Skip to content

Latest commit

 

History

History
248 lines (216 loc) · 10.2 KB

chap-03.md

File metadata and controls

248 lines (216 loc) · 10.2 KB
pagetitle author
3. Relations

3. Relations

Equivalence Relations

1. Define two points ( x 0 , y 0 ) and ( x 1 , y 1 ) of the plane to be equivalent if y 0 x 0 2 = y 1 x 1 2 . Check that this is an equivalence relation and describe the equivalence classes.

Proof. $\quad$ It is easily seen that the relation is reflexive, symmetric, and transitive. Each equivalence class is a parabola given by x x 2 + c .$\quad\square$

2. Let C be a relation on a set A . If A 0 A , define the restriction of C to A 0 to be the relation C ( A 0 × A 0 ) . Show that the restriction of an equivalence relation is an equivalence relation.

Proof. $\quad$It is clear that: $$ \forall x\in A_0\forall y\in A_0((x,y)\in C \Leftrightarrow (x,y)\in C\cap (A_0\times A_0)). $$ Thus all the properties for an equivalence relation hold in C ( A 0 × A 0 ) .$\quad\square$

3. Here is a “proof” that every relation C that is both symmetric and transitive is also reflexive: “Since C is symmetric, a C b implies b C a . Since C is transitive, a C b and b C a together imply a C a , as desired.” Find the flaw in this argument.

Proof. $\quad$ Let C A × A be a relation. If C is symmetric and transitive, then: $$ \forall a\forall b(aCb\Rightarrow aCa). $$ If C is reflexive, then: $$ \forall a\in A(aCa). $$

4. Let f : A B be a surjective function. Let us define a relation on A by setting a 0 a 1 if f ( a 0 ) = f ( a 1 ) .
 $\quad$(a) Show that this is an equivalence relation.
 $\quad$(b) Let A be the set of equivalence classes. Show there is a bijective correspondence of A with B .

Proof. $\quad$(a) f ( a ) = f ( a ) . f ( a ) = f ( b ) and f ( b ) = f ( c ) f ( a ) = f ( c ) . f ( a ) = f ( b ) f ( b ) = f ( a ) .
 $\quad$(b) Let g : A B be a function given by x f ( a ) where a x , and let x , y A and a , b x . Since a b , f ( x ) = f ( y ) , thus g is well-defined. If x y and c x and d y , then f ( c ) f ( d ) , thus g ( x ) g ( y ) , so g is injective. Since f is surjective, for every d B there is c A such that f ( c ) = d , and since is an equivalence relation, there is z A such that c z ; g ( z ) = f ( c ) = d , so g is surjective. Thefefore, g is bijective.$\quad$

5. Let S and S be the following subsets of the plane: $$ \begin{array}{rl} S &={(x, y)\mid y = x + 1\text{ and }0 < x < 2},\ S' &={(x, y)\mid y - x\text{ is an integer}}. \end{array} $$  $\quad$(a) Show that S is an equivalence relation on the real line and S S . Describe the equivalence classes of S .
 $\quad$(b) Show that given any collection of equivalence relations on a set A , their intersection is an equivalence relation on A .
 $\quad$(c) Describe the equivalence relation T on the real line that is the intersection of all equivalence relations on the real line that contain S . Describe the equivalence classes of T .

Proof. $\quad$(a) x x = 0 . a b = n z b a = n . a b = n , b c = m for n , m z a c = n + m .
 $\quad$(b) Let R = R i i I be the collection of equivalence relations on A indexed by a nonempty set I . For all a A , since a R i for each i I , ( a , a ) i I R . If ( a , b ) i I R , then ( a , b ) R i for each i I , thus ( a , a ) , ( b , b ) , ( b , a ) R i for each i I , so ( a , a ) , ( b , b ) , ( b , a ) i I R . Similarly, if ( a , b ) , ( b , c ) i I R , then ( a , c ) i I R .
 $\quad$(c) A equivalence relation on the real line that contain S need more equations. y = x for the reflexivity, x = y + 1 for the symmetry. Thus 0 < x < 3 and 0 < y < 3 . x = y + 1 + 1 for the transitivity, thus in general, y x is an integer, 0 < x < 3 and 0 < y < 3 . T is the restriction of S to ( 0 , 3 ) . This definition is minimal with respect to the previous equations. T can be seen as the intersection of two equivalence relations,

