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Jim Hefferon
proofs
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73dc37e63558cdeb8ac5164c33ee3a45fdf73620...9b35ab9ca2fe1499b1a62b8df87b1783d517dc21
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Commits (1)
Through all exercises and answers, including infinity
· 9b35ab9c
Jim Hefferon
authored
Aug 04, 2018
9b35ab9c
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infinity.tex
infinity.tex
+50
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iblcompactmaxlength.pdf
output/iblcompactmaxlength.pdf
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infinity.tex
View file @
9b35ab9c
...
...
@@ 5,20 +5,9 @@ that if two finite sets correspond then they have the same number of elements.
\begin{df}
Two sets have the
\definend
{
same cardinality
}
(or are
\definend
{
equinumerous
}
)
if there is a
or are
\definend
{
equinumerous
}
if there is a
correspondence from one to the other.
We write
$
A
\sim
B
$
.
\end{df}
\begin{df}
A set is
\definend
{
finite
}
if it has~
$
n
$
elements for some
$
n
\in\N
$
,
that is, if it has the same cardinality as
$
\setbuilder
{
i
\in\N
}{
i< n
}
=
\set
{
0
,
1
,
\ldots
,n

1
}$
.
Otherwise the set is
\definend
{
infinite
}
.
% \end{df}\begin{df}
A set is
\definend
{
denumerable
}
if it has the same cardinality
as
$
\N
$
.
A set is
\definend
{
countable
}
if it is either finite or denumerable.
We write
$
A
\sim
B
$
.
\end{df}
\begin{problem}
\label
{
ex:EquinumeruousIsEquivalence
}
...
...
@@ 42,6 +31,17 @@ Prove that the relation~$\sim$ is an equivalence.
\end{answer}
\end{problem}
\begin{df}
A set is
\definend
{
finite
}
if it has~
$
n
$
elements for some
$
n
\in\N
$
,
that is, if it has the same cardinality as some initial segment
$
\setbuilder
{
i
\in\N
}{
i< n
}
=
\set
{
0
,
1
,
\ldots
,n

1
}$
of the natural numbers.
Otherwise the set is
\definend
{
infinite
}
.
% \end{df}\begin{df}
A set is
\definend
{
denumerable
}
if it has the same cardinality
as
$
\N
$
.
A set is
\definend
{
countable
}
if it is either finite or denumerable.
\end{df}
\begin{problem}
\label
{
ex:IntegersCountable
}
Prove.
\begin{exes}
...
...
@@ 49,12 +49,12 @@ Prove.
The set of integers is countable.
\end{exercise}
\begin{answer}
We will define a map
$
\map
{
f
}{
\N
}{
\Z
}$
and just note that it is clearly a
correspondence.
The map alternate between positive and negative values:
Consider the map
$
\map
{
f
}{
\N
}{
\Z
}$
that alternates between positive and negative values:
$
0
\mapsunder
{}
0
$
,
$
1
\mapsunder
{}
1
$
,
$
2
\mapsunder
{}

1
$
,
$
3
\mapsunder
{}
2
$
,
$
4
\mapsunder
{}

2
$
,
$
5
\mapsunder
{}
3
$
,
\ldots\,
.
More formally,
$
f
(
n
)=
n
/
2
$
if
$
n
$
is even and
$
f
(
n
)=(
n
+
1
)/
2
$
if~
$
n
$
is odd.
More formally,
$
f
(
n
)=
n
/
2
$
if
$
n
$
is even and
$
f
(
n
)=(
n
+
1
)/
2
$
if~
$
n
$
is odd.
It is clearly a correspondence.
\end{answer}
\begin{exercise}
The set~
$
\N\times\N
$
is countable.
...
...
@@ 64,13 +64,23 @@ Prove.
$
\map
{
f
}{
\N
}{
\N\times\N
}$
and only note that it is a correspondence,
without going through the verification details.
The function alternates between columns in the same way that
Jackie Chan, when up against multiple bad guys, alternates among his
attackers.
Thus
$
0
\mapsunder
{}
(
0
,
0
)
$
,
$
1
\mapsunder
{}
(
0
,
1
)
$
,
$
2
\mapsunder
{}
(
1
,
0
)
$
,
The function alternates between columns:
$
0
\mapsunder
{}
(
0
,
0
)
$
,
$
1
\mapsunder
{}
(
0
,
1
)
$
,
$
2
\mapsunder
{}
(
1
,
0
)
$
,
$
3
\mapsunder
{}
(
0
,
2
)
$
,
$
4
\mapsunder
{}
(
1
,
1
)
$
,
$
5
\mapsunder
{}
(
2
,
0
)
$
,
$
6
\mapsunder
{}
(
0
,
3
)
$
,
\ldots
.
\begin{center}
\begin{tabular}
{
rcccccccccc
}
\textit
{
Input
}
&$
0
$
&$
1
$
&$
2
$
&$
3
$
&$
4
$
&$
5
$
&$
6
$
&$
7
$
&$
8
$
&$
\cdots
$
\\
\cline
{
211
}
\rule
{
0em
}{
2.25ex
}
\textit
{
Output
}
&$
\seq
{
0
,
0
}$
&$
\seq
{
0
,
1
}$
&$
\seq
{
1
,
0
}$
&$
\seq
{
0
,
2
}$
&$
\seq
{
1
,
1
}$
&$
\seq
{
2
,
0
}$
&$
\seq
{
0
,
3
}$
&$
\seq
{
1
,
2
}$
&$
\seq
{
2
,
1
}$
&$
\cdots
$
\end{tabular}
\end{center}
\remark
a sketch helps here.
The values associated with successive arguments trace out the
diagonals of the array~
$
\N\times\N
$
.
...
...
@@ 170,8 +180,8 @@ Prove.
\begin{problem}
\label
{
ex:OntoFromNatsToSetEquivToCountable
}
Prove that the following are equivalent for a set~
$
A
$
:
(i)~
$
A
$
is countable
(ii)~
$
A
$
is empty or there is an onto function from
$
\N
$
to~
$
A
$
(i)~
$
A
$
is countable
,
(ii)~
$
A
$
is empty or there is an onto function from
$
\N
$
to~
$
A
$
,
(iii)~there is a onetoone function from
$
A
$
to~
$
\N
$
.
\begin{answer}
We will show that (i) implies~(ii), which implies~(iii), and which
...
...
@@ 180,28 +190,32 @@ implies~(i).
First the (i) implies~(ii) proof.
If~
$
A
$
is countable then there are two cases: either~
$
A
$
is finite
or~
$
A
$
is denumerable.
The case that~
$
A
$
is denumerable by definition has a
correspondence between~
$
\N
$
and~
$
A
$
.
If~
$
A
$
is finite then it is either empty or nonempty.
The empty case is trivial so suppose that
$
A
$
is finite and nonempty
and let
$
a
$
be an element of~
$
A
$
.
Then there is a correspondence~
$
\map
{
f
}{
\set
{
0
,
1
,
\ldots
,A

