Characteristic function / Indicator function
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I really don’t understand the characteristic function in terms of sets. I get that it maps a set to the set {0,1} and you can use this to count the number of elements in that set. But,set operations like unions and intersections, I don’t get how it’s used, also proving the Inclusion-Exclusion principle. Could someone explain this to me, can’t find a site that explains it properly.
elementary-number-theory elementary-set-theory
edited Jan 6 at 13:39
Andrés E. Caicedo
65.4k8158249
asked Jan 6 at 13:38
DraxlerDraxler
122
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add a comment |
$begingroup$
I really don’t understand the characteristic function in terms of sets. I get that it maps a set to the set {0,1} and you can use this to count the number of elements in that set. But,set operations like unions and intersections, I don’t get how it’s used, also proving the Inclusion-Exclusion principle. Could someone explain this to me, can’t find a site that explains it properly.
elementary-number-theory elementary-set-theory
edited Jan 6 at 13:39
Andrés E. Caicedo
65.4k8158249
asked Jan 6 at 13:38
DraxlerDraxler
122
$endgroup$
add a comment |
$begingroup$
I really don’t understand the characteristic function in terms of sets. I get that it maps a set to the set {0,1} and you can use this to count the number of elements in that set. But,set operations like unions and intersections, I don’t get how it’s used, also proving the Inclusion-Exclusion principle. Could someone explain this to me, can’t find a site that explains it properly.
elementary-number-theory elementary-set-theory
edited Jan 6 at 13:39
Andrés E. Caicedo
65.4k8158249
asked Jan 6 at 13:38
DraxlerDraxler
122
$endgroup$
I really don’t understand the characteristic function in terms of sets. I get that it maps a set to the set {0,1} and you can use this to count the number of elements in that set. But,set operations like unions and intersections, I don’t get how it’s used, also proving the Inclusion-Exclusion principle. Could someone explain this to me, can’t find a site that explains it properly.
elementary-number-theory elementary-set-theory
elementary-number-theory elementary-set-theory
edited Jan 6 at 13:39
Andrés E. Caicedo
65.4k8158249
asked Jan 6 at 13:38
DraxlerDraxler
122
edited Jan 6 at 13:39
Andrés E. Caicedo
65.4k8158249
asked Jan 6 at 13:38
DraxlerDraxler
122
edited Jan 6 at 13:39
Andrés E. Caicedo
65.4k8158249
edited Jan 6 at 13:39
Andrés E. Caicedo
65.4k8158249
edited Jan 6 at 13:39
Andrés E. Caicedo
65.4k8158249
65.4k8158249
asked Jan 6 at 13:38
DraxlerDraxler
122
asked Jan 6 at 13:38
DraxlerDraxler
122
asked Jan 6 at 13:38
DraxlerDraxler
122
122
add a comment |
add a comment |
2 Answers
2
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oldest
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The characteristic (or : indicator) function for a subset $A$ of $X$ is the function :
$1_A ; X to { 0,1 } text { such that : for every } x in X : 1_A(x)=1 text { iff } x in A$.
But a function is a set of pairs, i.e. a subset of the cartesian product.
Thus :
$1_A = { (z,b) mid z in A text { and } b in { 0,1 } }$
and : $1_A subseteq X times { 0,1 }$.
answered Jan 6 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
65.9k449114
$endgroup$
add a comment |
$begingroup$
For definition of characteristic function see the answer of Mauro.
If $A,B$ are sets then $mathbf1_{Acap B}=mathbf1_Amathbf1_B$.
This because:$$xin Acap Biff xin Awedge xin Biffmathbf1_A(x)=1wedgemathbf1_B(x)=1iff mathbf1_Amathbf1_B=1$$
Concerning unions we have: $$mathbf1_{Acup B}=mathbf1_A+mathbf1_B-mathbf1_{Acap B}=mathbf1_A+mathbf1_B-mathbf1_{A}mathbf1_{B}$$
Observe that by substitution of argument $x$ both sides we get $1$ as result if and only if $xin Acup B$, and $0$ otherwise. So the functions on LHS and RHS are the same.
The principle of inclusion/exclusion rests on the equality:$$mathbf1_{bigcup_{i=1}^n A_i}=sum_{i=1}^nmathbf1_{A_i}-sum_{1leq i<jleq n}mathbf1_{A_icap A_j}+cdots+(-1)^nmathbf1_{A_1capcdotscap A_n}tag1$$
For a proof of $(1)$ see this answer.
