# How to Prove It - Solutions

## Chapter - 4, Relations

### Summary

• Antisymmetric Relation:
• Suppose $R$ is a relation on a set $A$. Then $R$ is said to be antisymmetric if $\forall x \in A \forall y \in A((xRy \land yRx) \to x = y)$.
• For example:
1. $x \le y$, where $x,y \in \mathbb R$.
2. $x \subseteq y$, where $x,y \subseteq A$.
• It can be seen from the above examples, that $y$ is atleast as $x$ . Thus antisymmetric relation gives a sense of ordering.
• Partial Order: Consider $R$ a relation defined on set $A$. $R$ will be called partial order on $A$, if $R$ is reflexive, transitive and antisymmetric.
• Total Order: Consider $R$ a relation defined on set $A$. $R$ will be called total order, if $R$ is partial order and $\forall x \in A \forall y \in A(xRy \lor yRx)$. As it can be seen from the definition, every two elements of the set $A$ are in $R$. Also, since $R$ is antisymmetric, everyone of these elements/objects have a sense of ordering with respect to every other element/object.
• If $R$ is a partial order on a set $A$. Suppose $B \subseteq A$:
• Smallest and Minimal element(s):
• Suppose $b \in B$. If $\forall x \in B(bRx)$, then $b$ is called $R$-smallest element of $B$. Also, if $\lnot \exists x \in B(xRb \land x \ne b)$ is called an $R$-minimal element of $B$.
• If $B$ has a smallest element then this element is unique.
• If $b$ is the smallest element of B, then $b$ is also a minimal element of $B$, and it is the only minimal element.
• If $R$ is a total order and $b$ is a minimal element of $B$, then $b$ is the smallest element of $B$.
• Largest and Maximal element(s):
• If $\forall x \in B(xRb)$, then $b$ is called $R$-largest element of $B$. Also, if $\lnot \exists x \in B(bRx \land x \ne b)$ is called an $R$-maximal element of $B$.
• Other properties of largest and maximal are similar to smallest and minimal as above.
• Lower Bound and Upper Bound:
• Suppose $R$ is a partial order on $A$, $B \subseteq A$, and $a \in A$. Then $a$ is called a lower bound for $B$ if $\forall x \in B(aRx)$. Similarly, it is an upper bound for $B$ if $\forall x \in B(xRa)$.
• Thus there can be multiple lower bounds or upper bounds for a set.
• Note that lower/upper bound need not be part of the set $B$. However, smallest/largest element of $B$ must be the part of set $B$.
• Least Upper Bound and Greatest Lower Bound:
Suppose $R$ is a partial order on $A$ and $B \subseteq A$. Let $U$ be the set of all upper bounds for $B$, and let $L$ be the set of all lower bounds. If $U$ has a smallest element, then this smallest element is called the least upper bound of $B$. If $L$ has a largest element, then this largest element is called the greatest lower bound of $B$. The phrases least upper bound and greatest lower bound are sometimes abbreviated l.u.b. and g.l.b.
• Suppose $A$ is a set, $\mathcal F \subseteq P (A)$, and $\mathcal F \ne \phi$. Then the least upper bound of $\mathcal F$ (in the subset partial order) is $\cup \mathcal F$ and the greatest lower bound of $\mathcal F$ is $\cap \mathcal F$.

Soln1

(a)

Reflexive: True Transitive: True Antisymmetric: True

Thus it is partial order. It is not total order since $(a,c) \notin R$.

(b)

Reflexive: True Transitive: True Antisymmetric: False, Because $(-1,1) \in R \land (1,-1) \in R$ but $-1 \ne 1$.

Thus it is not partial order.

(c)

Reflexive: True Transitive: True Antisymmetric: True

Thus it is partial order. It is not total order because $(-1,1) \notin R$.

Soln2

(a)

Reflexive: True Transitive: True Antisymmetric: True

Thus it is partial order. It is also total order because every two english words, say $x$ and $y$, then either $(x,y) \in R$ or $(y,x) \in R$.

(b)

Reflexive: True Transitive: True Antisymmetric: False. For example $(apple,ape) \in R \land (ape,apple) \in R$ but $apple != ape$.

(c)

Reflexive: True Transitive: True Antisymmetric: False. Suppose $x$ and $y$ are two different countries with same population. Then $(x,y) \in R \land (y,x) \in R$ but $x != y$.

Soln3

(a)

$R = \{ (1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(4,4) \}, B = \{2,3,4\}$.

Minimal elements = $\{ 2 \}$.
Smallest element = $2$
Greatest Lower Bound = 2.

Maximal elements = $\{3, 4 \}$.
Largest = N.A.
Least Upper Bound = N.A.

(b)

$% $.

Minimal elements = $\{ 1 \}$.
Smallest element = $1$
Greatest Lower Bound = 1.

Maximal elements = N.A.
Largest = N.A.
Least Upper Bound = 2

(c)

Minimal elements = $\{ \phi \}$.
Smallest element = $\phi$
Greatest Lower Bound = $\phi$.

Maximal elements = $\{x \in \mathcal P(\mathbb N) \, \vert \, \text{x contains exactly 5 elements.} \}$
Largest = N.A.
Least Upper Bound = N.A.

Soln4

($\to$)Suppose $R$ is symmetric and antisymmetric. Suppose $(x,y)$ in $R$. Since $R$ is symmetric $xRy \land yRx$. Since $xRy \land yRx$ and $R$ is antisymmetric, it follows that $x = y$. Thus $(x,y) \in i_A$. Since $(x,y)$ is arbitrary, it follows that $R \subseteq i_A$.