1
}}{
A
}$
,
The empty set is part of statement~(ii)
so suppose that
$
A
$
is finite and nonempty, and let
$
a
$
be an element of~
$
A
$
.
Then there is a correspondence from some initial segment of the natural
numbers~
$
\map
{
f
}{
\set
{
0
,
1
,
\ldots
,A

1
}}{
A
}$
,
so define
$
\map
{
\hat
{
f
}}{
\N
}{
A
}$
extending~
$
f
$
by
$
\hat
{
f
}
(
i
)=
f
(
i
)
$
if
$
i
\in\set
{
0
,
1
,
\ldots
,A

1
}$
, and
$
\hat
{
f
}
(
i
)=
a
$
otherwise.
The extension~
$
\hat
{
f
}$
is onto because
$
f
$
is onto.
The other case, that~
$
A
$
is denumerable, by definition has a
correspondence between~
$
\N
$
and~
$
A
$
.
$
\hat
{
f
}
(
i
)=
a
$
if
$
i>A

1
$
.
This extension~
$
\hat
{
f
}$
is onto because
$
f
$
is onto.
Next is the (ii) implies~(iii) proof.
Next is the proof that statement~(ii) implies~(iii).
By~(ii) either
$
A
$
is the empty set
or there is an onto function from
$
\N
$
to~
$
A
$
.
If~
$
A
$
is empty then the empty function is a map with domain~
$
A
$
and
codomain~
$
\N
$
that is
vacuously onetoone
.
The other possibility is that
there is an onto function~
$
\map
{
f
}{
\N
}{
A
}$
.
codomain~
$
\N
$
that is
onetoone, vacuously
.
If
$
A
$
is nonempty then
there is an onto function~
$
\map
{
f
}{
\N
}{
A
}$
.
To get a map~
$
g
$
with domain~
$
A
$
and codomain~
$
\N
$
that is onetoone
define~
$
g
(
a
)
$
to be the least number
$
n
\in\N
$
such that~
$
f
(
n
)=
a
$
;
at least one such number exists because~
$
f
$
is onto.
Clearly~
$
g
$
is onetoone.
Finally
for
the (iii) implies~(i) proof.
Let~
$
\map
{
f
}{
A
}{
\N
}$
be onetoone
.
Finally
,
the (iii) implies~(i) proof.
Assume that there is a onetoone function
$
\map
{
f
}{
A
}{
\N
}$
.
Let
$
B
=
\range
(
A
)
\subseteq\N
$
.
Then the function~
$
\map
{
\hat
{
f
}}{
A
}{
B
}$
given by~
$
\hat
{
f
}
(
a
)=
f
(
a
)
$
is a
correspondence (it only eliminates unused elements of the codomain).
...
...
@@ 215,7 +229,7 @@ Recall that~$B\subseteq\N$ so we can define a map
$
\map
{
g
}{
\N
}{
B
}$
by:
take
$
g
(
j
)
$
to be the
$
j
$
th largest element of~
$
B
$
.
This map is onto because of the assumption that~
$
B
$
is not finite, and it
is onetoone by construction
(
if
$
g
(
j
_
0
)=
g
(
j
_
1
)
$
then the
$
j
_
0
$
th largest
is onetoone by construction
:~
if
$
g
(
j
_
0
)=
g
(
j
_
1
)
$
then the
$
j
_
0
$
th largest
element of~
$
B
$
is the same as the
$
j
_
1
$
th largest element, which is only
possible if
$
j
_
0
=
j
_
1
$
).
\end{answer}
...
...
output/iblcompactmaxlength.pdf
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