#import "@youwen/zen:0.1.0": *

#show: zen.with(
  title: "Homework 4",
  author: "Youwen Wu",
)

#set heading(numbering: none)

#set par(first-line-indent: 0pt, spacing: 1em)

#let nonzero = $ZZ_(!=0)$

Problems:

2.2: \#1dfh, 2bdf, 5, 6d, 7q, 8h, 9b, 10e, 11b, 13a, 15b, 16

2.3: \#1bhLmn, 2bhLmn, 3, 12, 16d, 18bd

= 2.2

*1d.*

$A - (B - C) = {1,3,5,7,9}$

*1f.*

$A union (C sect D) = {1,2,3,5,7,8,9}$

*1h.*

${1,5,7}$

*2b.*

$[2,8)$

*2d.*

$[3,6]$

*2f.*

$(6,8)$

*5.*

Only $C$ and $D$ are disjoint.

*6d.*

$A = {1, 2}, B = {3, 4}, C = {2, 3}$

*7q.*

Claim: if $A subset.eq B$, then $A union C subset.eq B union C$.

#proof[
  Suppose $A subset.eq B$. Therefore $forall a in A, a in B$. This implies that
  $forall a in A, a in A union C$, and $forall a in A, a in B union C$.
  Note that $forall c in C$, both $c in A union C$ and $c in B union C$. Since
  $forall x in A union C$, either $x in A$ or $x in C$, and either way $x in B$
  and $x in C$, then $A union B subset.eq A union C$.
]

*8h.*

Claim: $(A union B)^c = A^c sect B^c$.

#proof[
  For all $x$ in $(A union B)^c$, $x$ is not in $A union B$. Therefore $x$ is
  not in $A$ and $x$ is not in $B$. In other words $(A union B)^c$ is comprised of all $x$ not in $A$ and not in $B$, which are all the elements in both $A^c$ and $B^c$, which is $A^c sect B^c$.
]

*9b.*

Claim: If $A subset.eq B union C$ and $A sect B = emptyset$, then $A subset.eq C$.

#proof[
  Suppose $A subset.eq B union C$. Then $forall a in A$, either $a in B$ or $a
  in C$. However $a$ is never in $B$, as $A$ and $B$ are disjoint and $a in B$
  would be a contradiction of this fact. So in fact the only case is $a in C$.
  Therefore we have $forall a in A, a in C$, which is the definition of $A
  subset.eq C$.
]

*10e.*

Claim: if $A union B subset.eq C union D$, $A sect B = emptyset$, and $C subset.eq A$, then $B subset.eq D$.

#proof[
  Suppose $C subset.eq A$. Then $forall c in C, c in A$. Since $A$ and $B$ are
  disjoint, $forall b in B, b in.not C$. Because we assume $A union B subset.eq
  C union D$, $forall x in A union B, x in C union D$. Then this implies
  $forall a in A$ and $forall b in B$, either $a in C$, or $b in C$, or $a in
  D$, or $b in D$. However we established earlier that $forall b in B, b in.not
  C$. This means that $forall b in B, b in D$.
  Therefore $B subset.eq D$.
]

*11b.*

Claim: if $A sect C subset.eq B sect C$, then $A subset.eq B$.

Counterexample: $A = {1,2,3}, B = {2,3,4}, C = {2,3,4}$.

*13a.*

(I wrote a Vim macro to compute these because it was so annoying).
$
  A times B = \
  {(1,a), (1, e), (1, k), (1, n), (1, r), (3,a), (3, e), (3, k), (3, n), (3, r), (5,a), (5, e), (5, k), (5, n), (5, r)} \
  B times A = \
  {(a,1), (e, 1), (k, 1), (n, 1), (r, 1), (a,3), (e, 3), (k, 3), (n, 3), (r, 3), (a,5), (e, 5), (k, 5), (n, 5), (r, 5)} \
$

*15b.*

Claim: $A times emptyset = emptyset$.

#proof[
  Let there be a set $X$ such that $A times emptyset = X$. Then any element in
  $X$ is an ordered pair $(a,b)$ such that $a in A$ and $b in emptyset$.
  But $b in emptyset$ is a contradiction no ordered pair can exist. Therefore
  $X$ contains no elements and $X = emptyset$, so $A times emptyset =
  emptyset$.
]

*16a.*

$
  A = {1}, B = {2}, C = {3}, D = {4}
$

*16b.*

$
  A = {1}, B = {2}, C = {2}
$

*16c.*

$
  A = {1}, B = {2}, C = {3}
$

= 2.3

*1b.*

$
  union.big_(A in cal(A)) A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} \
  sect.big_(A in cal(A)) A = emptyset
$

*1h.*

$
  union.big_(A in cal(A)) A = [-pi, infinity) \
  sect.big_(A in cal(A)) A = [-pi, 0]
$

*1L.*

$
  union.big_(L in cal(l)) L = (-infinity, infinity) \
  sect.big_(L in cal(l)) L = emptyset
$

*1m.*

$
  union.big_(A in cal(A)) A = (-infinity, infinity) - ZZ \
  sect.big_(A in cal(A)) A = emptyset
$

*1n.*

$
  union.big_(D in cal(D)) D = (-infinity, 1) \
  sect.big_(D in cal(D)) D = (-1, 0]
$

*2b.*

No.

*2h.*

No.

*2L.*

Yes.

*2m.*

Yes.

*2n.*

No.

*3a.*

Claim: let $cal(A)$ be a family of sets, for every set B in the family
$cal(A)$, $B subset.eq union.big _(A in cal(A)) A$

#proof[
  By the definition of the union of sets, $forall b in B$, $b in union.big _(A
  in cal(A)) A$. Therefore $B subset.eq union.big _(A in cal(A))$.
]

*3b.*

Claim: let $cal(A)$ be a nonempty family of sets and $B$ be a set. Then if $A
subset.eq B$ for all $A in cal(A)$, then $union.big _(A in cal(A)) A subset.eq
B$.

#proof[
  Suppose $A subset.eq B$ for all $A in cal(A)$. Then $forall a in A, a in B$
  for all $A in cal(A)$. Then note that $forall  in union.big _(A in cal(A))
  A$, there exists some $A$ with $x in A$. This implies that any element of
  $union.big _(A in cal(A)) A$ is an element of some $A in cal(A)$. By our
  earlier assumption this implies that any element in $union.big _(A in cal(A))
  A$ is an element of $B$. This is the definition of $union.big _(A in cal(A))
  A subset.eq B$.
]

*12a.*

$
  cal(A) = {{1, x} : x in {2, 3, dots.c, 20}}
$

*12b.*

$
  cal(B) = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}}
$

*12c.*

$
  cal(l) = {{x} : x in {1, 2, 3, dots.c, 20}}
$

*16d.*

Claim: $sect.big _(i = 1) ^infinity A_i subset.eq sect.big _(i=k) ^m A_i$.

#proof[
  Note that

  $
    sect.big_(i = 1)^infinity A_i = (sect.big_(i = 1)^(k-1) A_i) sect (sect.big_(i = k)^m A_i) sect (sect.big_(i = m + 1)^infinity A_i)
  $

  This implies that the set $sect.big _(i=1) ^infinity A_i$ contains only
  elements belonging to (by the definition of the set intersection) $sect.big
  _(i=k) ^m A_i$. Therefore $sect.big _(i=1) ^infinity A_i$ is indeed a subset
  of $sect.big _(i=k) ^m A_i$.
]

*18b.*

$
  {(-infinity ,infinity), (-infinity, 1]}
$

*18d.*

${{1, 2, 3}, {1, 2}, {1}, emptyset}$