($\leftarrow$)Suppose $R \subseteq i_A$. Suppose $(x,y) \in R$, thus $(x,y) \in i_A$. Since $(x,y) \in i_A$, it follows $(x,y) \in i_A \land (y,x) \in i_A$. Thus $R$ is symmetric. Now to prove antisymmetric, suppose $(x,y) \in R \land (y,x) \in R$. Since $R \subseteq i_A$, it follows $(x,y) \in i_A$. since $(x,y) \in i_A$, it follows that $x = y$. Thus $R$ is antisymmetric.

Soln5

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

Reflexive:

Suppose $x$ is arbitrary element of $B$. Since $B \subseteq A$, $x \in A$. Since $R$ is reflexive, if follows that $(x,x) \in R$. Since $x \in B$, it follows $(x,x) \in B \times B$. Thus if $x \in B$, then $(x,x) \in R \cap (B \times B)$. Since $x$ is arbitrary, we can conclude that $R \cap (B \times B)$ is reflexive.

Transitive:

Suppose $x,y,z$ are arbitrary elements of $B$ such that $(x,y) \in R \cap (B \times B) \land (y,z) \in R \cap (B \times B)$. Thus $(x,y) \in R \land (y,z) \in R$. Since $R$ is transitive, it follows $(x,z) \in R$. Since $x \in B \land z \in B$, thus $(x,z) \in B \times B$. Thus if $(x,y) \in R \cap (B \times B) \land (y,z) \in R \cap (B \times B)$, then $(x,z) \in R \cap (B \times B)$. Since $x,y,z$, are arbitrary, we can conclude that $R \cap (B \times B)$ is transitive.

Antisymmetric:

Suppose $x,y$ are arbitrary elements of $B$ such that $(x,y) \in R \cap (B \times B) \land (y,x) \in R \cap (B \times B)$. Thus $(x,y) \in R \land (y,x) \in R$. Since $R$ is antisymmetric, it follows $x = y$. Thus if $(x,y) \in R \cap (B \times B) \land (y,x) \in R \cap (B \times B)$, then $x = y$. Since $(x,y)$, is arbitrary, we can conclude that $R \cap (B \times B)$ is antisymmetric.

Since $R \cap (B \times B)$ is reflexive, transitive, antisymmetric, it follows that $R \cap (B \times B)$ is partial order on $B$.

Soln6

(a) Suppose $R_1$ and $R_2$ are partial orders on $A$.

Suppose $x \in A$. Since $R_1$ is reflexive, $(x,x) \in R_1$. Similarly since $R_2$ is reflexive, $(x,x) \in R_2$. Thus $(x,x) \in R_1 \cap R_2$. Since $x$ is arbitrary, $R_1 \cap R_2$ is reflexive.

Suppose $x,y,z$ are arbitrary elements of $A$ such that $(x,y) \in R_1 \cap R_2 \land (y,z) \in R_1 \cap R_2$. It follows that $(x,y) \in R_1 \land (y,z) \in R_1$. Since $R_1$ is transitive, it follows $(x,z) \in R_1$. Similarly since $(x,y) \in R_2 \land (y,z) \in R_2$ and $R_2$ is transitive, it follows that $(x,z) \in R_2$. Thus we have $(x,z) \in R_1 \cap R_2$. Since $x,y,z$ are arbitrary, it follows that $R_1 \cap R_2$ is transitive.

Suppose $x,y$ are arbitrary elements of $A$ such that $(x,y) \in R_1 \cap R_2 \land (y,x) \in R_1 \cap R_2$. Thus $(x,y) \in R_1 \land (y,x) \in R_1$. Since $R_1$ is antisymmetric, it follows $x = y$. Thus if $(x,y) \in R_1 \cap R_2 \land (y,x) \in R_1 \cap R_2$, then $x = y$. Since $(x,y)$, is arbitrary, we can conclude that $R_1 \cap R_2$ is antisymmetric.

Thus $R_1 \cap R_2$ is partial order on $A$.

(b) False. Counter example:

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

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

As we can see $R_1$ and $R_2$ are partial order. But $R_1 \cup R_2$ is not partial order because it is not transitive.

Soln7

Suppose $R_1$ is partial order on $A_1$ and $R_2$ is partial order on $A_2$.

(a)

Reflexive:

Suppose $x \in A_1 \cup A_2$. Two possible cases:

• Case 1: $x \in A_1$
Since $R_1$ is partial order, $(x,x) \in R_1$. Or we can also say $(x,x) \in R_1 \cup R_2$.

• Case 2: $x \in A_2$
Since $R_2$ is partial order, $(x,x) \in R_2$. Or we can also say $(x,x) \in R_1 \cup R_2$.

Thus from all cases, $R_1 \cup R_2$ is reflexive on $A_1 \cup A_2$.

Transitive:

Suppose $x,y,z$ are arbitrary elements of $A_1 \cup A_2$ such that $(x,y) \in R_1 \cup R_2 \land (y,z) \in R_1 \cup R_2$. We have following possible cases:

• Case 1: $(x,y) \in R_1 \land (y,z) \in R_1$:
Since $R_1$ is transitive, it follows $(x,z) \in R_1$. We can also say $(x,z) \in R_1 cup R_2$.

