Exercise for the break Draw, for four sets, a diagram of sets, which shows all possible intersecting sets.
An abstract and
Exercises Let
L
A
{\displaystyle {}LA}
G
A
{\displaystyle {}GA}
R
A
{\displaystyle {}RA}
G
A
∖
R
A
{\displaystyle {}GA\setminus RA}
(
L
A
∩
G
A
)
∪
(
L
A
∩
R
A
)
{\displaystyle {}{\left(LA\cap GA\right)}\cup {\left(LA\cap RA\right)}}
R
A
∖
(
G
A
∪
R
A
)
{\displaystyle {}RA\setminus {\left(GA\cup RA\right)}}
R
A
∖
(
G
A
∪
L
A
)
{\displaystyle {}RA\setminus {\left(GA\cup LA\right)}}
(
R
A
∖
G
A
)
∩
(
(
L
A
∪
G
A
)
∖
(
G
A
∩
R
A
)
)
{\displaystyle {}{\left(RA\setminus GA\right)}\cap {\left({\left(LA\cup GA\right)}\setminus {\left(GA\cap RA\right)}\right)}}
Determine for the sets
M
=
{
a
,
b
,
c
,
d
,
e
}
,
N
=
{
a
,
c
,
e
}
,
P
=
{
b
}
,
R
=
{
b
,
d
,
e
,
f
}
{\displaystyle M=\{a,b,c,d,e\},\,N=\{a,c,e\},\,P=\{b\},\,R=\{b,d,e,f\}}
the following sets.
M
∩
N
{\displaystyle {}M\cap N}
M
∩
N
∩
P
∩
R
{\displaystyle {}M\cap N\cap P\cap R}
M
∪
R
{\displaystyle {}M\cup R}
(
N
∪
P
)
∩
R
{\displaystyle {}{\left(N\cup P\right)}\cap R}
N
∖
R
{\displaystyle {}N\setminus R}
(
M
∪
P
)
∖
(
R
∖
N
)
{\displaystyle {}{\left(M\cup P\right)}\setminus {\left(R\setminus N\right)}}
(
(
P
∪
R
)
∩
N
)
∩
R
{\displaystyle {}{\left({\left(P\cup R\right)}\cap N\right)}\cap R}
(
R
∖
P
)
∩
(
M
∖
N
)
{\displaystyle {}{\left(R\setminus P\right)}\cap {\left(M\setminus N\right)}}
Sketch the following subsets of
R
2
{\displaystyle {}\mathbb {R} ^{2}}
{
(
x
,
y
)
∣
x
=
5
}
{\displaystyle {}{\left\{(x,y)\mid x=5\right\}}}
{
(
x
,
y
)
∣
x
≥
4
and
y
=
3
}
{\displaystyle {}{\left\{(x,y)\mid x\geq 4{\text{ and }}y=3\right\}}}
{
(
x
,
y
)
∣
y
2
≥
2
}
{\displaystyle {}{\left\{(x,y)\mid y^{2}\geq 2\right\}}}
{
(
x
,
y
)
∣
|
x
|
=
3
and
|
y
|
≤
2
}
{\displaystyle {}{\left\{(x,y)\mid \vert {x}\vert =3{\text{ and }}\vert {y}\vert \leq 2\right\}}}
{
(
x
,
y
)
∣
3
x
≥
y
and
5
x
≤
2
y
}
{\displaystyle {}{\left\{(x,y)\mid 3x\geq y{\text{ and }}5x\leq 2y\right\}}}
{
(
x
,
y
)
∣
x
y
=
0
}
{\displaystyle {}{\left\{(x,y)\mid xy=0\right\}}}
{
(
x
,
y
)
∣
x
y
=
1
}
{\displaystyle {}{\left\{(x,y)\mid xy=1\right\}}}
{
(
x
,
y
)
∣
x
y
≥
1
and
y
≥
x
3
}
{\displaystyle {}{\left\{(x,y)\mid xy\geq 1{\text{ and }}y\geq x^{3}\right\}}}
{
(
x
,
y
)
∣
0
=
0
}
{\displaystyle {}{\left\{(x,y)\mid 0=0\right\}}}
{
(
x
,
y
)
∣
0
=
1
}
{\displaystyle {}{\left\{(x,y)\mid 0=1\right\}}}
Let
A
,
B
{\displaystyle {}A,\,B}
C
{\displaystyle {}C}
A
∖
(
B
∩
C
)
=
(
A
∖
B
)
∪
(
A
∖
C
)
.
