
-$+\begin{equation*} x/r = \{y \mid (x,y) \in r x/r = \{y \mid (x,y) \in r -$+\end{equation*}
The set $\{ x/r \mid x \in A \}$ is a partition: The set $\{ x/r \mid x \in A \}$ is a partition:
* each set non-empty   * each set non-empty
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* their union is $A$   * their union is $A$
Conversely: each collection of sets $P$ that is a partition defines equivalence class by Conversely: each collection of sets $P$ that is a partition defines equivalence class by
-$+\begin{equation*} r = \{ (x,y) \mid \exists c \in P. x \in c \land y \in c \} r = \{ (x,y) \mid \exists c \in P. x \in c \land y \in c \} -$+\end{equation*}

Congruence: equivalence that agrees with some set of operations.  ​ Congruence: equivalence that agrees with some set of operations.  ​
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**Example:​** an example function $f : A \to B$ for $A = \{a,b,c\}$, $B=\{1,​2,​3\}$ is **Example:​** an example function $f : A \to B$ for $A = \{a,b,c\}$, $B=\{1,​2,​3\}$ is
-$+\begin{equation*} f = \{ (a,3), (b,2), (c,3) \} f = \{ (a,3), (b,2), (c,3) \} -$+\end{equation*}

Definition of function, injectivity,​ surjectivity. Definition of function, injectivity,​ surjectivity.
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Function update operator takes a function $f : A \to B$ and two values $a_0 \in A$, $b_0 \in B$ and creates a new function $f[a_0 \mapsto b_0]$ that behaves like $f$ in all points except at $a_0$, where it has value $b_0$. ​ Formally, Function update operator takes a function $f : A \to B$ and two values $a_0 \in A$, $b_0 \in B$ and creates a new function $f[a_0 \mapsto b_0]$ that behaves like $f$ in all points except at $a_0$, where it has value $b_0$. ​ Formally,
-$+\begin{equation*} f[a_0 \mapsto b_0](x) = \left\{\begin{array}{l} f[a_0 \mapsto b_0](x) = \left\{\begin{array}{l} b_0, \mbox{ if } x=a_0 \\ b_0, \mbox{ if } x=a_0 \\ f(x), \mbox{ if } x \neq a_0 f(x), \mbox{ if } x \neq a_0 -$+\end{equation*}

==== Domain and Range of Relations and Functions ==== ==== Domain and Range of Relations and Functions ====

For relation $r \subseteq A \times B$ we define domain and range of $r$: For relation $r \subseteq A \times B$ we define domain and range of $r$:
-$+\begin{equation*} dom(r) = \{ x.\ \exists y. (x,y) \in r \} dom(r) = \{ x.\ \exists y. (x,y) \in r \} -$ +\end{equation*}
-$+\begin{equation*} ran(r) = \{ y.\ \exists x. (x,y) \in r \} ran(r) = \{ y.\ \exists x. (x,y) \in r \} -$+\end{equation*}
Clearly, $dom(r) \subseteq A$ and $ran(r) \subseteq B$. Clearly, $dom(r) \subseteq A$ and $ran(r) \subseteq B$.

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**Partial function** $f : A \hookrightarrow B$ is relation $f \subseteq A \times B$ such that **Partial function** $f : A \hookrightarrow B$ is relation $f \subseteq A \times B$ such that
-$+\begin{equation*} \forall x \in A. \exists^{\le 1} y.\ (x,y)\in f \forall x \in A. \exists^{\le 1} y.\ (x,y)\in f -$+\end{equation*}

Generalization of function update is override of partial functions, $f \oplus g$ Generalization of function update is override of partial functions, $f \oplus g$
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The following properties follow from the definitions:​ The following properties follow from the definitions:​
-$+\begin{equation*} (S \bullet r_1) \bullet r_2 = S \bullet (r_1 \circ r_2) (S \bullet r_1) \bullet r_2 = S \bullet (r_1 \circ r_2) -$ +\end{equation*}
-$+\begin{equation*} S \bullet r = ran(\Delta_S \circ r) S \bullet r = ran(\Delta_S \circ r) -$+\end{equation*}

===== Further references ===== ===== Further references =====

-  * [[:Gallier Logic Book]], Chapter 2
* [[sav08:​discrete_mathematics_by_rosen|Discrete Mathematics by Rosen]]   * [[sav08:​discrete_mathematics_by_rosen|Discrete Mathematics by Rosen]]
+  * [[:Gallier Logic Book]], Chapter 2