• Case 2: $(x,y) \in R_2 \land (y,z) \in R_2$:
Since $R_2$ is transitive, it follows $(x,z) \in R_2$. We can also say $(x,z) \in R_1 cup R_2$.

• Case 3: $(x,y) \in R_1 \land (y,z) \in R_2$:
• Case 4: $(x,y) \in R_2 \land (y,z) \in R_1$:
Both cases(3 & 4) are not possible, since both of them require $y \in A_1 \cap A_ 2$. But $A_1 \cap A_2 = \phi$, so these cases do not exist.

Thus from all cases, $R_1 \cup R_2$ is transitive on $A_1 \cup A_2$.

Antisymmetric:

Suppose $x,y$ are arbitrary elements of $A_1 \cup A_2$ such that $(x,y) \in R_1 \cup R_2 \land (y,x) \in R_1 \cup R_2$. we have following possible cases:

• Case 1: $(x,y) \in R_1 \land (y,x) \in R_1$:
Since $R_1$ is antisymmetric, it follows $x = y$.

• Case 2: $(x,y) \in R_2 \land (y,x) \in R_2$:
Since $R_2$ is antisymmetric, it follows $x = y$.

• Case 3: $(x,y) \in R_1 \land (y,x) \in R_2$:
• Case 4: $(x,y) \in R_2 \land (y,x) \in R_1$:
Both cases(3 & 4) are not possible, since both of them require $y \in A_1 \cap A_ 2$. But $A_1 \cap A_2 = \phi$, so these cases do not exist.

Thus from both cases, $R_1 \cup R_2$ is antisymmetric on $A_1 \cup A_2$.

As we can see now, $R_1 \cup R_2$ is reflexive, transitive and antisymmetric, it follows that $R_1 \cup R_2$ is partial order on $A_1 \cup A_2$.

(b)

Reflexive:

Suppose $x \in A_1 \cup A_2$. Two possible cases:

• Case 1: $x \in A_1$
Since $R_1$ is partial order, $(x,x) \in R_1$. Or we can also say $(x,x) \in R_1 \cup R_2 \cup (A_1 \times A_2)$.

• Case 2: $x \in A_2$
Since $R_2$ is partial order, $(x,x) \in R_2$. Or we can also say $(x,x) \in R_1 \cup R_2 \cup (A_1 \times A_2)$.

Thus from all cases, $R_1 \cup R_2 \cup (A_1 \times A_2)$ is reflexive on $A_1 \cup A_2$.

Transitive:

Suppose $x,y,z$ are arbitrary elements of $A_1 \cup A_2$ such that $(x,y) \in (R_1 \cup R_2) \cup (A_1 \times A_2) \land (y,z) \in (R_1 \cup R_2) \cup (A_1 \times A_2)$. We have following possible cases:

• Case 1: $(x,y) \in (R_1 \cup R_2) \land (y,z) \in (R_1 \cup R_2)$:
Since $R_1 \cup R_2$ is transitive(proved in part (a)), it follows $(x,z) \in R_1 \cup R_2$. We can also say $(x,z) \in (R_1 \cup R_2) \cup (A_1 \times A_2)$.

• Case 2: $(x,y) \in A_1 \times A_2 \land (y,z) \in A_1 \times A_2$:
Thus $y \in A_1 \land y \in A_2$. This is not possible since $A_1 \cap A_2 = \phi$.

• Case 3: $(x,y) \in (R_1 \cup R_2) \land (y,z) \in A_1 \times A_2$:
Here two further cases are possible:
• Case a: $(x,y) \in R_1 \land (y,z) \in A_1 \times A_2$:
Thus $x \in A_1 \land y \in A_1 \land z \in A_2$. Clearly $(x,z) \in A_1 \times A_2$. Or we can also say $(x,z) \in (R_1 \cup R_2) \cup (A_1 \times A_2)$.

• Case b: $(x,y) \in R_2 \land (y,z) \in A_1 \times A_2$:
Thus $y \in A_2 \land y \in A_1$. Clearly $(x,z) \in A_1 \times A_2$. This is not possible since $A_1 \cap A_2 = \phi$.

• Case 4: $(x,y) \in A_1 \times A_2 \land (y,z) \in (R_1 \cup R_2)$:
Here two further cases are possible:
• Case a: $(x,y) \in A_1 \times A_2 \land (y,z) \in R_1$:
Thus $y \in A_1 \land y \in A_2$. This is not possible since $A_1 \cap A_2 = \phi$.

• Case b: $(x,y) \in A_1 \times A_2 \land (y,z) \in R_2$:
Thus $x \in A_1 \land y \in A_2 \land z \in A_2$. Clearly $(x,z) \in A_1 \times A_2$. Or we can also say $(x,z) \in (R_1 \cup R_2) \cup (A_1 \times A_2)$.

Thus from all cases, $R_1 \cup R_2 \cup (A_1 \times A_2)$ is transitive on $A_1 \cup A_2$.

Antisymmetric:

Suppose $x,y$ are arbitrary elements of $A_1 \cup A_2$ such that $(x,y) \in (R_1 \cup R_2) \cup (A_1 \times A_2) \land (y,x) \in (R_1 \cup R_2) \times (A_1 \cup A_2)$. we have following possible cases:

• Case 1: $(x,y) \in R_1 \cup R_2 \land (y,x) \in R_1 \cup R_2$:
Since we know from last part that $R_1 \cup R_2 is antisymmetric, thus$ x = y .

