Mixing
$text{Show that }F^{I}=F^{I}_{text{fin}} iff I text{ finite, where }I text{ is a set and } Ftext{ is a...
$begingroup$
$F^{I} text{ is the vectorspace of functions which map from }I text{ to } F text{ and }F^{I}_{text{fin}}:= { f:Irightarrow F |text{ } overset{exists}{ninmathbb{N}_0,}overset{exists}{I_fsubseteq I}:text{#} I_f = n text{ and} begin{cases}f(i)=0text{ , for each }iin Ibackslash I_f\f(i)neq 0text{ , for each }iin I_fend{cases}}$
The context is that I have already proved that $B:={e_i:iin I}$ is linear independant in $F^{I}$ where
$$forall_{iin I}e_i:Irightarrow F, e_{i}(j):=delta_{ij}:=begin{cases} 1,text{ if }i=j\0,text{ if }ineq jend{cases}$$
And now I want to Show that for $F_{text{fin}}^{I}$ $B$ is a base. For the special case it is noteworthy that $F_{text{fin}}^{I}$ and $F^{I}$ describes the same vectorspace and thus have the same base if $I$ is finite.
linear-algebra
asked Jan 18 at 21:18
RM777RM777
38012
$endgroup$
add a comment |
$begingroup$
$F^{I} text{ is the vectorspace of functions which map from }I text{ to } F text{ and }F^{I}_{text{fin}}:= { f:Irightarrow F |text{ } overset{exists}{ninmathbb{N}_0,}overset{exists}{I_fsubseteq I}:text{#} I_f = n text{ and} begin{cases}f(i)=0text{ , for each }iin Ibackslash I_f\f(i)neq 0text{ , for each }iin I_fend{cases}}$
The context is that I have already proved that $B:={e_i:iin I}$ is linear independant in $F^{I}$ where
$$forall_{iin I}e_i:Irightarrow F, e_{i}(j):=delta_{ij}:=begin{cases} 1,text{ if }i=j\0,text{ if }ineq jend{cases}$$
And now I want to Show that for $F_{text{fin}}^{I}$ $B$ is a base. For the special case it is noteworthy that $F_{text{fin}}^{I}$ and $F^{I}$ describes the same vectorspace and thus have the same base if $I$ is finite.
linear-algebra
asked Jan 18 at 21:18
RM777RM777
38012
$endgroup$
add a comment |
$begingroup$
$F^{I} text{ is the vectorspace of functions which map from }I text{ to } F text{ and }F^{I}_{text{fin}}:= { f:Irightarrow F |text{ } overset{exists}{ninmathbb{N}_0,}overset{exists}{I_fsubseteq I}:text{#} I_f = n text{ and} begin{cases}f(i)=0text{ , for each }iin Ibackslash I_f\f(i)neq 0text{ , for each }iin I_fend{cases}}$
The context is that I have already proved that $B:={e_i:iin I}$ is linear independant in $F^{I}$ where
$$forall_{iin I}e_i:Irightarrow F, e_{i}(j):=delta_{ij}:=begin{cases} 1,text{ if }i=j\0,text{ if }ineq jend{cases}$$
And now I want to Show that for $F_{text{fin}}^{I}$ $B$ is a base. For the special case it is noteworthy that $F_{text{fin}}^{I}$ and $F^{I}$ describes the same vectorspace and thus have the same base if $I$ is finite.
linear-algebra
asked Jan 18 at 21:18
RM777RM777
38012
$endgroup$
$F^{I} text{ is the vectorspace of functions which map from }I text{ to } F text{ and }F^{I}_{text{fin}}:= { f:Irightarrow F |text{ } overset{exists}{ninmathbb{N}_0,}overset{exists}{I_fsubseteq I}:text{#} I_f = n text{ and} begin{cases}f(i)=0text{ , for each }iin Ibackslash I_f\f(i)neq 0text{ , for each }iin I_fend{cases}}$
The context is that I have already proved that $B:={e_i:iin I}$ is linear independant in $F^{I}$ where
$$forall_{iin I}e_i:Irightarrow F, e_{i}(j):=delta_{ij}:=begin{cases} 1,text{ if }i=j\0,text{ if }ineq jend{cases}$$
And now I want to Show that for $F_{text{fin}}^{I}$ $B$ is a base. For the special case it is noteworthy that $F_{text{fin}}^{I}$ and $F^{I}$ describes the same vectorspace and thus have the same base if $I$ is finite.
linear-algebra
linear-algebra
asked Jan 18 at 21:18
RM777RM777
38012
asked Jan 18 at 21:18
RM777RM777
38012
edited Jan 20 at 16:10
RM777
asked Jan 18 at 21:18
RM777RM777
38012
asked Jan 18 at 21:18
RM777RM777
38012
asked Jan 18 at 21:18
RM777RM777
38012
38012
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We do not even need that $F$ is a field. If $I$ is finite, then we certainly have $F^I = F_{mathrm{fin}}^I$. The inclusion $F_{mathrm{fin}}^I subset F^I$ always holds (by definition). To see the other inclusion, let $fin F^I$. Then
$$
lbrace iin I,vert , f(i)neq 0rbracesubset I,
$$
and since the RHS is finite, the LHS must also be finite. Hence $fin F_{mathrm{fin}}^I$ by definition.
