how to prove it - ch4, sec4.4(Ordering Relations)

This section talks about mathematical meaning of ordering elements of sets, finding smallest, finding minimal element of a set with respect to a particular relation.

Antisymmetric Relation: Intuitively, its opposite of a symmetric relation. Formally, Suppose R is a relation on set A. Then R is antisymmetric if $\forall x\in A\forall y\in A(xRy\wedge yRx\to x=y)$

So, in an antisymmetric relation, For two different x and y if $xRy$ then $(y,x)\notin R$. It, in some sense, is saying that "y is atleast as large as x". And, this though motivates to think about the "ordering".

Suppose R is a relation on a set A.
Partial ordering: R is called a partial ordering on A(or just partial ordering if A is clear from the context) if it is reflexive, transitive and antisymmetric.

Total ordering: R is called a total ordering on A if it is partial order and $\forall x\in A\forall y\in A(xRy\vee yRx)$

So, in case of total ordering, its possible to compare any two elements of A as one of the two ordered pairs formed by them must be in the relation.


Suppose R is a partial ordering on A and $B\subseteq A$.

Smallest and Minimal: Then $b\in B$ is called the R-smallest(or just smallest if R is clear from the context) element of B if $\forall x\in B\left(bRx\right)$.
It is called an R-minimal(or just minimal) if $\neg \exists x\in B(xRb\wedge x\ne b)$

Largest and Maximal: These are exactly opposite of Smallest and Minimal respectively. $b\in B$ is called the R-largest element of B if $\forall x\in B\left(xRb\right)$.
It is called an R-maximal if $\neg \exists x\in B(bRx\wedge x\ne b)$

lower bound and upper bound: Then $a\in A$ is called a lower bound for B if $\forall x\in B\left(aRx\right)$. Similarly, it is called an upper bound if $\forall x\in B\left(xRa\right)$

The only difference between a lower bound and smallest element of B is that smallest element has to be in B while its not mandatory for lower bound.


Least Upper Bound(lub) and Greatest Least Bound(glb): Simply speaking, smallest from the set of all upper bounds of B is called lub and largest from the set of all lower bounds of B is called the glb. LUB is also called "supremum"(sup) and GLB is also called "infimum"(inf).


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Ex-1
(a) Reflexive: Yes
Transitive: Yes
Antisymmetric: Yes

Hence, it is partial order.

It is not total order as none of (a,c) or (c,a) are in R.

(b) Reflexive: Yes
Transitive: Yes
Antisymmetric: No as (-4,4) and (4,-4) both are in R

Hence, it is not partial order.


(c) Reflexive: Yes
Transitive: Yes
Antisymmetric: Yes

Hecne, it is partial order.

It is not total order as none of (4,-4) or (-4,4) are in R


Ex-2
(a) Reflexive: Yes
Transitive: Yes
Antisymmetric: Yes

It is partial order. It is also total order.

(b) Reflexive: Yes
Transitive: Yes
Antisymmetric: Yes

It is partial order. It is also total order.

(c) Reflexive: Yes
Transitive: Yes
Antisymmetric: Yes

It is partial order. It is also total order.


Ex-3
(a) Set of Minimal elements = {2}
Set of Maximal elements = {3,4}
smallest = 2
largest = does not exist

least upper bound = does not exist
greatest lower bound = 2

(a) Set of Minimal elements = {1}
Set of Maximal elements = does not exist
smallest = 1
largest = does not exist

least upper bound = 2
greatest lower bound = 1

(c) Set of Minimal elements = {$\varnothing$}
Set of Maximal elements = {$x\in P\left(N\right)$ | x has exactly 5 elements}
smallest = $\varnothing$
largest = does not exist

least upper bound = does not exist
greatest lower bound = $\varnothing$


Ex-4 ($\to$) Suppose R is both symmetric and antisymmetric. Let x,y be arbitrary elements of A s.t. $(x,y)\in R$. Since R is symmetric, so $(y,x)\in R$. Since R is antisymmetric, so $x=y$. Thus, $(x,y)\in i_A$. Since x,y are arbitrary, so $R\subseteq i_A$

($\leftarrow$) Suppose $R\subseteq i_A$. Let x,y be arbitrary elements of A s.t. $(x,y)\in R$. It follows from the assumption that $(x,y)\in i_A$, hence $x=y$. Thus $(y,x)\in R$. Since x,y are arbitrary, it follows that R is both symmetric and antisymmetric


Ex-5 Suppose R is a partial order on A and $B\subseteq A$.