• Case 2: $(x,y) \in A_1 \times A_2 \land (y,x) \in A_1 \times A_2$:
Thus $y \in A_1 \land y \in A_2$. This is not possible since $A_1 \cap A_2 = \phi$.

• Case 3: $(x,y) \in R_1 \cup R_2 \land (y,x) \in A_1 \times A_2$:
Here two further cases are possible:
• Case a: $(x,y) \in R_1 \land (y,x) \in A_1 \times A_2$:
Thus $x \in A_1 \land x \in A_2$. This is not possible since $A_1 \cap A_2 = \phi$.

• Case b: $(x,y) \in R_2 \land (y,x) \in A_1 \times A_2$:
Thus $y \in A_2 \land y \in A_2$. This is not possible since $A_1 \cap A_2 = \phi$.

• Case 4: $(x,y) \in A_1 \times A_2 \land (y,x) \in R_1 \cup R_2$:
Here two further cases are possible:
• Case a: $(x,y) \in A_1 \times A_2 \land (y,x) \in R_1$:
Thus $y \in A_2 \land y \in A_1$. This is not possible since $A_1 \cap A_2 = \phi$.

• Case b: $(x,y) \in A_1 \times A_2 \land (y,x) \in R_2$:
Thus $x \in A_1 \land x \in A_2$. This is not possible since $A_1 \cap A_2 = \phi$.

Thus from all cases, $R_1 \cup R_2 \cup (A_1 \times A_2)$ is antisymmetric on $A_1 \cup A_2$.

Thus from $R_1 \cup R_2 \cup (A_1 \times A_2)$ is partial order.

(c)

Suppose $R_1$ and $R_2$ are total order on $A_1$ and $A_2$ respectively.

$R_1 \cup R_2$ is not total order on $A_1 \cup A_2$. Counter example: Suppose $x \in A_1 \land y \in A_2$, then $x,y \text{are both} \in A_1 \cup A_2$ but $(x,y) \notin R_1$, and also $(x,y) \notin R_2$ because $A_1 \cap A_2 = \phi$.

Now suppose $(x,y) \in A_1 \cup A_2$, then we have following possible cases:

• Case 1: $x \in A_1 \land y \in A_1$:
Since $R_1$ is total order on $A_1$, $(x,y) \in R_1$ or $(y,x) \in R_1$. Thus we can also say $(x,y) \in R_1 \cup R_2 \cup (A_1 \times A_2)$ or $(y,x) \in R_1 \cup R_2 \cup (A_1 \times A_2)$.

• Case 2: $x \in A_2 \land y \in A_2$:
Since $R_2$ is total order on $A_2$, $(x,y) \in R_2$ or $(y,x) \in R_2$. Thus we can also say $(x,y) \in R_1 \cup R_2 \cup (A_1 \times A_2)$ or $(y,x) \in R_1 \cup R_2 \cup (A_1 \times A_2)$.

• Case 3: $x \in A_1 \land y \in A_2$:
Thus $(x,y) \in A_1 \times A_2$. Thus we can also say $(x,y) \in R_1 \cup R_2 \cup (A_1 \times A_2)$ or $(y,x) \in R_1 \cup R_2 \cup (A_1 \times A_2)$.

• Case 4: $x \in A_2 \land y \in A_1$:
Thus $(y,x) \in A_1 \times A_2$. Thus we can also say $(x,y) \in R_1 \cup R_2 \cup (A_1 \times A_2)$ or $(y,x) \in R_1 \cup R_2 \cup (A_1 \times A_2)$.

Thus from all the cases, we can conclude that $R_1 \cup R_2 \cup (A_1 \times A_2)$ is total order.

Soln8

Reflexive: Suppose $(a,b) \in A \times B$. Since $R$ is reflexive, it follows $(a,a) \in R$. Similarly since $S$ is reflexive, it follows that $(b,b) \in S$. Thus $((a,b),(a,b)) \in T$. Since $(a,b)$ is arbitrary, it follows that $T$ is reflexive.

Transitive:

Suppose $(a,b),(c,d),(e,f) \in A \times B$ such that $((a,b),(c,d)) \in T \land ((c,d),(e,f)) \in T$. Since $((a,b),(c,d)) \in T$, it follows $(a,c) \in R \land (c,e) \in R$. Since $R$ is transitive, $(a,e) \in R$. Similarly, since $S$ is transitive, and $(b,d) \in S \land (d,f) \in A$, follows that $(b,f) \in S$. Now since $(a,e) \in R \land (b,f) \in S$, it follows that $((a,b),(e,f)) \in T$. Since $(a,b),(c,d),(e,f)$ is arbitrary, we can conclude that $T$ is transitive.

Antisymmetric:

Suppose $(a,b),(c,d) \in A \times B$ such that $((a,b),(c,d)) \in T \land ((c,d),(a,b)) \in T$. Thus $(a,c) \in R \land (c,a) \in R$. Since $R$ is antisymmetric, it follows that $a = c$. Similarly since $(b,d) \in S \land (d,b) \in S$ and $S$ is transitive, it follows that $b = d$. Thus $(a,b) = (c,d)$. Since $(a,b), (c,d)$ are arbitrary, it follows that $T$ is partial order.

Thus $T$ is partial order.