Extra open brace or missing close brace

$S'\cap
{(x,y)\mid\text{either }0&lt;x&lt;3$
and 0 < y < 3 , or ( x 0 or x 3 ) and ( y 0 or
Extra close brace or missing open brace

$y\ge 3)}$
.

Order Relations

6. Define a relation on the plane by setting $$ (x_0, y_0) < (x_1, y_1) $$ if either y 0 x 0 2 < y 1 x 1 2 , or y 0 x 0 2 = y 1 x 1 2 and x 0 < x 1 . Show that this is an order relation on the plane, and describe it geometrically.

Proof. $\quad$It is easily seen that comparability, nonreflexivity and transitivity hold for the given relation. Geometrically, if ( x 0 , y 0 ) < ( x 1 , y 1 ) , then ( x 0 , y 0 ) lies in y = x 2 + c for some c R and ( x 1 , y 1 ) lies in y = x 2 + d for some d R and either c < d , or c = d and x 0 < x 1 .$\quad\square$

7. Show that the restriction of an order relation is an order relation.

Proof. $\quad$Let A and A 0 be sets such that A 0 A and let C be an order relation on A . It is clear that: $$ \forall x\in A_0\forall y\in A_0((x,y)\in C \Leftrightarrow (x,y)\in C\cap (A_0\times A_0)). $$ Thus all the properties for an order relation hold in C ( A 0 × A 0 ) .$\quad\square$

8. Check that the relation defined in Example 7 is an order relation.

Proof. $\quad$From Example 7, "Define x C y if x 2 < y 2 , or if x 2 = y 2 and x < y ."
 $\quad$Clear.$\quad\square$

9. Check that the dictionary order is an order relation.

Proof. $\quad$Clear.$\quad\square$

10. $\quad$(a) Show that the map f : ( 1 , 1 ) R of Example 9 is order preserving.
 $\quad$(b) Show that the equation g ( y ) = 2 y / [ 1 + ( 1 + 4 y 2 ) 1 / 2 ] defines a function g : R ( 1 , 1 ) that is both a left and a right inverse for f .

Proof. $\quad$(a) From Example 9, f is given by x x / ( 1 x 2 ) . Suppose that y > x . f ( y ) f ( x ) = y / ( 1 y 2 ) x / ( 1 x 2 ) = ( y y x 2 x + x y 2 ) / ( ( 1 y 2 ) ( 1 x 2 ) ) = ( y x + y x ( y x ) ) / ( ( 1 y 2 ) ( 1 x 2 ) ) . Since y x > 0 and | y x | > | x y ( y x ) | , f ( x ) < f ( y ) . Thus f is order preserving; thus injective, and also neither upper-bounded nor lower-bounded; thus surjective. Therefore, ( 1 , 1 ) and R have the same order type.
 $\quad$(b) Brute-force is enough. ;-)$\quad\square$

11. Show that an element in an ordered set has at most one immediate successor and at most one immediate predecessor. Show that a subset of an ordered set has at most one smallest element and at most one largest element.