For any suitable measure $mu$ we can take expectation on both sides of $(1)$ resulting in:$$muleft(bigcup_{i=1}^n A_iright)=sum_{i=1}^nmu(A_i)-sum_{1leq i<jleq n}mu(A_icap A_j)+cdots+(-1)^nmu(A_1capcdotscap A_n)$$
answered Jan 6 at 14:23
drhabdrhab
101k545136
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add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
The characteristic (or : indicator) function for a subset $A$ of $X$ is the function :
$1_A ; X to { 0,1 } text { such that : for every } x in X : 1_A(x)=1 text { iff } x in A$.
But a function is a set of pairs, i.e. a subset of the cartesian product.
Thus :
$1_A = { (z,b) mid z in A text { and } b in { 0,1 } }$
and : $1_A subseteq X times { 0,1 }$.
answered Jan 6 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
65.9k449114
$endgroup$
add a comment |
$begingroup$
The characteristic (or : indicator) function for a subset $A$ of $X$ is the function :
$1_A ; X to { 0,1 } text { such that : for every } x in X : 1_A(x)=1 text { iff } x in A$.
But a function is a set of pairs, i.e. a subset of the cartesian product.
Thus :
$1_A = { (z,b) mid z in A text { and } b in { 0,1 } }$
and : $1_A subseteq X times { 0,1 }$.
answered Jan 6 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
65.9k449114
$endgroup$
add a comment |
$begingroup$
The characteristic (or : indicator) function for a subset $A$ of $X$ is the function :
$1_A ; X to { 0,1 } text { such that : for every } x in X : 1_A(x)=1 text { iff } x in A$.
But a function is a set of pairs, i.e. a subset of the cartesian product.
Thus :
$1_A = { (z,b) mid z in A text { and } b in { 0,1 } }$
and : $1_A subseteq X times { 0,1 }$.
answered Jan 6 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
65.9k449114
$endgroup$
The characteristic (or : indicator) function for a subset $A$ of $X$ is the function :
$1_A ; X to { 0,1 } text { such that : for every } x in X : 1_A(x)=1 text { iff } x in A$.
But a function is a set of pairs, i.e. a subset of the cartesian product.
Thus :
$1_A = { (z,b) mid z in A text { and } b in { 0,1 } }$
and : $1_A subseteq X times { 0,1 }$.
answered Jan 6 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
65.9k449114
answered Jan 6 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
65.9k449114
answered Jan 6 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
65.9k449114
answered Jan 6 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
65.9k449114
65.9k449114
add a comment |
add a comment |
$begingroup$
For definition of characteristic function see the answer of Mauro.
If $A,B$ are sets then $mathbf1_{Acap B}=mathbf1_Amathbf1_B$.
This because:$$xin Acap Biff xin Awedge xin Biffmathbf1_A(x)=1wedgemathbf1_B(x)=1iff mathbf1_Amathbf1_B=1$$
Concerning unions we have: $$mathbf1_{Acup B}=mathbf1_A+mathbf1_B-mathbf1_{Acap B}=mathbf1_A+mathbf1_B-mathbf1_{A}mathbf1_{B}$$
Observe that by substitution of argument $x$ both sides we get $1$ as result if and only if $xin Acup B$, and $0$ otherwise. So the functions on LHS and RHS are the same.
The principle of inclusion/exclusion rests on the equality:$$mathbf1_{bigcup_{i=1}^n A_i}=sum_{i=1}^nmathbf1_{A_i}-sum_{1leq i<jleq n}mathbf1_{A_icap A_j}+cdots+(-1)^nmathbf1_{A_1capcdotscap A_n}tag1$$
For a proof of $(1)$ see this answer.
For any suitable measure $mu$ we can take expectation on both sides of $(1)$ resulting in:$$muleft(bigcup_{i=1}^n A_iright)=sum_{i=1}^nmu(A_i)-sum_{1leq i<jleq n}mu(A_icap A_j)+cdots+(-1)^nmu(A_1capcdotscap A_n)$$
answered Jan 6 at 14:23
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
For definition of characteristic function see the answer of Mauro.
If $A,B$ are sets then $mathbf1_{Acap B}=mathbf1_Amathbf1_B$.