Total Order:

$T$ is not total order even if $R$ and $S$ are full order. Suppose $(x,y) \in R \land (y,x) \notin R$ and $(a,b) \in S \land (b,a) \notin S$. Then neither $((y,a),(x,b)) \in T$ and nor $((x,b),(y,a)) \in T$ is true.

Soln9

Note: Proof of transitivity can be done in a shorter way as suggested in the comment below.

Reflexive: Suppose $(a,b) \in A \times B$. Since $R$ is reflexive, it follows $(a,a) \in R$. Similarly since $S$ is reflexive, it follows that $(b,b) \in S$. Thus $((a,b),(a,b)) \in L$. Since $(a,b)$ is arbitrary, it follows that $T$ is reflexive.

Transitive:

Suppose $(a,b),(c,d),(e,f) \in A \times B$ such that $((a,b),(c,d)) \in L \land ((c,d),(e,f)) \in L$. Since $((a,b),(c,d)) \in L$, it follows $(a,c) \in R \land (c,e) \in R$. Since $R$ is transitive, $(a,e) \in R$. Now we have following cases:

Case 1: $a = c \land c = e$ Since $((a,b),(c,d)) \in L \land a = c$, it follows $(b,d) \in S$. Similarly since $((c,d),(e,f)) \in L \land c = e$, it follows $(d,f) \in S$. Now since $(a,e) \in R$ and $a = e$ and $(b,f) in S$, it follows $((a,b),(e,f)) \in L$.

Case 2: $a = c \land c \ne e$ Since $a = c \land c \ne e$, it follows $a \ne e$. Since $(a,e) \in R$, it follows $((a,b),(e,f)) \in L$.

Case 3: $a \ne c \land c = e$ Since $a \ne c \land c = e$, it follows $a \ne e$. Since $(a,e) \in R$, it follows $((a,b),(e,f)) \in L$.

Case 4: $a \ne c \land c \ne e$ Since $a \ne c \land c \ne e$, it may be either $a = e$, or $a \ne e$. Suppose $a = e$, then since $(c,e) \in R$, it follows $(c,a) \in R$. Thus $(a,c) \in R \land (c,a) \in R$ and $R$ is antisymmetric, it follows $a = c$. But we have $a \ne c$. Thus our assumption $a = e$ is not correct. Or $a \ne e$. Now since $(a,e) \in R \land a \ne e$, it follows $((a,b),(e,f)) \in L$.

Thus from all cases $L$ is transitive.

Antisymmetric:

Suppose $(a,b),(c,d) \in A \times B$ such that $((a,b),(c,d)) \in L \land ((c,d),(a,b)) \in L$. Since $((a,b),(c,d)) \in L$, it follows $(a,c) \in R$. Similarly since $((c,d),(a,b)) \in L$, it follows $(c,a) \in R$. Since $(a,c) \in R \land (c,a) \in R$ and $R$ is antisymmetric, it follows $a = c$. Since $a = c$ and $((a,b),(c,d)) \in L$, it follows $(b,d) \in S$. Also, since $((c,d),(a,b)) \in L$ and $a = c$, it follows $(d,b) \in S$. Since $S$ is antisymmetric, it follows $b = d$. Thus $a = c \land b = d$, or we can say $(a,b) = (c,d)$. Thus $L$ is antisymmetric.

Thus $L$ is partial order.

Total Order:

Suppose $R$ and $S$ are total orders.

Now suppose $(x,y),(a,b)$ are in $A \times B$. Thus $a$ and $x$ are in $A$ and $y$ and $b$ are in $B$. Since $R$ is total order on $A$ and $S$ is total order on $B$. Thus we have $(x,a) \in R \lor (a,x) \in R$ and $(y,b) \in S \lor (b,y) \in S$. Thus we have following cases:

• Case 1: $(x,a) \in R \land (y,b) \in S$.
• Case a: $x = a$
Thus $((x,y),(a,b)) \in L$. We can also say $((x,y),(a,b)) \in L \lor ((a,b),(x,y)) \in L$.
• Case b: $x \ne a$
Thus $((x,y),(a,b)) \in L$. We can also say $((x,y),(a,b)) \in L \lor ((a,b),(x,y)) \in L$.
• Case 2: $(x,a) \in R \land (b,y) \in S$.
• Case a: $x = a$
Since $x = a$, if follows $(a,x) \in R$. Thus $((a,b),(x,y)) \in L$. We can also say $((x,y),(a,b)) \in L \lor ((a,b),(x,y)) \in L$.

• Case b: $x \ne a$
Thus $((x,y),(a,b)) \in L$. We can also say $((x,y),(a,b)) \in L \lor ((a,b),(x,y)) \in L$.

• Case 3: $(a,x) \in R \land (y,b) \in S$.
• Case a: $x = a$
Since $x = a$, if follows $(x,a) \in R$. Thus $((x,y),(a,b)) \in L$. We can also say $((x,y),(a,b)) \in L \lor ((a,b),(x,y)) \in L$.

• Case b: $x \ne a$
Thus $((a,b),(x,y)) \in L$. We can also say $((x,y),(a,b)) \in L \lor ((a,b),(x,y)) \in L$.

• Case 4: $(a,x) \in R \land (b,y) \in S$.
• Case a: $x = a$
Thus $((a,b),(x,y)) \in L$. We can also say $((x,y),(a,b)) \in L \lor ((a,b),(x,y)) \in L$.

• Case b: $x \ne a$
Thus $((a,b),(x,y)) \in L$. We can also say $((x,y),(a,b)) \in L \lor ((a,b),(x,y)) \in L$.

