Mixing
How can one rigorously determine the cardinality of an infinite dimensional vector space?
$begingroup$
Suppose $V$ is a vector space over a scalar field $F$. If $dim(V)=n$, then $|V|=|F|^n$. How can I rigorously determine the cardinality of $V$ when $V$ is infinite dimensional?
My thought was that if $mathscr{B}$ is an ordered basis for $V$, then the cardinality of $V$ is given by the set of functions from $mathscr{B}to F$, by identifying elements of $V$ with their $mathscr{B}$-coordinate vector. However, I feel that we should only count functions with finite support since infinite sums don't make sense.
Is this correct? If so, how does one find the cardinality of ${fcolonmathscr{B}to Fmid mathrm{supp }(f)<infty}$, in terms of say $|F|$ and $|mathscr{B}|$? Thanks.
linear-algebra cardinals
asked Sep 11 '12 at 18:41
NastassjaNastassja
685413
$endgroup$
|
show 1 more comment
$begingroup$
Suppose $V$ is a vector space over a scalar field $F$. If $dim(V)=n$, then $|V|=|F|^n$. How can I rigorously determine the cardinality of $V$ when $V$ is infinite dimensional?
My thought was that if $mathscr{B}$ is an ordered basis for $V$, then the cardinality of $V$ is given by the set of functions from $mathscr{B}to F$, by identifying elements of $V$ with their $mathscr{B}$-coordinate vector. However, I feel that we should only count functions with finite support since infinite sums don't make sense.
Is this correct? If so, how does one find the cardinality of ${fcolonmathscr{B}to Fmid mathrm{supp }(f)<infty}$, in terms of say $|F|$ and $|mathscr{B}|$? Thanks.
linear-algebra cardinals
asked Sep 11 '12 at 18:41
NastassjaNastassja
685413
$endgroup$
1
$begingroup$
Can you find the number of functions from $mathscr{B}$ to $F$ with support of size at most $n$?
$endgroup$
– Chris Eagle
Sep 11 '12 at 18:48
$begingroup$
@ChrisEagle Wouldn't that require choosing $n$ vectors in $mathscr{B}$ to send to nonzero elements of $F$? That seems like it would already be very large since $mathscr{B}$ is infinite.
$endgroup$
– Nastassja
Sep 11 '12 at 19:10
$begingroup$
@Nastassja But what would the infinite cardinal be?
$endgroup$
– Alex Becker
Sep 11 '12 at 19:10
1
$begingroup$
@Nastassja The point is that the set of functions from $mathscr B$ to $F$ with support at most $n$ is a union of at most $|mathscr B|^n$ copies of $F^n$. This lets you calculate the cardinality using cardinal arithmetic.
$endgroup$
– Alex Becker
Sep 11 '12 at 19:19
$begingroup$
@AlexBecker Thanks. May I check if I did this right? Since $B$ is infinite, $|B|^n=|B|$ for all $n$. Also, $|B||F|^n=max{|B|,|F|^n}=max{|B|,|F|}$ regardless of whether $F$ is finite or infinite. Doing this for all $n$, the cardinality of $V$ is $max{|B|,|F|}cdotaleph_0=max{|B|,|F|}$ anyway since $|B|geqaleph_0$?
$endgroup$
– Nastassja
Sep 11 '12 at 19:28
|
show 1 more comment
$begingroup$
Suppose $V$ is a vector space over a scalar field $F$. If $dim(V)=n$, then $|V|=|F|^n$. How can I rigorously determine the cardinality of $V$ when $V$ is infinite dimensional?
My thought was that if $mathscr{B}$ is an ordered basis for $V$, then the cardinality of $V$ is given by the set of functions from $mathscr{B}to F$, by identifying elements of $V$ with their $mathscr{B}$-coordinate vector. However, I feel that we should only count functions with finite support since infinite sums don't make sense.
Is this correct? If so, how does one find the cardinality of ${fcolonmathscr{B}to Fmid mathrm{supp }(f)<infty}$, in terms of say $|F|$ and $|mathscr{B}|$? Thanks.
linear-algebra cardinals
asked Sep 11 '12 at 18:41
NastassjaNastassja
685413
$endgroup$
Suppose $V$ is a vector space over a scalar field $F$. If $dim(V)=n$, then $|V|=|F|^n$. How can I rigorously determine the cardinality of $V$ when $V$ is infinite dimensional?
My thought was that if $mathscr{B}$ is an ordered basis for $V$, then the cardinality of $V$ is given by the set of functions from $mathscr{B}to F$, by identifying elements of $V$ with their $mathscr{B}$-coordinate vector. However, I feel that we should only count functions with finite support since infinite sums don't make sense.