This because:$$xin Acap Biff xin Awedge xin Biffmathbf1_A(x)=1wedgemathbf1_B(x)=1iff mathbf1_Amathbf1_B=1$$
Concerning unions we have: $$mathbf1_{Acup B}=mathbf1_A+mathbf1_B-mathbf1_{Acap B}=mathbf1_A+mathbf1_B-mathbf1_{A}mathbf1_{B}$$
Observe that by substitution of argument $x$ both sides we get $1$ as result if and only if $xin Acup B$, and $0$ otherwise. So the functions on LHS and RHS are the same.
The principle of inclusion/exclusion rests on the equality:$$mathbf1_{bigcup_{i=1}^n A_i}=sum_{i=1}^nmathbf1_{A_i}-sum_{1leq i<jleq n}mathbf1_{A_icap A_j}+cdots+(-1)^nmathbf1_{A_1capcdotscap A_n}tag1$$
For a proof of $(1)$ see this answer.
For any suitable measure $mu$ we can take expectation on both sides of $(1)$ resulting in:$$muleft(bigcup_{i=1}^n A_iright)=sum_{i=1}^nmu(A_i)-sum_{1leq i<jleq n}mu(A_icap A_j)+cdots+(-1)^nmu(A_1capcdotscap A_n)$$
answered Jan 6 at 14:23
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
For definition of characteristic function see the answer of Mauro.
If $A,B$ are sets then $mathbf1_{Acap B}=mathbf1_Amathbf1_B$.
This because:$$xin Acap Biff xin Awedge xin Biffmathbf1_A(x)=1wedgemathbf1_B(x)=1iff mathbf1_Amathbf1_B=1$$
Concerning unions we have: $$mathbf1_{Acup B}=mathbf1_A+mathbf1_B-mathbf1_{Acap B}=mathbf1_A+mathbf1_B-mathbf1_{A}mathbf1_{B}$$
Observe that by substitution of argument $x$ both sides we get $1$ as result if and only if $xin Acup B$, and $0$ otherwise. So the functions on LHS and RHS are the same.
The principle of inclusion/exclusion rests on the equality:$$mathbf1_{bigcup_{i=1}^n A_i}=sum_{i=1}^nmathbf1_{A_i}-sum_{1leq i<jleq n}mathbf1_{A_icap A_j}+cdots+(-1)^nmathbf1_{A_1capcdotscap A_n}tag1$$
For a proof of $(1)$ see this answer.
For any suitable measure $mu$ we can take expectation on both sides of $(1)$ resulting in:$$muleft(bigcup_{i=1}^n A_iright)=sum_{i=1}^nmu(A_i)-sum_{1leq i<jleq n}mu(A_icap A_j)+cdots+(-1)^nmu(A_1capcdotscap A_n)$$
answered Jan 6 at 14:23
drhabdrhab
101k545136
$endgroup$
For definition of characteristic function see the answer of Mauro.
If $A,B$ are sets then $mathbf1_{Acap B}=mathbf1_Amathbf1_B$.
This because:$$xin Acap Biff xin Awedge xin Biffmathbf1_A(x)=1wedgemathbf1_B(x)=1iff mathbf1_Amathbf1_B=1$$
Concerning unions we have: $$mathbf1_{Acup B}=mathbf1_A+mathbf1_B-mathbf1_{Acap B}=mathbf1_A+mathbf1_B-mathbf1_{A}mathbf1_{B}$$
Observe that by substitution of argument $x$ both sides we get $1$ as result if and only if $xin Acup B$, and $0$ otherwise. So the functions on LHS and RHS are the same.
The principle of inclusion/exclusion rests on the equality:$$mathbf1_{bigcup_{i=1}^n A_i}=sum_{i=1}^nmathbf1_{A_i}-sum_{1leq i<jleq n}mathbf1_{A_icap A_j}+cdots+(-1)^nmathbf1_{A_1capcdotscap A_n}tag1$$
For a proof of $(1)$ see this answer.
For any suitable measure $mu$ we can take expectation on both sides of $(1)$ resulting in:$$muleft(bigcup_{i=1}^n A_iright)=sum_{i=1}^nmu(A_i)-sum_{1leq i<jleq n}mu(A_icap A_j)+cdots+(-1)^nmu(A_1capcdotscap A_n)$$
answered Jan 6 at 14:23
drhabdrhab
101k545136
edited Jan 6 at 14:38
answered Jan 6 at 14:23
drhabdrhab
101k545136
answered Jan 6 at 14:23
drhabdrhab
101k545136
answered Jan 6 at 14:23
drhabdrhab
101k545136
101k545136
add a comment |
add a comment |
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