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What will happen when an empty set involved in a expression?
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Let $g(x)=h(x)+sum_kinmathcalKf(x,k)$ where $mathcalK$ is a set with finite number of elements. My question is when $mathcalK$ is an empty set, then what is $g(x)$? Is $g(x)$ will be undefined or $g(x)=h(x)$? I am confused in the scenario where an expression involving a set. Any help is appreciated.
calculus real-analysis abstract-algebra
asked Jul 15 at 11:07
Dave
19919
add a comment |Â
up vote
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Let $g(x)=h(x)+sum_kinmathcalKf(x,k)$ where $mathcalK$ is a set with finite number of elements. My question is when $mathcalK$ is an empty set, then what is $g(x)$? Is $g(x)$ will be undefined or $g(x)=h(x)$? I am confused in the scenario where an expression involving a set. Any help is appreciated.
calculus real-analysis abstract-algebra
asked Jul 15 at 11:07
Dave
19919
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $g(x)=h(x)+sum_kinmathcalKf(x,k)$ where $mathcalK$ is a set with finite number of elements. My question is when $mathcalK$ is an empty set, then what is $g(x)$? Is $g(x)$ will be undefined or $g(x)=h(x)$? I am confused in the scenario where an expression involving a set. Any help is appreciated.
calculus real-analysis abstract-algebra
asked Jul 15 at 11:07
Dave
19919
Let $g(x)=h(x)+sum_kinmathcalKf(x,k)$ where $mathcalK$ is a set with finite number of elements. My question is when $mathcalK$ is an empty set, then what is $g(x)$? Is $g(x)$ will be undefined or $g(x)=h(x)$? I am confused in the scenario where an expression involving a set. Any help is appreciated.
calculus real-analysis abstract-algebra
asked Jul 15 at 11:07
Dave
19919
asked Jul 15 at 11:07
Dave
19919
asked Jul 15 at 11:07
Dave
19919
asked Jul 15 at 11:07
Dave
19919
19919
add a comment |Â
add a comment |Â
1 Answer
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active
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up vote
4
down vote
accepted
Any sum over the empty set
$$sum_x in varnothing F(x)$$
is equal to zero by definition, so $g(x) = h(x)$ if $mathcalK = varnothing$.
Answer to the comment: $displaystyle prod_x in varnothing F(x)$ is defined as $1$, $max varnothing$ and $min varnothing$ don't exist. In specific situations we may abuse the notation and write $max varnothing = -infty$, $min varnothing = +infty$.
Generally suppose we have an associative, commutative binary operation $ast$ on elements of some set $S$. For a finite, non-empty $A = a_1, ldots, a_n subseteq S$ and arbitrary function $F : S to S$ we can define
$$DeclareMathOperator*opLARGE ast op limits_x in A F(x) = F(a_1) ast F(a_2) ast ldots ast F(a_n).$$
It's then reasonable to define $op limits_x in varnothing F(x)$ to be the neutral element of $ast$ (provided that one exists), because in that case the following rule, which holds for all non-empty finite sets $A subseteq S$, will continue to hold with $A$ empty:
Suppose $A subseteq S$ is finite and $b in S setminus A$. Then
$$op limits_x in A cup b F(x) = left[ op limits_x in A F(x) right] ast F(b).$$
That justifies the definition of the product over the empty set to be $1$. As for $min$ and $max$ as operations on $mathbbR$, they don't have neutral elements, so they are not defined on the empty set. However, we can extend the order $(mathbbR, leqslant)$ to $overlinemathbbR = mathbbR cup -infty, +infty $ so that $-infty < a < +infty$ for every $a in mathbbR$. Then the extended order $overlinemathbbR$ has the greatest element $+infty$, which becomes the neutral element of $min$, so it makes sense to define $min varnothing = +infty$ and similarly, $max varnothing = -infty$.
answered Jul 15 at 11:18
Adayah
6,575822
What about other operations over an empty set? For example, $max$, $min$, $prod$.
– Dave
Jul 15 at 11:44
1
@Dave If $s=max S$ where $S$ is a set then $sin S$. This is not possible for $S=varnothing$ so $maxvarnothing$ is not defined. Same story for $min$. Further $sum_xinvarnothingx$ and $prod_xinvarnothingx$ both correspond with neutral elements wrt certain operations. The first neutral wrt what commonly is called addition (hence is often denoted as $0$), the second neutral wrt what is commonly called multiplication (hence is often denoted as $1$).
– drhab
Jul 15 at 12:29
@Dave I've edited my post to answer your question.
– Adayah
Jul 15 at 12:54
If $max S$ is defined then always $max Sin S$. So I disagree with $minvarnothing=+infty$ and $maxvarnothing=-infty$. We have situations where $infvarnothing=+infty$ and $supvarnothing=-infty$.