Proof. $\quad$Let S be an ordered set, and let a , b , c S . If a has immediate successors, b and c , then by comparability, b = c ; otherwise b < c or b > c , a contradiction. Similarly to immediate predecessor, smallest element, and largest element.$\quad\square$

12. Let $\mathbb{Z}+$ denote the set of positive integers. Consider the following order relations on $\mathbb{Z}+\times\mathbb{Z}_+$:
 $\quad$(i) The dictionary order.
 $\quad$(ii) ( x 0 , y 0 ) < ( x 1 , y 1 ) if either x 0 y 0 < x 1 y 1 , or x 0 y 0 = x 1 y 1 and y 0 < y 1 .
 $\quad$(iii) ( x 0 , y 0 ) < ( x 1 , y 1 ) if either x 0 + y 0 < x 1 + y 1 , or x 0 + y 0 = x 1 + y 1 and y 0 < y 1 .
 $\quad$In these order relations, which elements have immediate predecessors? Does the set have a smallest element? Show that all three order types are different.

Proof. $\quad$(i) ( x , 1 ) for every $x\in\mathbb{Z}+$ has no immediate predecessor. $(1,1)$ is the smallest.
 $\quad$(ii) $(x,1)$ and $(1,y)$ for every $x,y\in\mathbb{Z}
+$ have no immediate predecessor. No smallest element.
 $\quad$(iii) ( 1 , 1 ) has no immediate predecessor. ( 1 , 1 ) is the smallest.
 $\quad$ Suppose f ( b ) = c , where f is bijective order preserving function, and b has immediate predecessor a , and c has no immediate predecessor, then there is no f ( a ) . They all have different order types.$\quad\square$

13. Prove the following:
Theorem. If an ordered set A has the least upper bound property, then it has the greatest lower bound property.

Proof. $\quad$Let S A be bounded below, and let

Extra open brace or missing close brace

$T={x\in A\mid x$
is a lower bound of
Extra close brace or missing open brace

$S}$
be nonempty. T has a least upper bound t , and clearly t is a greatest lower bound of S .$\quad\square$

14. If C is a relation on a set A , define a new relation D on A by letting ( b , a ) D if ( a , b ) C .
 $\quad$(a) Show that C is symmetric if and only if C = D .
 $\quad$(b) Show that if C is an order relation, D is also an order relation.
 $\quad$(c) Prove the converse of the theorem in Exercise 13.

Proof. $\quad$(a) Clear.
 $\quad$(b) "If y D x and z D y , then z D x ." implies "If z D y D x , then z D x ". D is transitive; the other properties are obvious.
 $\quad$(c) Let S A be bounded above and let

Extra open brace or missing close brace

$T={x\in A\mid x$
is a upper bound of
Extra close brace or missing open brace

$S}$
be nonempty. T has a greatest lower bound t , and clearly t is a least upper bound of S .$\quad\square$

15. Assume that the real line has the least upper bound property.
 $\quad$(a) Show that the sets $$ \begin{array}{ll} \left[0,1\right] &={x\mid 0\le x\le 1},\ \left[0,1\right) &={x\mid 0\le x <1} \end{array} $$ have the least upper bound property.
 $\quad$(b) Does [ 0 , 1 ] × [ 0 , 1 ] in the dictionary order have the least upper bound property? What about [ 0 , 1 ] × [ 0 , 1 ) ? What about [ 0 , 1 ) × [ 0 , 1 ] ?

Proof. $\quad$(a) Let S [ 0 , 1 ] . Since S of R is bound above, S has a least upper bound m . Clearly m [ 0 , 1 ] , thus [ 0 , 1 ] also has the least upper bound property. Let T [ 0 , 1 ) . Since T of R is bounded above, T has a least upper bound n such that n [ 0 , 1 ] . If n = 1 , then T of [ 0 , 1 ) is not bounded above. Thus if T of [ 0 , 1 ) is bounded, then n [ 0 , 1 ) . Therefore, [ 0 , 1 ) has the least upper bound property.
 $\quad$(b) [ 0 , 1 ] × [ 0 , 1 ] and [ 0 , 1 ) × [ 0 , 1 ] have the least upper bound property. 0 × [ 0 , 1 ) [ 0 , 1 ] × [ 0 , 1 ) has no least upper bound.$\quad\square$