Thus from all the cases, $((x,y),(a,b)) \in L \lor ((a,b),(x,y)) \in L$. Since $(x,y),(a,b)$ are arbitrary, we can conclude that $L$ is total order.

Soln10

Suppose $x,y$ are arbitrary elements of $A$.

($\to$) Suppose $xRy$. Suppose $a \in P_x$. Since $a \in P_x$, it follows $aRx$. Since $aRx$ and $xRy$ and $R$ is partial order, it follows $aRy$. Thus $a \in P_y$. Since $a$ is arbitrary, it follows that $P_x \subseteq P_y$.

($\leftarrow$)Suppose $P_x \subseteq P_y$. Since $R$ is partial order, xRx. Thus $x \in P_x$. Since $P_x \subseteq P_y$, it follows $x \in P_y$, thus $xRy$.

Since $x$ and $y$ are arbitrary elements, we can conclude that $\forall x \in A \forall y \in A (xRy \leftrightarrow P_x \subseteq P_y)$.

Soln11

Minimal elements: $\{ x \in B \, \vert \, x \text{ is prime} \}$. Smallest element: N.A.

Soln12

We shall prove this by contradiction. Suppose $% $ has a minimal element, say $M$. Thus $% $. Also since $M$ is minimal, it follows $\lnot \exists X \in F (X \subseteq M )$. Since $X \ne \phi$, suppose $x_0$ is an element in $M$. Then $M$ contains all elements of set $% $. Now consider the set $% $. Clearly $N \in \mathcal F$. Also all the elements that exist in $N$ are greater than $x_0$, thus $N \subseteq M$. Also, since $x_0 \in M \land x_0 \notin N$, $M \ne N$. Thus $N \in \mathcal F \land N \subseteq M$. But it contradicts our assumption that $\lnot \exists X \in F (X \subseteq M )$. Thus we can conclude that there does not exist any minimal element in $\mathcal F$.

Soln13

Suppose $R$ is partial order on $A$.

Reflexive:

Suppose $x \in A$. Since $R$ is partial order, $(x,x) \in R$. Thus $(x,x) \in R^{-1}$. Since $x$ is arbitrary, we can conclude that $R^{-1}$ is reflexive.

Transitive:

Suppose $x, y, z$ are arbitrary elements of $A$ such that $(x,y) \in R^{-1} \land (y,z) \in R^{-1}$. It follows that $(y,x) \in R \land (z,y) \in R$. Since $R$ is transitive, it follows that $(z,x) \in R$. Thus $(x,z) in R^{-1}$. Since $x,y,z$ are arbitrary, we can conclude the $R^{-1}$ is transitive.

Antisymmetric:

Suppose $x, y$ are arbitrary elements of $A$ such that $(x,y) \in R^{-1} \land (y,x) \in R^{-1}$. It follows that $(y,x) \in R \land (x,y) \in R$. Since $R$ is antisymmetric, it follows that $y = x$. Since $x,y$ are arbitrary, we can conclude the $R^{-1}$ is antisymmetric.

Thus $R$ is reflexive, transitive and antisymmetric, we can conclude that $R$ is partial order.

Total Order:

Suppose $R$ is total order. Suppose $x,y$ are arbitrary elements of $A$. Since $R$ is total order we have two possible cases:

• Case 1: $(x,y) \in R$.
Thus $(y,x) \in R^{-1}$. We can also say that $(x,y) \in R^{-1} \lor (y,x) \in R^{-1}$.

• Case 2: $(y,x) \in R$.
Thus $(x,y) \in R^{-1}$. We can also say that $(x,y) \in R^{-1} \lor (y,x) \in R^{-1}$.

Since $x,y$ are arbitrary, we can conclude that $R^{-1}$ is total order.

Soln14

(a)

Suppose $b$ is the $R$-largest element of $B$, iff
$\forall x \in B(xRb)$ iff
$\forall x \in B((x,b) \in R))$ iff
$\forall x \in B((b,x) \in R^{-1})$ iff
$\forall x \in B(b R^{-1} x)$ iff
$x$ is the $R^{-1}$-smallest element of $B$.

(b)

Suppose $b$ is the $R$-maximal element of $B$. iff
$\lnot \exists x \in B(bRx \land b \ne x)$. iff
$\lnot \exists x \in B(x R^{-1} b \land b \ne x)$. iff
$x$ is the $R^{-1}$-minimal element of $B$.

Soln15

(a)

Suppose $b$ is the $R_11$-smallest element of $B$. Suppose $x$ is arbitrary element in $B$. Since $b$ is $R_1$-smallest element in $B$, it follows that $(b,x) \in R_1$. Since $R_1 \subseteq R_2$, it follows $(b,x) \in R_2$. Since $x$ is arbitrary, it follows $\forall x \in B(b R_2 x)$. Thus $b$ is also the $R_2$-smallest element of $B$.

(b)

Suppose $b$ is an $R_2$-minimal element of $B$. Suppose $x$ is arbitrary element in $B$ such that $x \ne b$. Then $(x,b) \notin R_2$. Since $R_1 \subseteq R_2$, it follows that such element $(b,x)$ must also not exist in $R_1$. Thus $\lnot \exists x \in B(x R_1 b \land x != b$. It follows that $b$ is the $R_1$-minimal element of $B$.