Is this correct? If so, how does one find the cardinality of ${fcolonmathscr{B}to Fmid mathrm{supp }(f)<infty}$, in terms of say $|F|$ and $|mathscr{B}|$? Thanks.
linear-algebra cardinals
linear-algebra cardinals
asked Sep 11 '12 at 18:41
NastassjaNastassja
685413
asked Sep 11 '12 at 18:41
NastassjaNastassja
685413
asked Sep 11 '12 at 18:41
NastassjaNastassja
685413
asked Sep 11 '12 at 18:41
NastassjaNastassja
685413
asked Sep 11 '12 at 18:41
NastassjaNastassja
685413
685413
1
$begingroup$
Can you find the number of functions from $mathscr{B}$ to $F$ with support of size at most $n$?
$endgroup$
– Chris Eagle
Sep 11 '12 at 18:48
$begingroup$
@ChrisEagle Wouldn't that require choosing $n$ vectors in $mathscr{B}$ to send to nonzero elements of $F$? That seems like it would already be very large since $mathscr{B}$ is infinite.
$endgroup$
– Nastassja
Sep 11 '12 at 19:10
$begingroup$
@Nastassja But what would the infinite cardinal be?
$endgroup$
– Alex Becker
Sep 11 '12 at 19:10
1
$begingroup$
@Nastassja The point is that the set of functions from $mathscr B$ to $F$ with support at most $n$ is a union of at most $|mathscr B|^n$ copies of $F^n$. This lets you calculate the cardinality using cardinal arithmetic.
$endgroup$
– Alex Becker
Sep 11 '12 at 19:19
$begingroup$
@AlexBecker Thanks. May I check if I did this right? Since $B$ is infinite, $|B|^n=|B|$ for all $n$. Also, $|B||F|^n=max{|B|,|F|^n}=max{|B|,|F|}$ regardless of whether $F$ is finite or infinite. Doing this for all $n$, the cardinality of $V$ is $max{|B|,|F|}cdotaleph_0=max{|B|,|F|}$ anyway since $|B|geqaleph_0$?
$endgroup$
– Nastassja
Sep 11 '12 at 19:28
|
show 1 more comment
1
$begingroup$
Can you find the number of functions from $mathscr{B}$ to $F$ with support of size at most $n$?
$endgroup$
– Chris Eagle
Sep 11 '12 at 18:48
$begingroup$
@ChrisEagle Wouldn't that require choosing $n$ vectors in $mathscr{B}$ to send to nonzero elements of $F$? That seems like it would already be very large since $mathscr{B}$ is infinite.
$endgroup$
– Nastassja
Sep 11 '12 at 19:10
$begingroup$
@Nastassja But what would the infinite cardinal be?
$endgroup$
– Alex Becker
Sep 11 '12 at 19:10
1
$begingroup$
@Nastassja The point is that the set of functions from $mathscr B$ to $F$ with support at most $n$ is a union of at most $|mathscr B|^n$ copies of $F^n$. This lets you calculate the cardinality using cardinal arithmetic.
$endgroup$
– Alex Becker
Sep 11 '12 at 19:19
$begingroup$
@AlexBecker Thanks. May I check if I did this right? Since $B$ is infinite, $|B|^n=|B|$ for all $n$. Also, $|B||F|^n=max{|B|,|F|^n}=max{|B|,|F|}$ regardless of whether $F$ is finite or infinite. Doing this for all $n$, the cardinality of $V$ is $max{|B|,|F|}cdotaleph_0=max{|B|,|F|}$ anyway since $|B|geqaleph_0$?
$endgroup$
– Nastassja
Sep 11 '12 at 19:28
1
1
$begingroup$
Can you find the number of functions from $mathscr{B}$ to $F$ with support of size at most $n$?
$endgroup$
– Chris Eagle
Sep 11 '12 at 18:48
$begingroup$
Can you find the number of functions from $mathscr{B}$ to $F$ with support of size at most $n$?
$endgroup$
– Chris Eagle
Sep 11 '12 at 18:48
$begingroup$
@ChrisEagle Wouldn't that require choosing $n$ vectors in $mathscr{B}$ to send to nonzero elements of $F$? That seems like it would already be very large since $mathscr{B}$ is infinite.
$endgroup$
– Nastassja
Sep 11 '12 at 19:10
$begingroup$
@ChrisEagle Wouldn't that require choosing $n$ vectors in $mathscr{B}$ to send to nonzero elements of $F$? That seems like it would already be very large since $mathscr{B}$ is infinite.