Let x be an arbitrary element of B. Then $(x,x)\in BXB$, also $(x,x)\in AXA$, so $(x,x)\in R$. Thus $(x,x)\in R\cap \left(BXB\right)$. Since x is arbitrary, so $R\cap \left(BXB\right)$ is reflexive on B

Let x,y,z be arbitrary elements of B s.t. $(x,y)\in R\cap \left(BXB\right)$ and $(y,z)\in R\cap \left(BXB\right)$. It follows that $(x,y)\in R$ and $(y,z)\in R$. Since R is a partial order on A, so R is transitive and hence $(x,z)\in R$. Also $(x,z)\in BXB$. Thus $(x,z)\in R\cap \left(BXB\right)$. Since x,y,z are arbitrary so $R\cap \left(BXB\right)$ is transitive on B

Let x,y be arbitrary elements of B s.t. $(x,y)\in R\cap \left(BXB\right)$. Then $(x,y)\in R$. Since R is partial order and hence antisymmetric on A, so $(x,y)\in R\wedge (y,x)\in R\to x=y$. So $(x,y)\in (R\cap (BXB\left)\right)\wedge (y,x)\in (R\cap (BXB\left)\right)\to x=y$. Since x,y are arbitrary so $R\cap \left(BXB\right)$ is antisymmetric on B

Thus $R\cap \left(BXB\right)$ is partial order on B.


Ex-6 Since $R_1$ and $R_2$ both are partial order on A. So both are reflexive, transitive and antisymmetric on A.
(a) Yes.

Let x be an arbitrary element of A. It follows that $(x,x)\in R_1$ and also $(x,x)\in R_2$. Then $(x,x)\in R_1\cap R_2$. Since x is arbitrary, so $R_1\cap R_2$ is reflexive on A

Let x,y be arbitrary elements of A s.t. $(x,y)\in R_1\cap R_2$. Then $(x,y)\in R_1$ and $(x,y)\in R_2$. Since $R_1$ and $R_2$, both are antisymmetric so, $(x,y)\in R_1\wedge (y,x)\in R_1\to x=y$ and $(x,y)\in R_2\wedge (y,x)\in R_2\to x=y$. Thus $(x,y)\in R_1\cap R_2\wedge (y,x)\in R_1\cap R_2\to x=y$. Since x,y are arbitrary so $R_1\cap R_2$ is antisymmetric on A

Let x,y,z be arbitrary elements of A s.t. $(x,y)\in R_1\cap R_2$ and $(y,z)\in R_1\cap R_2$. Then $(x,y)\in R_1$ and $(y,z)\in R_1$, $(x,y)\in R_2$ and $(y,z)\in R_2$. Since both are transitive on A, so $(x,z)\in R_1$ and $(x,z)\in R_2$. So $(x,z)\in R_1\cap R_2$. Since x,y,z are arbitrary so $R_1\cap R_2$ is transitive on A

Hence $R_1\cap R_2$ is Partial order on A

(b) No. Here is a counterexample.
A = {1,2,3}

$R_1$ = {(1,1), (2,2), (3,3), (1,2)}

$R_2$ = {(1,1), (2,2), (3,3), (2,3)}

$R_1\cup R_2$ = {(1,1), (2,2), (3,3), (1,2), (2,3)}

clearly its not transitive as $(1,2)\in R_1\cup R_2$ and $(2,3)\in R_1\cup R_2$, but $(1,3)\notin R_1\cup R_2$. Hence $R_1\cup R_2$ is not partial order on A.


Ex-7 Suppose $R_1$ is partial order on $A_1$, $R_2$ is partial order on $A_2$ and $A_1\cap A_2=\varnothing$
(a) Let x be an arbitrary element of $A_1\cup A_2$. So either $x\in A_1$ or $x\in A_2$. So either $(x,x)\in R_1$ or $(x,x)\in R_2$. Then $(x,x)\in R_1\cup R_2$. Since x is arbitrary so $R_1\cup R_2$ is reflexive on $A_1\cup A_2$

Let x,y be arbitrary elements of $A_1\cup A_2$ s.t. $(x,y)\in R_1\cup R_2$. Let us consider following exhaustive set of cases
Case#1: $x\in A_1$ and $y\in A_1$
Then $(x,y)\in R_1\wedge (y,x)\in R_1\to x=y$. Thus, $(x,y)\in R_1\cup R_2\wedge (y,x)\in R_1\cup R_2\to x=y$

Case#2: $x\in A_2$ and $y\in A_2$
by a similar argument as above case, $(y,x)\in R_1\cup R_2$

Case#3: $x\in A_1$ and $y\in A_2$
Then $(x,y)\notin R_1$ and $(x,y)\notin R_2$, hence $(x,y)\notin R_1\cup R_2$. That is contradictory to our assumption, hence this case is not possible

Case#4: $x\in A_2$ and $y\in A_1$
This case is also not possible for the same reason as above.