{\displaystyle {}A\setminus {\left(B\cap C\right)}={\left(A\setminus B\right)}\cup {\left(A\setminus C\right)}\,.}
Let
A
,
B
{\displaystyle {}A,B}
C
{\displaystyle {}C}
A
∪
∅
=
A
,
{\displaystyle {}A\cup \emptyset =A\,,}
A
∩
∅
=
∅
,
{\displaystyle {}A\cap \emptyset =\emptyset \,,}
A
∩
B
=
B
∩
A
,
{\displaystyle {}A\cap B=B\cap A\,,}
A
∪
B
=
B
∪
A
,
{\displaystyle {}A\cup B=B\cup A\,,}
A
∩
(
B
∩
C
)
=
(
A
∩
B
)
∩
C
,
{\displaystyle {}A\cap (B\cap C)=(A\cap B)\cap C\,,}
A
∪
(
B
∪
C
)
=
(
A
∪
B
)
∪
C
,
{\displaystyle {}A\cup (B\cup C)=(A\cup B)\cup C\,,}
A
∩
(
B
∪
C
)
=
(
A
∩
B
)
∪
(
A
∩
C
)
,
{\displaystyle {}A\cap (B\cup C)=(A\cap B)\cup (A\cap C)\,,}
A
∪
(
B
∩
C
)
=
(
A
∪
B
)
∩
(
A
∪
C
)
,
{\displaystyle {}A\cup (B\cap C)=(A\cup B)\cap (A\cup C)\,,}
A
∖
(
B
∪
C
)
=
(
A
∖
B
)
∩
(
A
∖
C
)
.
{\displaystyle {}A\setminus (B\cup C)=(A\setminus B)\cap (A\setminus C)\,.}
Let
M
{\displaystyle {}M}
N
{\displaystyle {}N}
disjoint sets ,MDLD/disjoint sets
and
x
∈
M
{\displaystyle {}x\in M}
M
∖
{
x
}
{\displaystyle {}M\setminus \{x\}}
N
∪
{
x
}
{\displaystyle {}N\cup \{x\}}
M
∪
N
=
(
M
∖
{
x
}
)
∪
(
N
∪
{
x
}
)
{\displaystyle {}M\cup N={\left(M\setminus \{x\}\right)}\cup {\left(N\cup \{x\}\right)}\,}
holds.
Sketch the set
M
=
{
(
x
,
y
)
∈
R
2
∣
4
x
−
7
y
=
3
}
{\displaystyle {}M={\left\{(x,y)\in \mathbb {R} ^{2}\mid 4x-7y=3\right\}}}
N
=
{
(
x
,
y
)
∈
R
2
∣
3
x
+
2
y
=
5
}
{\displaystyle {}N={\left\{(x,y)\in \mathbb {R} ^{2}\mid 3x+2y=5\right\}}}
Determine the intersection
M
∩
N
{\displaystyle {}M\cap N}
We consider the two sets
E
=
{
(
x
y
z
)
∈
R
3
∣
−
3
x
+
2
y
−
6
z
=
0
}
{\displaystyle {}E={\left\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in \mathbb {R} ^{3}\mid -3x+2y-6z=0\right\}}\,}
and
F
=
{
(
x
y
z
)
∈
R
3
∣
7
x
−
5
y
−
4
z
=
0
}
.
{\displaystyle {}F={\left\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in \mathbb {R} ^{3}\mid 7x-5y-4z=0\right\}}\,.}
Find a description of the intersection
G
:=
E
∩
F
=
{
(
x
y
z
)
∈
R
3
∣
−
3
x
+
2
y
−
6
z
=
0
and
7
x
−
5
y
−
4
z
=
0
}
,
{\displaystyle {}G:=E\cap F={\left\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in \mathbb {R} ^{3}\mid -3x+2y-6z=0{\text{ and }}7x-5y-4z=0\right\}}\,,}
similar to
Example 1.2
.
Show that the set
{
(
x
,
y
)
∈
Z
2
∣
3
x
+
5
y
=
6
}
{\displaystyle {\left\{(x,y)\in \mathbb {Z} ^{2}\mid 3x+5y=6\right\}}}
is not empty.
Show that the set
{
(
x
,
y
)
∈
Z
2
∣
6
x
+
9
y
=
5
}
{\displaystyle {\left\{(x,y)\in \mathbb {Z} ^{2}\mid 6x+9y=5\right\}}}
is empty.
Describe, for every combination
(including the case where the product is taken with itself),
the product set of the following geometric sets .
A circle
K
{\displaystyle {}K}
A line segment
I
{\displaystyle {}I}
A line
G
{\displaystyle {}G}
A parabola
P
{\displaystyle {}P}
Which product set can we realize as a subset in space?
Let
M
{\displaystyle {}M}
N
{\displaystyle {}N}
A
⊆
M
{\displaystyle {}A\subseteq M}
B
⊆
N
{\displaystyle {}B\subseteq N}
(
A
×
N
)
∩
(
M
×
B
)
=
A
×
B
.
{\displaystyle {}{\left(A\times N\right)}\cap {\left(M\times B\right)}=A\times B\,.}
Let
A
{\displaystyle {}A}
B
{\displaystyle {}B}
disjoint sets ,MDLD/disjoint sets
and let
C
{\displaystyle {}C}
C
×
(
A
⊎
B
)
=
(
C
×
A
)
⊎
(
C
×
B
)
.