$endgroup$
– Nastassja
Sep 11 '12 at 19:10
$begingroup$
@Nastassja But what would the infinite cardinal be?
$endgroup$
– Alex Becker
Sep 11 '12 at 19:10
$begingroup$
@Nastassja But what would the infinite cardinal be?
$endgroup$
– Alex Becker
Sep 11 '12 at 19:10
1
1
$begingroup$
@Nastassja The point is that the set of functions from $mathscr B$ to $F$ with support at most $n$ is a union of at most $|mathscr B|^n$ copies of $F^n$. This lets you calculate the cardinality using cardinal arithmetic.
$endgroup$
– Alex Becker
Sep 11 '12 at 19:19
$begingroup$
@Nastassja The point is that the set of functions from $mathscr B$ to $F$ with support at most $n$ is a union of at most $|mathscr B|^n$ copies of $F^n$. This lets you calculate the cardinality using cardinal arithmetic.
$endgroup$
– Alex Becker
Sep 11 '12 at 19:19
$begingroup$
@AlexBecker Thanks. May I check if I did this right? Since $B$ is infinite, $|B|^n=|B|$ for all $n$. Also, $|B||F|^n=max{|B|,|F|^n}=max{|B|,|F|}$ regardless of whether $F$ is finite or infinite. Doing this for all $n$, the cardinality of $V$ is $max{|B|,|F|}cdotaleph_0=max{|B|,|F|}$ anyway since $|B|geqaleph_0$?
$endgroup$
– Nastassja
Sep 11 '12 at 19:28
$begingroup$
@AlexBecker Thanks. May I check if I did this right? Since $B$ is infinite, $|B|^n=|B|$ for all $n$. Also, $|B||F|^n=max{|B|,|F|^n}=max{|B|,|F|}$ regardless of whether $F$ is finite or infinite. Doing this for all $n$, the cardinality of $V$ is $max{|B|,|F|}cdotaleph_0=max{|B|,|F|}$ anyway since $|B|geqaleph_0$?
$endgroup$
– Nastassja
Sep 11 '12 at 19:28
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $V$ is a vector space over
$F$ and $V$ has a basis $B$.
From the definition of a basis every $vin V$ can be written as a unique sum of basis elements and scalars. That is, there is a finite subset of $Btimes (Fsetminus{0})$ whose sum is $v$, and if we require that this set is a function on its domain, then this set is unique.
This gives a well-defined injection from $V$ into finite subsets of $Btimes(Fsetminus{0})$. Assuming the axiom of choice we have that,
$$|V|leqleft|[Btimes(Fsetminus{0})]^{<omega}right|=|Btimes F|=max{|B|,|F|}leq|V|implies|V|=max{|B|,|F|}.$$
answered Sep 11 '12 at 19:02
Asaf Karagila♦Asaf Karagila
306k33437768
$endgroup$
$begingroup$
The question asks what is the cardinality of $V$, given the dimension of $V$ and the cardinality of the scalar field.
$endgroup$
– Chris Eagle
Sep 11 '12 at 19:06
$begingroup$
Thanks, but I don't see how this applies. I'm already assuming a basis is known to exist, and trying to compute the cardinality of the vector space.
$endgroup$
– Nastassja
Sep 11 '12 at 19:17
1
$begingroup$
Writing and revising while drinking and using iPhone keyboard is just hellish!! :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 20:02
$begingroup$
Thanks Asaf, I think this is a much neater presentation than what I said above.
$endgroup$
– Nastassja
Sep 11 '12 at 21:10
$begingroup$
@Nastassja: Well to be fair I just finished my M.Sc. thesis and I had to write something like that there... :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 21:12
|
show 3 more comments
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
Suppose that $V$ is a vector space over
$F$ and $V$ has a basis $B$.
From the definition of a basis every $vin V$ can be written as a unique sum of basis elements and scalars. That is, there is a finite subset of $Btimes (Fsetminus{0})$ whose sum is $v$, and if we require that this set is a function on its domain, then this set is unique.
This gives a well-defined injection from $V$ into finite subsets of $Btimes(Fsetminus{0})$. Assuming the axiom of choice we have that,
$$|V|leqleft|[Btimes(Fsetminus{0})]^{<omega}right|=|Btimes F|=max{|B|,|F|}leq|V|implies|V|=max{|B|,|F|}.$$
answered Sep 11 '12 at 19:02
Asaf Karagila♦Asaf Karagila
306k33437768
$endgroup$
$begingroup$
The question asks what is the cardinality of $V$, given the dimension of $V$ and the cardinality of the scalar field.