Thus $(x,y)\in R_1\cup R_2\wedge (y,x)\in R_1\cup R_2\to x=y$. Since x,y are arbitrary so $R_1\cup R_2$ is antisymmetric on $A_1\cup A_2$

Let x,y,z be arbitrary elements of $A_1\cup A_2$ s.t. $(x,y)\in R_1\cup R_2$ and $(y,z)\in R_1\cup R_2$. Let us consider following exhaustive set of cases
Case#1: $(x,y)\in R_1$ and $(y,z)\in R_1$
Then $(x,z)\in R_1$ and hence $(x,z)\in R_1\cup R_2$

Case#2: $(x,y)\in R_2$ and $(y,z)\in R_2$
Then $(x,z)\in R_2$ and hence $(x,z)\in R_1\cup R_2$

Case#3: $(x,y)\in R_1$ and $(y,z)\in R_2$
Since $A_1\cap A_2=\varnothing$, so $y$ can not belong to both $A_1$ and $A_2$ and hence this case is not possible

Case#4: $(x,y)\in R_2$ and $(y,z)\in R_3$
For same reason as above, this case is also not possible.

Thus $(x,z)\in R_1\cup R_2$. Since x,y,z is arbitrary so $R_1\cup R_2$ is transitive on $A_1\cup A_2$.

Hence $R_1\cup R_2$ is partial order on $A_1\cup A_2$

(b) It can be proven by using the similar arguments as in part(a)

(c) Suppose $R_1$ and $R_2$ are total orders. Let $x\in A_1$ and $y\in A_2$. Then neither (x,y) nor (y,x) will be in $R_1\cup R_2$. And hence its not total order. On the other hand, $(x,y)\in A_{1}X{A}_2$, so $R_1\cup R_2\cup A_{1}X{A}_2$ is total order.


Ex-8 Let x be an arbitrary element of A and y be an arbitrary element of B. Since R is partial order, so xRx and similarly ySy. Thus $\left(\right(x,y),(x,y\left)\right)\in T$. Since x,y is arbitrary, so T is reflexive in (A X B)

Let x,v be arbitrary elements of A and y,w be arbitrary elements of B. Suppose $\left(\right(x,y),(v,w\left)\right)\in T$ and $\left(\right(v,w),(x,y)\in T$. Since $\left(\right(x,y),(v,w\left)\right)\in T$, so $(x,v)\in R$ and $(y,w)\in S$. Also since $\left(\right(v,w),(x,y\left)\right)\in T$, so $(v,x)\in R$ and $(w,y)\in S$. Since R and S are antisymmetric, it follows that x = v and y = w. Thus, (x,y) = (v,w). Since x,y,v,w are arbitrary so T is antisymmetric

Let x,y,z be arbitrary elements of A and u,v,w be arbitrary elements of B s.t. $\left(\right(x,u),(y,v\left)\right)\in T$ and $\left(\right(y,v),(z,w\left)\right)\in T$. It follows that $(x,y)\in R$ and $(u,v)\in S$, $(y,z)\in R$ and $(v,w)\in S$. Since R and S both are transitive, hence $(x,z)\in R$ and $(u,w)\in S$. Thus $\left(\right(x,u),(z,w\left)\right)\in T$. Since x,y,z,u,v,w are arbitrary so T is transitive.

Hence T is partial order in A X B

Suppose R and S is total order. Let $(x,u)\in AXB$ and $(y,v)\in AXB$ and x,y be arbitrary elements of A and u,v be arbitrary elements of B. It follows that either $(x,y)\in R$ or $(y,x)\in R$ and either $(u,v)\in S$ or $(v,u)\in S$. Let us consider following case,

Case#: $(x,y)\in R$ and $(v,u)\in S$
Clearly, neither $\left(\right(x,u),(y,v\left)\right)\in T$ nor $\left(\right(y,v),(x,u\left)\right)\in T$. And, hence T is not total order.