Soln16

Suppose $b$ is the $R$-largest element of $B$. Suppose $b$ is not a maximal element of $B$. Thus there exist an element $x \in B$ such that $bRx$. But it contradicts with the given that $b$ is the largest element of $B$. Thus assumption $b$ is not a maximal element is not correct. Hence, $b$ is a maximal element of $B$.

Suppose $c$ is also a maximal element of $B$. Thus $\lnot \exists x \in B(cRx \land c \ne x)$, or $\forall x \in B(cRx \to c = x)$. Also since $b$ is largest element of $B$, $cRb$. Thus $c = b$. Thus we can conclude that $b$ is the only maximal element.

Soln17 (Using the hint from book as not able to do it myself).

False. Counter example:

Suppose $A = \mathbb R \times \mathbb R$, and let $R = \{((x, y), (x', y')) \in A \times A \, \vert \, x \le x' and y \le y' \}$.

This can be proved directly using excercise - 8 that it is partial order.

Now, suppose $B = \{ (0, 0) \} \cup (\{1\} \times \mathbb R)$.

Thus, $R$ contains elements like …. $((1,-3),(1,0))$, $((1,-2),(1,0))$, $((1,-1),(1,0))$, $((1,0),(1,0))$, $((0,0),(0,0))$, $((0,0),(1,0))$, $((0,0),(1,1))$, $((0,0),(1,2))$, $((0,0),(1,3))$, $((0,0),(1,3))$,…..

Thus it can be noticed here, how might this example is developed. $B$ got created in such a way that $R$ looks like it will contain two minimals but because other minimal moves towards infinity, we only get one defined minimal. There is no element such that $x R (0,0)$ is defined, so $(0,0)$ is the only minimal.

Now since elements like $((0,0),(1,-1)), ((0,0),(1,-2)), ((0,0),(1,-3))$ …. are not present in $R$(which can be seen that it is because of the particular choice of B), $(0,0)$ is not the smallest element as an element $s$ is smallest iff $\forall x \in B(s R x)$. Thus in our example, this is not the case that for all values, say $(x,y)$, of $B$, $(0,0) R (x,y)$.

Soln18

Note: There are errors in this solution. Refer to the below comment for correct solution.

(a)

($\to$)Suppose $x$ is upper bound on $B_1$. Suppose $b_2$ is arbitrary element of $B_2$. Since it is given that $\forall x \in B_2 \exists y \in B_1(x R y)$, if suppose for $m \in B_1$, we have $\forall x \in B_2 (x R m)$. Since $b_2 \in B_2$, it follows $b_2 R m$. But since $x$ is upper bound for $B_1$ and $m \in B_1$, it follows $(m R x )$. Since $R$ is partial order on $A$ and $(b_2 R m) \land (m R x)$, it follows $b_2 R x$. Since $b_2$ is arbitrary element of $B_2$, it follows that $x$ is upper bound for $B_2$.

($\leftarrow$)Suppose $x$ is upper bound on $B_2$. Suppose $b_1$ is arbitrary element of $B_1$. Since it is given that $\forall x \in B_1 \exists y \in B_2(x R y)$, if suppose for $m \in B_2$, we have $\forall x \in B_1 (x R m)$. Since $b_1 \in B_1$, it follows $b_1 R m$. But since $x$ is upper bound for $B_2$ and $m \in B_2$, it follows $(m R x )$. Since $R$ is partial order on $A$ and $(b_1 R m) \land (m R x)$, it follows $b_1 R x$. Since $b_1$ is arbitrary element of $B_1$, it follows that $x$ is upper bound for $B_1$.

(b)

We will prove this by contra-positive. Suppose $m$ is a maximal element of $B_1$. Similarly, suppose $n$ is a maximal element of $B_2$.

Since it is given that $\forall x \in B_1 \exists y \in B_2(x R y)$, suppose for the value $b_2 \in B_2$, we have $\forall x \in B_1 (x R b_2)$. Similarly, since it is given that $\forall x \in B_2 \exists y \in B_1(x R y)$, suppose for the value $b_1 \in B_2$, we have $\forall x \in B_2 (x R b_1)$. Now since $b_1 \in B_1$ and $\forall x \in B_1 (x R b_2)$, it follows that $b_1 R b_2$. Similarly, since $b_2 \in B_2$ and $\forall x \in B_2 (x R b_1)$, it follows that $b_2 R b_1$. Thus $b_1 R b_2 \land b_2 R b_1$ and since $R$ is partial order, it follows that $b_1 = b_2$.

Since $m$ is a maximal element of $B_1$, it follows $b_1 R m$. Also since $\forall x \in B_1 (x R b_2)$, it follows, $m R b_2$. Since $b_2 = b_1$, it follows $m R b_1$. But $m$ is a maximal element of $B_1$, thus $m = b_1 = b_2$.

Since $n$ is a maximal element of $B_2$, it follows $b_2 R n$. Also since $\forall x \in B_2 (x R b_1)$, it follows, $n R b_1$. Since $b_2 = b_1$, it follows $n R b_2$. But $n$ is a maximal element of $B_2$, thus $n = b_2 = b_1$.

Thus $m = n = b_1 = b_2$. Thus $B_1$ and $B_2$ are not disjoint. Thus by contra-positive, if $B_1$ and $B_2$ are disjoint, then there is no element $b_1 \in B_1$ and $b_2 \in B_2$ such that $b_1 = b_2$. And if there is no such element, then $B_1$ and $B_2$ does not have a maximal element.