$endgroup$
– Chris Eagle
Sep 11 '12 at 19:06
$begingroup$
Thanks, but I don't see how this applies. I'm already assuming a basis is known to exist, and trying to compute the cardinality of the vector space.
$endgroup$
– Nastassja
Sep 11 '12 at 19:17
1
$begingroup$
Writing and revising while drinking and using iPhone keyboard is just hellish!! :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 20:02
$begingroup$
Thanks Asaf, I think this is a much neater presentation than what I said above.
$endgroup$
– Nastassja
Sep 11 '12 at 21:10
$begingroup$
@Nastassja: Well to be fair I just finished my M.Sc. thesis and I had to write something like that there... :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 21:12
|
show 3 more comments
$begingroup$
Suppose that $V$ is a vector space over
$F$ and $V$ has a basis $B$.
From the definition of a basis every $vin V$ can be written as a unique sum of basis elements and scalars. That is, there is a finite subset of $Btimes (Fsetminus{0})$ whose sum is $v$, and if we require that this set is a function on its domain, then this set is unique.
This gives a well-defined injection from $V$ into finite subsets of $Btimes(Fsetminus{0})$. Assuming the axiom of choice we have that,
$$|V|leqleft|[Btimes(Fsetminus{0})]^{<omega}right|=|Btimes F|=max{|B|,|F|}leq|V|implies|V|=max{|B|,|F|}.$$
answered Sep 11 '12 at 19:02
Asaf Karagila♦Asaf Karagila
306k33437768
$endgroup$
$begingroup$
The question asks what is the cardinality of $V$, given the dimension of $V$ and the cardinality of the scalar field.
$endgroup$
– Chris Eagle
Sep 11 '12 at 19:06
$begingroup$
Thanks, but I don't see how this applies. I'm already assuming a basis is known to exist, and trying to compute the cardinality of the vector space.
$endgroup$
– Nastassja
Sep 11 '12 at 19:17
1
$begingroup$
Writing and revising while drinking and using iPhone keyboard is just hellish!! :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 20:02
$begingroup$
Thanks Asaf, I think this is a much neater presentation than what I said above.
$endgroup$
– Nastassja
Sep 11 '12 at 21:10
$begingroup$
@Nastassja: Well to be fair I just finished my M.Sc. thesis and I had to write something like that there... :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 21:12
|
show 3 more comments
$begingroup$
Suppose that $V$ is a vector space over
$F$ and $V$ has a basis $B$.
From the definition of a basis every $vin V$ can be written as a unique sum of basis elements and scalars. That is, there is a finite subset of $Btimes (Fsetminus{0})$ whose sum is $v$, and if we require that this set is a function on its domain, then this set is unique.
This gives a well-defined injection from $V$ into finite subsets of $Btimes(Fsetminus{0})$. Assuming the axiom of choice we have that,
$$|V|leqleft|[Btimes(Fsetminus{0})]^{<omega}right|=|Btimes F|=max{|B|,|F|}leq|V|implies|V|=max{|B|,|F|}.$$
answered Sep 11 '12 at 19:02
Asaf Karagila♦Asaf Karagila
306k33437768
$endgroup$
Suppose that $V$ is a vector space over
$F$ and $V$ has a basis $B$.
From the definition of a basis every $vin V$ can be written as a unique sum of basis elements and scalars. That is, there is a finite subset of $Btimes (Fsetminus{0})$ whose sum is $v$, and if we require that this set is a function on its domain, then this set is unique.
This gives a well-defined injection from $V$ into finite subsets of $Btimes(Fsetminus{0})$. Assuming the axiom of choice we have that,
$$|V|leqleft|[Btimes(Fsetminus{0})]^{<omega}right|=|Btimes F|=max{|B|,|F|}leq|V|implies|V|=max{|B|,|F|}.$$
answered Sep 11 '12 at 19:02
Asaf Karagila♦Asaf Karagila
306k33437768
edited Jul 11 '14 at 20:59
answered Sep 11 '12 at 19:02
Asaf Karagila♦Asaf Karagila
306k33437768
answered Sep 11 '12 at 19:02
Asaf Karagila♦Asaf Karagila
306k33437768
answered Sep 11 '12 at 19:02
Asaf Karagila♦Asaf Karagila
306k33437768
306k33437768
$begingroup$
The question asks what is the cardinality of $V$, given the dimension of $V$ and the cardinality of the scalar field.