{\displaystyle {}C\times (A\uplus B)=(C\times A)\uplus (C\times B)\,.}
Let
A
{\displaystyle {}A}
B
{\displaystyle {}B}
disjoint sets .MDLD/disjoint sets
Show the equality
(
A
⊎
B
)
×
(
A
⊎
B
)
=
(
A
×
A
)
⊎
(
A
×
B
)
⊎
(
B
×
A
)
⊎
(
B
×
B
)
.
{\displaystyle {}(A\uplus B)\times (A\uplus B)=(A\times A)\uplus (A\times B)\uplus (B\times A)\uplus (B\times B)\,.}
Hand-in-exercises Sketch the following subsets in
R
2
{\displaystyle {}\mathbb {R} ^{2}}
{
(
x
,
y
)
∣
|
2
x
|
=
5
and
|
y
|
≥
3
}
{\displaystyle {}{\left\{(x,y)\mid \vert {2x}\vert =5{\text{ and }}\vert {y}\vert \geq 3\right\}}}
{
(
x
,
y
)
∣
−
3
x
≥
2
y
and
4
x
≤
−
5
y
}
{\displaystyle {}{\left\{(x,y)\mid -3x\geq 2y{\text{ and }}4x\leq -5y\right\}}}
{
(
x
,
y
)
∣
y
2
−
y
+
1
≤
4
}
{\displaystyle {}{\left\{(x,y)\mid y^{2}-y+1\leq 4\right\}}}
{
(
x
,
y
)
∣
x
y
=
2
or
x
2
+
y
2
=
1
}
{\displaystyle {}{\left\{(x,y)\mid xy=2{\text{ or }}x^{2}+y^{2}=1\right\}}}
Sketch the set
M
=
{
(
x
,
y
)
∈
R
2
∣
−
5
x
+
2
y
=
6
}
{\displaystyle {}M={\left\{(x,y)\in \mathbb {R} ^{2}\mid -5x+2y=6\right\}}}
N
=
{
(
x
,
y
)
∈
R
2
∣
7
x
−
5
y
=
4
}
{\displaystyle {}N={\left\{(x,y)\in \mathbb {R} ^{2}\mid 7x-5y=4\right\}}}
Determine den intersection
M
∩
N
{\displaystyle {}M\cap N}
Does the "subtraction rule“ hold for the union of sets, i.e., can we infer from
A
∪
C
=
B
∪
C
{\displaystyle {}A\cup C=B\cup C}
A
=
B
{\displaystyle {}A=B}
Prove the following
(set-theoretical versions of)
syllogisms of Aristotle. Let
A
,
B
,
C
{\displaystyle {}A,B,C}
Modus Barbara:
B
⊆
A
{\displaystyle {}B\subseteq A}
C
⊆
B
{\displaystyle {}C\subseteq B}
C
⊆
A
{\displaystyle {}C\subseteq A}
Modus Celarent:
B
∩
A
=
∅
{\displaystyle {}B\cap A=\emptyset }
C
⊆
B
{\displaystyle {}C\subseteq B}
C
∩
A
=
∅
{\displaystyle {}C\cap A=\emptyset }
Modus Darii:
B
⊆
A
{\displaystyle {}B\subseteq A}
C
∩
B
≠
∅
{\displaystyle {}C\cap B\neq \emptyset }
C
∩
A
≠
∅
{\displaystyle {}C\cap A\neq \emptyset }
Modus Ferio:
B
∩
A
=
∅
{\displaystyle {}B\cap A=\emptyset }
C
∩
B
≠
∅
{\displaystyle {}C\cap B\neq \emptyset }
C
⊈
A
{\displaystyle {}C\not \subseteq A}
Modus Baroco:
B
⊆
A
{\displaystyle {}B\subseteq A}
B
⊈
C
{\displaystyle {}B\not \subseteq C}
A
⊈
C
{\displaystyle {}A\not \subseteq C}
Let
M
{\displaystyle {}M}
N
{\displaystyle {}N}
A
1
,
A
2
⊆
M
{\displaystyle {}A_{1},A_{2}\subseteq M}
B
1
,
B
2
⊆
N
{\displaystyle {}B_{1},B_{2}\subseteq N}
(
A
1
×
B
1
)
∩
(
A
2
×
B
2
)
=
(
A
1
∩
A
2
)
×
(
B
1
∩
B
2
)
.
{\displaystyle {}{\left(A_{1}\times B_{1}\right)}\cap {\left(A_{2}\times B_{2}\right)}={\left(A_{1}\cap A_{2}\right)}\times {\left(B_{1}\cap B_{2}\right)}\,.}
Let
A
{\displaystyle {}A}
B
{\displaystyle {}B}
A
⊆
B
{\displaystyle {}A\subseteq B}
A
∩
B
=
A
{\displaystyle {}A\cap B=A}
A
∪
B
=
B
{\displaystyle {}A\cup B=B}
A
∖
B
=
∅
{\displaystyle {}A\setminus B=\emptyset }
There exists a set
C
{\displaystyle {}C}
B
=
A
∪
C
{\displaystyle {}B=A\cup C}
There exists a set
D
{\displaystyle {}D}
A
=
B
∩
D
{\displaystyle {}A=B\cap D}