– drhab
Jul 15 at 12:55
@drhab I wrote that $max$ and $min$ of $varnothing$ formally don't exist and in certain situations we abuse the notation and write $min varnothing = infty$ and $max varnothing = -infty$. After all, it's only a convention, and if you would like me to, I can think of some realistic situation where it would be useful. Also, by definition $inf varnothing = max a in mathbbR : (forall b in varnothing) , a leqslant b = max mathbbR$ so $max mathbbR notin mathbbR$ if we assume (which you agreed with in some situations :p) that $inf varnothing = +infty$.
– Adayah
Jul 15 at 13:15
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Any sum over the empty set
$$sum_x in varnothing F(x)$$
is equal to zero by definition, so $g(x) = h(x)$ if $mathcalK = varnothing$.
Answer to the comment: $displaystyle prod_x in varnothing F(x)$ is defined as $1$, $max varnothing$ and $min varnothing$ don't exist. In specific situations we may abuse the notation and write $max varnothing = -infty$, $min varnothing = +infty$.
Generally suppose we have an associative, commutative binary operation $ast$ on elements of some set $S$. For a finite, non-empty $A = a_1, ldots, a_n subseteq S$ and arbitrary function $F : S to S$ we can define
$$DeclareMathOperator*opLARGE ast op limits_x in A F(x) = F(a_1) ast F(a_2) ast ldots ast F(a_n).$$
It's then reasonable to define $op limits_x in varnothing F(x)$ to be the neutral element of $ast$ (provided that one exists), because in that case the following rule, which holds for all non-empty finite sets $A subseteq S$, will continue to hold with $A$ empty:
Suppose $A subseteq S$ is finite and $b in S setminus A$. Then
$$op limits_x in A cup b F(x) = left[ op limits_x in A F(x) right] ast F(b).$$
That justifies the definition of the product over the empty set to be $1$. As for $min$ and $max$ as operations on $mathbbR$, they don't have neutral elements, so they are not defined on the empty set. However, we can extend the order $(mathbbR, leqslant)$ to $overlinemathbbR = mathbbR cup -infty, +infty $ so that $-infty < a < +infty$ for every $a in mathbbR$. Then the extended order $overlinemathbbR$ has the greatest element $+infty$, which becomes the neutral element of $min$, so it makes sense to define $min varnothing = +infty$ and similarly, $max varnothing = -infty$.
answered Jul 15 at 11:18
Adayah
6,575822
What about other operations over an empty set? For example, $max$, $min$, $prod$.
– Dave
Jul 15 at 11:44
1
@Dave If $s=max S$ where $S$ is a set then $sin S$. This is not possible for $S=varnothing$ so $maxvarnothing$ is not defined. Same story for $min$. Further $sum_xinvarnothingx$ and $prod_xinvarnothingx$ both correspond with neutral elements wrt certain operations. The first neutral wrt what commonly is called addition (hence is often denoted as $0$), the second neutral wrt what is commonly called multiplication (hence is often denoted as $1$).
– drhab
Jul 15 at 12:29
@Dave I've edited my post to answer your question.
– Adayah
Jul 15 at 12:54
If $max S$ is defined then always $max Sin S$. So I disagree with $minvarnothing=+infty$ and $maxvarnothing=-infty$. We have situations where $infvarnothing=+infty$ and $supvarnothing=-infty$.
– drhab
Jul 15 at 12:55
@drhab I wrote that $max$ and $min$ of $varnothing$ formally don't exist and in certain situations we abuse the notation and write $min varnothing = infty$ and $max varnothing = -infty$. After all, it's only a convention, and if you would like me to, I can think of some realistic situation where it would be useful. Also, by definition $inf varnothing = max a in mathbbR : (forall b in varnothing) , a leqslant b = max mathbbR$ so $max mathbbR notin mathbbR$ if we assume (which you agreed with in some situations :p) that $inf varnothing = +infty$.
– Adayah
Jul 15 at 13:15
 |Â
show 1 more comment
up vote
4
down vote
accepted
Any sum over the empty set
$$sum_x in varnothing F(x)$$
is equal to zero by definition, so $g(x) = h(x)$ if $mathcalK = varnothing$.
Answer to the comment: $displaystyle prod_x in varnothing F(x)$ is defined as $1$, $max varnothing$ and $min varnothing$ don't exist. In specific situations we may abuse the notation and write $max varnothing = -infty$, $min varnothing = +infty$.
Generally suppose we have an associative, commutative binary operation $ast$ on elements of some set $S$. For a finite, non-empty $A = a_1, ldots, a_n subseteq S$ and arbitrary function $F : S to S$ we can define
$$DeclareMathOperator*opLARGE ast op limits_x in A F(x) = F(a_1) ast F(a_2) ast ldots ast F(a_n).$$
It's then reasonable to define $op limits_x in varnothing F(x)$ to be the neutral element of $ast$ (provided that one exists), because in that case the following rule, which holds for all non-empty finite sets $A subseteq S$, will continue to hold with $A$ empty:
Suppose $A subseteq S$ is finite and $b in S setminus A$. Then
$$op limits_x in A cup b F(x) = left[ op limits_x in A F(x) right] ast F(b).$$
That justifies the definition of the product over the empty set to be $1$. As for $min$ and $max$ as operations on $mathbbR$, they don't have neutral elements, so they are not defined on the empty set. However, we can extend the order $(mathbbR, leqslant)$ to $overlinemathbbR = mathbbR cup -infty, +infty $ so that $-infty < a < +infty$ for every $a in mathbbR$. Then the extended order $overlinemathbbR$ has the greatest element $+infty$, which becomes the neutral element of $min$, so it makes sense to define $min varnothing = +infty$ and similarly, $max varnothing = -infty$.
answered Jul 15 at 11:18
Adayah
6,575822
What about other operations over an empty set? For example, $max$, $min$, $prod$.