Ex-9 Let x be arbitrary element of A and y be arbitrary element of B. So $(x,x)\in R$ and $(y,y)\in S$. Thus $\left(\right(x,y),(x,y\left)\right)\in L$. Since x,y are arbitrary so L is reflexive in A X B

Let x,y be arbitrary elements of A and u,v be arbitrary elements of B. Suppose $\left(\right(x,u),(y,v\left)\right)\in L$ and $\left(\right(y,v),(x,u\left)\right)\in L$. Then $(x,y)\in R$ and also $(y,x)\in R$. Hence x = y. And then $(u,v)\in S$ and $(v,u)\in S$, then u = v. Thus (x,u) = (y,v). Since x,y,v,u are arbitrary so L is antisymmetric

Let x,y,z be arbitrary elements of A and u,v,w be arbitrary elements of B s.t. $\left(\right(x,u),(y,v\left)\right)\in L$ and $\left(\right(y,v),(z,w\left)\right)\in L$. It follows that $(x,y)\in R$ and $(y,z)\in R$, then $(x,z)\in R$. Now consider following cases
Case#1: $x\ne z$
Then $\left(\right(x,u),(z,w\left)\right)\in L$

Case#2: x = z
Since $(x,y)\in R$ and $(y,z)\in R$. Since x = z, so $(y,x)\in R$ as well. Since R is antisymmetric, so y = x. It follows that $(u,v)\in S$ and $(v,w)\in S$, then $(u,w)\in S$. Then $\left(\right(x,u),(z,w\left)\right)\in L$

Thus $\left(\right(x,u),(z,w\left)\right)\in L$. Since x,y,z,u,v,w are arbitrary so L is transitive. Hence L is partial order on A X B

Suppose R and S are total order. Let (x,u) and (y,v) be arbitrary elements of A X B. It follows that either $(x,y)\in R$ or $(y,x)\in R$ and either $(u,v)\in S$ or $(v,u)\in S$. Let us consider the case when $(x,y)\in R$ and $(v,u)\in S$ and x = y, then It is not necessary that one of the two ordered pairs formed by (x,y) and (y,v) are in L. Hence L is not total order.


Ex-10 Let x and y be arbitrary elements of A.
($\to$) Suppose $(x,y)\in R$. Let u be an arbitrary element of $P_x$. It follows that $(u,x)\in R$. R is partial order on A and hence transitive, so $(u,y)\in R$. Thus $u\in P_y$. Since u is arbitrary, so $P_x\subseteq P_y$.

($\leftarrow$) Suppose $P_x\subseteq P_y$. Clearly, $x\in P_x$ because $(x,x)\in R$. Then, it follows from our assumption that $x\in P_y$ also, so $(x,y)\in R$.


Ex-11 D = {$(x,y)\in Z^{+}$ X $Z^{+}$ | x divides y}
B = {$x\in Z\mid x>1$}

D-Minimal elements of B are all the prime numbers = {2,3,5,7,...}

As there is no element in B that divides every other element in B, so there does not exist a smallest element.


Ex-12 The implicit relation in this exercise is the subset-of relation.

given: B = {x" x =" {$Ex-13\; Let\; x\; be\; an\; arbitrary\; element\; of\; A.\; Then\; ">x"style="font-weight:bold;">Ex-13LetxbeanarbitraryelementofA.Then$(x,x) \in R$,so$(x,x) \in R^{-1}$also.Sincexisarbitrary,so$R^{-1}$isreflexiveinA$

Let x,y be arbitrary elements of A s.t. $(x,y)\in R^{-1}$ and $(y,x)\in R^{-1}$. Then $(y,x)\in R$ and also $(x,y)\in R$. But since R is antisymmetric, so x = y. Since x,y are arbitrary, so $R^{-1}$ is antisymmetric

Let x,y,z be arbitrary elements of A s.t. $(x,y)\in R^{-1}$ and $(y,z)\in R^{-1}$. Then $(y,x)\in R$ and $(z,y)\in R$. Since R is transitive, so $(z,x)\in R$. Thus $(x,z)\in R^{-1}$. Since x,y,z are arbitrary so $R^{-1}$ is transitive

Hence $R^{-1}$ is partial order in A.

Suppose R is total order. Let x,y be arbitrary elements of A. Then either $(x,y)\in R$ or $(y,x)\in R$. Thus either $(y,x)\in R^{-1}$ or $(x,y)\in R^{-1}$. Hence $R^{-1}$ is total order too.