$endgroup$
– Chris Eagle
Sep 11 '12 at 19:06
$begingroup$
Thanks, but I don't see how this applies. I'm already assuming a basis is known to exist, and trying to compute the cardinality of the vector space.
$endgroup$
– Nastassja
Sep 11 '12 at 19:17
1
$begingroup$
Writing and revising while drinking and using iPhone keyboard is just hellish!! :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 20:02
$begingroup$
Thanks Asaf, I think this is a much neater presentation than what I said above.
$endgroup$
– Nastassja
Sep 11 '12 at 21:10
$begingroup$
@Nastassja: Well to be fair I just finished my M.Sc. thesis and I had to write something like that there... :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 21:12
|
show 3 more comments
$begingroup$
The question asks what is the cardinality of $V$, given the dimension of $V$ and the cardinality of the scalar field.
$endgroup$
– Chris Eagle
Sep 11 '12 at 19:06
$begingroup$
Thanks, but I don't see how this applies. I'm already assuming a basis is known to exist, and trying to compute the cardinality of the vector space.
$endgroup$
– Nastassja
Sep 11 '12 at 19:17
1
$begingroup$
Writing and revising while drinking and using iPhone keyboard is just hellish!! :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 20:02
$begingroup$
Thanks Asaf, I think this is a much neater presentation than what I said above.
$endgroup$
– Nastassja
Sep 11 '12 at 21:10
$begingroup$
@Nastassja: Well to be fair I just finished my M.Sc. thesis and I had to write something like that there... :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 21:12
$begingroup$
The question asks what is the cardinality of $V$, given the dimension of $V$ and the cardinality of the scalar field.
$endgroup$
– Chris Eagle
Sep 11 '12 at 19:06
$begingroup$
The question asks what is the cardinality of $V$, given the dimension of $V$ and the cardinality of the scalar field.
$endgroup$
– Chris Eagle
Sep 11 '12 at 19:06
$begingroup$
Thanks, but I don't see how this applies. I'm already assuming a basis is known to exist, and trying to compute the cardinality of the vector space.
$endgroup$
– Nastassja
Sep 11 '12 at 19:17
$begingroup$
Thanks, but I don't see how this applies. I'm already assuming a basis is known to exist, and trying to compute the cardinality of the vector space.
$endgroup$
– Nastassja
Sep 11 '12 at 19:17
1
1
$begingroup$
Writing and revising while drinking and using iPhone keyboard is just hellish!! :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 20:02
$begingroup$
Writing and revising while drinking and using iPhone keyboard is just hellish!! :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 20:02
$begingroup$
Thanks Asaf, I think this is a much neater presentation than what I said above.
$endgroup$
– Nastassja
Sep 11 '12 at 21:10
$begingroup$
Thanks Asaf, I think this is a much neater presentation than what I said above.
$endgroup$
– Nastassja
Sep 11 '12 at 21:10
$begingroup$
@Nastassja: Well to be fair I just finished my M.Sc. thesis and I had to write something like that there... :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 21:12
$begingroup$
@Nastassja: Well to be fair I just finished my M.Sc. thesis and I had to write something like that there... :-)
$endgroup$
– Asaf Karagila♦
Sep 11 '12 at 21:12
|
show 3 more comments
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1
$begingroup$
Can you find the number of functions from $mathscr{B}$ to $F$ with support of size at most $n$?
$endgroup$
– Chris Eagle
Sep 11 '12 at 18:48
$begingroup$
@ChrisEagle Wouldn't that require choosing $n$ vectors in $mathscr{B}$ to send to nonzero elements of $F$? That seems like it would already be very large since $mathscr{B}$ is infinite.
$endgroup$
– Nastassja
Sep 11 '12 at 19:10
$begingroup$
@Nastassja But what would the infinite cardinal be?
$endgroup$
– Alex Becker
Sep 11 '12 at 19:10
1
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@Nastassja The point is that the set of functions from $mathscr B$ to $F$ with support at most $n$ is a union of at most $|mathscr B|^n$ copies of $F^n$. This lets you calculate the cardinality using cardinal arithmetic.
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– Alex Becker
Sep 11 '12 at 19:19
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@AlexBecker Thanks. May I check if I did this right? Since $B$ is infinite, $|B|^n=|B|$ for all $n$. Also, $|B||F|^n=max{|B|,|F|^n}=max{|B|,|F|}$ regardless of whether $F$ is finite or infinite. Doing this for all $n$, the cardinality of $V$ is $max{|B|,|F|}cdotaleph_0=max{|B|,|F|}$ anyway since $|B|geqaleph_0$?
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– Nastassja
Sep 11 '12 at 19:28