If $I$ is infinite, simply pick a nonzero element in $F$ (e.g. $1$) and consider the function which maps all elements of $I$ to $1$. This function does not have finite support, hence $F^I neq F_{mathrm{fin}}^I$, proving the contrapositive of the opposite implication, hence completing the proof.
answered Jan 18 at 21:32
o.h.o.h.
4616
$endgroup$
$begingroup$
Why is $F^I = F_{mathrm{fin}}^I$ if $I$ is finite?
$endgroup$
– RM777
Jan 18 at 21:34
1
$begingroup$
If $I$ has an integer-valued cardinality (e.g. is finite), then what we might refer to as the "support" of a function $fcolon Irightarrow F$ must have a (smaller) integer-valued cardinality (being a subset of $I$).
$endgroup$
– o.h.
Jan 18 at 21:39
add a comment |
$begingroup$
The space $F^I_{mathrm{fin}}$ can be more simply defined as
$$
F^I_{mathrm{fin}}=bigl{fin F^I: |{iin I:f(i)ne0}|<infty bigr}
$$
Take $uin F^I$ defined by $u(i)=1$, for every $iin I$. Then $unotin F^I_{mathrm{fin}}$, unless $I$ is finite.
answered Jan 18 at 21:45
egregegreg
183k1486205
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We do not even need that $F$ is a field. If $I$ is finite, then we certainly have $F^I = F_{mathrm{fin}}^I$. The inclusion $F_{mathrm{fin}}^I subset F^I$ always holds (by definition). To see the other inclusion, let $fin F^I$. Then
$$
lbrace iin I,vert , f(i)neq 0rbracesubset I,
$$
and since the RHS is finite, the LHS must also be finite. Hence $fin F_{mathrm{fin}}^I$ by definition.
If $I$ is infinite, simply pick a nonzero element in $F$ (e.g. $1$) and consider the function which maps all elements of $I$ to $1$. This function does not have finite support, hence $F^I neq F_{mathrm{fin}}^I$, proving the contrapositive of the opposite implication, hence completing the proof.
answered Jan 18 at 21:32
o.h.o.h.
4616
$endgroup$
$begingroup$
Why is $F^I = F_{mathrm{fin}}^I$ if $I$ is finite?
$endgroup$
– RM777
Jan 18 at 21:34
1
$begingroup$
If $I$ has an integer-valued cardinality (e.g. is finite), then what we might refer to as the "support" of a function $fcolon Irightarrow F$ must have a (smaller) integer-valued cardinality (being a subset of $I$).
$endgroup$
– o.h.
Jan 18 at 21:39
add a comment |
$begingroup$
We do not even need that $F$ is a field. If $I$ is finite, then we certainly have $F^I = F_{mathrm{fin}}^I$. The inclusion $F_{mathrm{fin}}^I subset F^I$ always holds (by definition). To see the other inclusion, let $fin F^I$. Then
$$
lbrace iin I,vert , f(i)neq 0rbracesubset I,
$$
and since the RHS is finite, the LHS must also be finite. Hence $fin F_{mathrm{fin}}^I$ by definition.
If $I$ is infinite, simply pick a nonzero element in $F$ (e.g. $1$) and consider the function which maps all elements of $I$ to $1$. This function does not have finite support, hence $F^I neq F_{mathrm{fin}}^I$, proving the contrapositive of the opposite implication, hence completing the proof.
answered Jan 18 at 21:32
o.h.o.h.
4616
$endgroup$
$begingroup$
Why is $F^I = F_{mathrm{fin}}^I$ if $I$ is finite?
$endgroup$
– RM777
Jan 18 at 21:34
1
$begingroup$
If $I$ has an integer-valued cardinality (e.g. is finite), then what we might refer to as the "support" of a function $fcolon Irightarrow F$ must have a (smaller) integer-valued cardinality (being a subset of $I$).
$endgroup$
– o.h.
Jan 18 at 21:39
add a comment |
$begingroup$
We do not even need that $F$ is a field. If $I$ is finite, then we certainly have $F^I = F_{mathrm{fin}}^I$. The inclusion $F_{mathrm{fin}}^I subset F^I$ always holds (by definition). To see the other inclusion, let $fin F^I$. Then
$$
lbrace iin I,vert , f(i)neq 0rbracesubset I,
$$
and since the RHS is finite, the LHS must also be finite. Hence $fin F_{mathrm{fin}}^I$ by definition.
If $I$ is infinite, simply pick a nonzero element in $F$ (e.g. $1$) and consider the function which maps all elements of $I$ to $1$. This function does not have finite support, hence $F^I neq F_{mathrm{fin}}^I$, proving the contrapositive of the opposite implication, hence completing the proof.
answered Jan 18 at 21:32
o.h.o.h.