Ex-14
(a) Let $b\in B$ and b is R-largest element of B iff
$\forall b\in B\left(xRb\right)$ iff
$\forall b\in B\left(b{R}^{-1}x\right)$ iff
b is the $R^{-1}$-smallest element of B

(b) Let $b\in B$ and b is R-maximal element of B iff
$\neg \exists x\in B\left(bRx\right)\wedge x\ne b$ iff
$\neg \exists x\in B\left(x{R}^{-1}b\right)\wedge x\ne b$ iff
b is the $R^{-1}$-minimal element of B


Ex-15
(a) Suppose b is the $R_1$-smallest element of B. Let x be arbitrary element of B. It follows that $\left(b{R}_{1}x\right)$. Since $R_1\subseteq R_2$, so $\left(b{R}_{2}x\right)$. Since x is arbitrary, so b is $R_2$-smallest element of B also.

(b) Suppose b is the $R_1$-minimal element of B. Then $\neg \exists x\in B\left(x{R}_{1}b\right)\wedge x\ne b$. Since $R_1\subseteq R_2$, so $\neg \exists x\in B\left(x{R}_{2}b\right)\wedge x\ne b$. Hence b is $R_2$-minimal element of B also.


Ex-16 Suppose b is the largest element of B. We'll prove by contradiction. Assume b is not the maximal element of B, then we can chose a particular element x in B s.t. $\left(bRx\right)$ and $x\ne b$. But b is the largest element of B so its not possible. Hence b is the maximal element of B.

Again we will use contradiction to prove that b is the only maximal element. Assume there exists another maximal element called c. Then, $\neg \exists x\in B\left(cRx\right)\wedge x\ne c$. Since b is largest element of B, so $\left(cRb\right)$. Thus, (c = b). Hence b is the unique maximal element of B.


Ex-17 Suppose b is the one and only minimal element of a subset(calling it B) of a partially ordered set with respect to a relation called R. Assume there exists a smallest element $c\in B$. Then $\left(cRb\right)$. But since b is the only minimal element, so (b = c). Hence b is the smallest element also. So if the smallest element *exists* then it is b. But it may not exist at all. So, the given statement is not true.

I cheated and looked for this counterexample in the hints section.

A = {R X R}
R = {((x,y),(u,v)) \in A X A | x \leq u \land y \leq v}
B = {(0,0)} $\cup$ ({1} X $R$)

Let us first prove that R is partial order on A.

Let x, y be arbitrary elements of A. clearly $\left(\right(x,y),(x,y\left)\right)\in R$ as ((x = x) and (y = y)). Thus R is reflexive on A.

Let x,y,u,v be arbitrary elements of A s.t. $\left(\right(x,y),(u,v\left)\right)\in R$ and $\left(\right(u,v),(x,y\left)\right)\in R$. Then $(x\le u)and(u\le x)$, $(y\le v)and(v\le y)$. So, (x = u) and (y = v) and hence (x,y) = (u,v). Thus R is antisymmetric on A

Let x,y,z,u,v,w be arbitrary elements of A s.t. $\left(\right(x,u),(y,v\left)\right)\in R$ and $\left(\right(y,v),(z,w\left)\right)\in R$. Then $x\le y$ and $y\le z$, hence $x\le z$. Also $u\le v$ and $v\le w$, hence $u\le w$. Thus $\left(\right(x,u),(z,w\left)\right)\in R$. Thus R is transitive.

Hence R is partial order on A.

clearly (0,0) is one and only minimal element of B but it is not the smallest as pairs like ((0,0),(1,-1)) are not in R.


Ex-18
(a) ($\to$) Suppose x is arbitrary element of A and x is an upper bound of $B_1$. Let y be an arbitrary element of $B_2$. Then we can chose a z in $B_1$ s.t. (yRz). Since x is an upper bound of $B_1$, so $zRx$. Since R is transitive, so $yRx$. Since y is arbitrary element of $B_2$, so x is an upper bound of $B_2$.

($\leftarrow$) It can be proven by using exactly a same statement as above.