Note: I am not confident of this proof. As from the givens it can be easily seen that $B_1$ and $B_2$ are not disjoint. Or the other cases might be that $B_1$ and $B_2$ are empty.

Soln19

(a) The proof is not correct. The problem with the proof is conclusion in both cases are wrong i.e.:

• Case 1: $bRx$. Since $x$ was arbitrary, we can conclude that $\forall x \in B(bRx)$, so $b$ is the smallest element of $R$.

Here the conclusion is wrong, because $bRx$ is not true for all values of $x$. it does not consider the other case, $xRb$. Similarly, in the other case $xRb$, we can not conclude that $xRb$ is true for all $x$ because it does not consider the case when $bRx$.

(b) Theorem is not correct. Counter example:

$A = \{1,2,3\}, B = \{1,2,3\}, R = \{(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)\}$. Clearly $B \subseteq A$ and $R$ is partial order on $A$. Note that $2 \in B$ but $2$ is niether $R$-largest element of $B$ nor the $R$-smallest element of $B$.

Soln20

(a) Suppose $b$ is smallest element of $B$. Thus $\forall x \in B(bRx)$. Thus the set of all lower bound elements of $B$ is $\forall x \in A(xRb)$ Since $B \subseteq A$, $b \in A$, and clearly in the set $\forall x \in A(xRb)$, $b$ is the greatest element. Thus $b$ is the greatest lower bound element of $B$.

(b) Suppose $b$ is largest element of $B$. Thus $\forall x \in B(xRb)$. Thus the set of all upper bound elements of $B$ is $\forall x \in A(bRx)$ Since $B \subseteq A$, $b \in A$, and clearly in the set $\forall x \in A(xRb)$, $b$ is the least element. Thus $b$ is the least upper bound element of $B$.

Soln21

(a) Suppose $U$ is the set of upper bound elements of $B$. Suppose $x \in U$ and suppose $y \in A$ such that $xRy$. Since $x \in U$, it follows that $\forall b \in B(bRx)$. Thus for any arbitrary element $b \in B$, $bRx$. Since $bRx \land xRy$, and $R$ is partial order, it follows $bRy$. Since $b$ is arbitrary, it follows $\forall b \in B(bRy)$. Thus $y$ is upper bound of $B$, or $y \in U$.

(b) Suppose $x$ is an arbitrary element of $B$. Suppose $y \in U$. Since $U$ is the set of upper bound of $B$, it follows that $xRy$. Since $y$ is arbitrary, it follows $\forall y \in U(xRy)$. Since $x \in A$(because $B \subseteq A$) and $\forall y \in U(xRy)$, it follows $x$ is a lower bound of $U$. Since $x$ is arbitrary, it follows every element of $B$ is lower bound element of $U$.

(c) Suppose $x$ is the greatest lower bound of $U$. Suppose $y$ is an arbitrary element of $B$. Then from part (b), $y$ is lower bound for $U$. Since $x$ is greatest lower bound of $U$, it follows $yRx$. Since $y$ is arbitrary, it follows that $\forall y \in B(yRx)$. Thus $x$ is upper bound of $B$. Thus $x \in U$. Since $x$ is lower bound of $U$, it follows $\forall y \in U(xRy)$. Thus $x$ is the lowest element in $U$. Since $U$ is the set of all upper bounds of $B$, it follows that $x$ is least upper bound of $B$.

Soln22

Suppose $B_1 \subseteq B_2$. Since $B_1 \subseteq B_2$, $x_2$ is also the upper bound of $B_1$. But since $x_1$ is the least upper bound of $B_1$, it must be smaller than all other upper bounds of $B_1$. Thus $x_1 R x_2$.

Soln23

• Proof of $\cup \mathcal F$ is l.u.b. of $\mathcal F$:

Suppose $X \in \mathcal F$. Suppose $x \in X$. Since $\forall X \in \mathcal F(x \in X \to x \in \cup \mathcal F)$, it follows that $x \in \cup \mathcal F$. Since $x$ is arbitrary, it follows $X \subseteq \cup \mathcal F$. Since $X$ is arbitrary, it follows that $\forall X \in \mathcal F(X \subseteq \cup \mathcal F)$, or $\cup \mathcal F$ is the largest subset element of $\mathcal F$. Thus all the upper bounds of $\mathcal F$ must be larger than $\cup \mathcal F$. Thus $\cup \mathcal F$ is the smallest element in all the uppper bounds of $\mathcal F$. Thus $\cup \mathcal F$ is the least upper bound of $\mathcal F$.

• Proof of $\cap \mathcal F$ is g.l.b. of $\mathcal F$:

Suppose $X \in \mathcal F$. Suppose $x \in \cap \mathcal F$. Then $\forall X \in \mathcal F(x \in X)$. Thus if $x \in \cap \mathcal F$, then $x \in X$. Since $x$ is arbitrary $\cap \mathcal F \subseteq X$. Since $X$ is arbitrary, it follows that $\forall X \in \mathcal F (\cap \mathcal F \subseteq X)$. Thus $\cap \mathcal F$ is the smallest element(wrt subset relation) in $\mathcal F$. Thus all the lower bounds of $\mathcal F$ must be lesser than $\cap \mathcal F$. Thus $\cap \mathcal F$ is the largest element in all the lower bounds of $\mathcal F$. Thus $\cap \mathcal F$ is the greatest lower bound of $\mathcal F$.