4616
$endgroup$
We do not even need that $F$ is a field. If $I$ is finite, then we certainly have $F^I = F_{mathrm{fin}}^I$. The inclusion $F_{mathrm{fin}}^I subset F^I$ always holds (by definition). To see the other inclusion, let $fin F^I$. Then
$$
lbrace iin I,vert , f(i)neq 0rbracesubset I,
$$
and since the RHS is finite, the LHS must also be finite. Hence $fin F_{mathrm{fin}}^I$ by definition.
If $I$ is infinite, simply pick a nonzero element in $F$ (e.g. $1$) and consider the function which maps all elements of $I$ to $1$. This function does not have finite support, hence $F^I neq F_{mathrm{fin}}^I$, proving the contrapositive of the opposite implication, hence completing the proof.
answered Jan 18 at 21:32
o.h.o.h.
4616
edited Jan 18 at 21:37
answered Jan 18 at 21:32
o.h.o.h.
4616
answered Jan 18 at 21:32
o.h.o.h.
4616
answered Jan 18 at 21:32
o.h.o.h.
4616
4616
$begingroup$
Why is $F^I = F_{mathrm{fin}}^I$ if $I$ is finite?
$endgroup$
– RM777
Jan 18 at 21:34
1
$begingroup$
If $I$ has an integer-valued cardinality (e.g. is finite), then what we might refer to as the "support" of a function $fcolon Irightarrow F$ must have a (smaller) integer-valued cardinality (being a subset of $I$).
$endgroup$
– o.h.
Jan 18 at 21:39
add a comment |
$begingroup$
Why is $F^I = F_{mathrm{fin}}^I$ if $I$ is finite?
$endgroup$
– RM777
Jan 18 at 21:34
1
$begingroup$
If $I$ has an integer-valued cardinality (e.g. is finite), then what we might refer to as the "support" of a function $fcolon Irightarrow F$ must have a (smaller) integer-valued cardinality (being a subset of $I$).
$endgroup$
– o.h.
Jan 18 at 21:39
$begingroup$
Why is $F^I = F_{mathrm{fin}}^I$ if $I$ is finite?
$endgroup$
– RM777
Jan 18 at 21:34
$begingroup$
Why is $F^I = F_{mathrm{fin}}^I$ if $I$ is finite?
$endgroup$
– RM777
Jan 18 at 21:34
1
1
$begingroup$
If $I$ has an integer-valued cardinality (e.g. is finite), then what we might refer to as the "support" of a function $fcolon Irightarrow F$ must have a (smaller) integer-valued cardinality (being a subset of $I$).
$endgroup$
– o.h.
Jan 18 at 21:39
$begingroup$
If $I$ has an integer-valued cardinality (e.g. is finite), then what we might refer to as the "support" of a function $fcolon Irightarrow F$ must have a (smaller) integer-valued cardinality (being a subset of $I$).
$endgroup$
– o.h.
Jan 18 at 21:39
add a comment |
$begingroup$
The space $F^I_{mathrm{fin}}$ can be more simply defined as
$$
F^I_{mathrm{fin}}=bigl{fin F^I: |{iin I:f(i)ne0}|<infty bigr}
$$
Take $uin F^I$ defined by $u(i)=1$, for every $iin I$. Then $unotin F^I_{mathrm{fin}}$, unless $I$ is finite.
answered Jan 18 at 21:45
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
The space $F^I_{mathrm{fin}}$ can be more simply defined as
$$
F^I_{mathrm{fin}}=bigl{fin F^I: |{iin I:f(i)ne0}|<infty bigr}
$$
Take $uin F^I$ defined by $u(i)=1$, for every $iin I$. Then $unotin F^I_{mathrm{fin}}$, unless $I$ is finite.
answered Jan 18 at 21:45
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
The space $F^I_{mathrm{fin}}$ can be more simply defined as
$$
F^I_{mathrm{fin}}=bigl{fin F^I: |{iin I:f(i)ne0}|<infty bigr}
$$
Take $uin F^I$ defined by $u(i)=1$, for every $iin I$. Then $unotin F^I_{mathrm{fin}}$, unless $I$ is finite.
answered Jan 18 at 21:45
egregegreg
183k1486205
$endgroup$
The space $F^I_{mathrm{fin}}$ can be more simply defined as
$$
F^I_{mathrm{fin}}=bigl{fin F^I: |{iin I:f(i)ne0}|<infty bigr}
$$
Take $uin F^I$ defined by $u(i)=1$, for every $iin I$. Then $unotin F^I_{mathrm{fin}}$, unless $I$ is finite.
answered Jan 18 at 21:45
egregegreg
183k1486205
answered Jan 18 at 21:45
egregegreg
183k1486205
answered Jan 18 at 21:45
egregegreg
183k1486205
answered Jan 18 at 21:45
egregegreg
183k1486205
183k1486205
add a comment |
add a comment |
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