(b) Suppose $B_1$ and $B_2$ are disjoint. We'll prove by contradiction. Let $b\in B_1$ be a maximal element of $B_1$. Since $b\in B_1$, so we can choose an element x in $B_2$ s.t. $bRx$. Since $B_1$ and $B_2$ are disjoint, so $b\ne x$. Since $x\in B_2$, so we can choose a $y\in B_1$ s.t. (xRy). Since (bRx) and (xRy), so $b\ne y$ as R is antisymmetric. Since R is transitive, so (bRy), which is a contradiction as b is a maximal element of B. Thus no maximal element exists for $B_1$. By similar argument, we say that no maximal element exists for $B_2$ also.


Ex-19
(a) Let us try to analyze the theorem, and in the process we'll try to find out what is wrong with the given proof.

given: R is a total order on A, $B\subseteq A$
goal: every element of B is either the smallest or the largest element of B i.e. $\forall b\in B$ (b is smallest) $\vee$ (b is largest).

Let b be arbitrary element of B.

given: R is a total order on A, $B\subseteq A$, b is arbitrary element of B
goal: either b is smallest or b is largest i.e. $\forall x\in B\left(bRx\right)$ $\vee$ $\forall x\in B\left(xRb\right)$

ah.. so flaw in the given proof is that it proves that $\forall x\in B\left(\right(bRx)\vee (xRb\left)\right)$ wherease what we need to prove is that $(\forall x\in B(bRx\left)\right)$ $\vee$ $(\forall x\in B(xRb\left)\right)$ and clearly, two statements are not equivalent

(b) Given theorem is incorrect and here is the counterexample.
A = {1,2,3}
R = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)}
B = {1,2,3}

We can see that, in this case R is total order on A but 2 is neither smallest nor largest element of B


Ex-20
(a) Suppose b is the smallest element of B. Let $a\in A$ be an arbitrary lower bound of B and x be an arbitrary element of B. Then (aRx). Let us choose x = b, so (aRb). So b is larger than a. Since a is arbitrary, so b is the greatest lower bound of B

(b) Suppose b is the largest element of B. Let $a\in A$ be an arbitrary upper bound of B and x be an arbitrary element of B. Then (xRa). Let us choose x = b, then (bRa). So b is smaller than a. Since a is arbitrary, so b is the least upper bound of B


Ex-21
(a) Suppose x be an arbitrary element of U and y is some element in A s.t. (xRy). Let b be an arbitrary element of B. Since x is an upper bound of B, so (bRx). Since (bRx) and (xRy), R is transitive so (bRy). Thus y is also an upper bound of B, hence $y\in U$.

(b) Let b be an arbitrary element of B and u be an arbitrary element of U. It follows that (bRu). Since u is arbitrary, so b is clearly a lower bound of U. Since b is arbitrary, so every element of B is a lower bound for U.

(c) Suppose x is the greatest lower bound of U. In part(b) we proved that every element of B is a lower bound of U, so x must be larger than all of them and hence x is an upper bound of B. But, all the elements of U are upper bounds of B and x is lower bound of U, so x must be the least upper bound of B.


Ex-22 Suppose $B_1\subseteq B_2$. Let $b_1$ be arbitrary element of $B_1$, then $\left(b_{1}R{x}_1\right)$. It follows from our assumption that $b_1\in B_2$. So $\left(b_{1}R{x}_2\right)$, hence $x_2$ is also an upper bound of $B_1$. Since $x_1$ is least upper bound of $B_1$, so it is smaller than any other upper bound of $B_1$, hence $\left(x_{1}R{x}_2\right)$.


Ex-23 There are two parts to be proven.

1. Least upper bound of $F$ is $\cup F$
$\cup F$ is the union of all the sets in $F$. so clearly every element of $F$ is a subset of $\cup F$, hence $\cup F$ is clearly an upper bound of $F$. Let X be an arbitrary element of $P\left(A\right)$ s.t. X is upper bound of $F$. Let f be an arbitrary element of $\cup F$, then we can choose a set B in $F$ s.t. $f\in B$. Since X is upper bound of $F$, so $B\subseteq X$ and hence $f\in X$. Since f is arbitrary, so $\cup F\subseteq X$. Since X is arbitrary so $\cup F$ is the least upper bound of $F$.

$\cap F$ is greatest lower bound of $F$ can also be proven using a similar argument. And, here it goes...

$\cap F$ is clearly a lower bound of $F$ as $\forall S\in F(\cap F\subseteq S)$
Let X be an arbitrary element of $P\left(A\right)$ s.t. X is lower bound of $F$.
That implies, $\forall S\in F(X\subseteq S)$
So, $X\subseteq \cap F$
Since X is arbitrary so $\cap F$ is the greatest lower bound of $F$