– Dave
Jul 15 at 11:44
1
@Dave If $s=max S$ where $S$ is a set then $sin S$. This is not possible for $S=varnothing$ so $maxvarnothing$ is not defined. Same story for $min$. Further $sum_xinvarnothingx$ and $prod_xinvarnothingx$ both correspond with neutral elements wrt certain operations. The first neutral wrt what commonly is called addition (hence is often denoted as $0$), the second neutral wrt what is commonly called multiplication (hence is often denoted as $1$).
– drhab
Jul 15 at 12:29
@Dave I've edited my post to answer your question.
– Adayah
Jul 15 at 12:54
If $max S$ is defined then always $max Sin S$. So I disagree with $minvarnothing=+infty$ and $maxvarnothing=-infty$. We have situations where $infvarnothing=+infty$ and $supvarnothing=-infty$.
– drhab
Jul 15 at 12:55
@drhab I wrote that $max$ and $min$ of $varnothing$ formally don't exist and in certain situations we abuse the notation and write $min varnothing = infty$ and $max varnothing = -infty$. After all, it's only a convention, and if you would like me to, I can think of some realistic situation where it would be useful. Also, by definition $inf varnothing = max a in mathbbR : (forall b in varnothing) , a leqslant b = max mathbbR$ so $max mathbbR notin mathbbR$ if we assume (which you agreed with in some situations :p) that $inf varnothing = +infty$.
– Adayah
Jul 15 at 13:15
 |Â
show 1 more comment
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Any sum over the empty set
$$sum_x in varnothing F(x)$$
is equal to zero by definition, so $g(x) = h(x)$ if $mathcalK = varnothing$.
Answer to the comment: $displaystyle prod_x in varnothing F(x)$ is defined as $1$, $max varnothing$ and $min varnothing$ don't exist. In specific situations we may abuse the notation and write $max varnothing = -infty$, $min varnothing = +infty$.
Generally suppose we have an associative, commutative binary operation $ast$ on elements of some set $S$. For a finite, non-empty $A = a_1, ldots, a_n subseteq S$ and arbitrary function $F : S to S$ we can define
$$DeclareMathOperator*opLARGE ast op limits_x in A F(x) = F(a_1) ast F(a_2) ast ldots ast F(a_n).$$
It's then reasonable to define $op limits_x in varnothing F(x)$ to be the neutral element of $ast$ (provided that one exists), because in that case the following rule, which holds for all non-empty finite sets $A subseteq S$, will continue to hold with $A$ empty:
Suppose $A subseteq S$ is finite and $b in S setminus A$. Then
$$op limits_x in A cup b F(x) = left[ op limits_x in A F(x) right] ast F(b).$$
That justifies the definition of the product over the empty set to be $1$. As for $min$ and $max$ as operations on $mathbbR$, they don't have neutral elements, so they are not defined on the empty set. However, we can extend the order $(mathbbR, leqslant)$ to $overlinemathbbR = mathbbR cup -infty, +infty $ so that $-infty < a < +infty$ for every $a in mathbbR$. Then the extended order $overlinemathbbR$ has the greatest element $+infty$, which becomes the neutral element of $min$, so it makes sense to define $min varnothing = +infty$ and similarly, $max varnothing = -infty$.
answered Jul 15 at 11:18
Adayah
6,575822
Any sum over the empty set
$$sum_x in varnothing F(x)$$
is equal to zero by definition, so $g(x) = h(x)$ if $mathcalK = varnothing$.
Answer to the comment: $displaystyle prod_x in varnothing F(x)$ is defined as $1$, $max varnothing$ and $min varnothing$ don't exist. In specific situations we may abuse the notation and write $max varnothing = -infty$, $min varnothing = +infty$.
Generally suppose we have an associative, commutative binary operation $ast$ on elements of some set $S$. For a finite, non-empty $A = a_1, ldots, a_n subseteq S$ and arbitrary function $F : S to S$ we can define
$$DeclareMathOperator*opLARGE ast op limits_x in A F(x) = F(a_1) ast F(a_2) ast ldots ast F(a_n).$$
It's then reasonable to define $op limits_x in varnothing F(x)$ to be the neutral element of $ast$ (provided that one exists), because in that case the following rule, which holds for all non-empty finite sets $A subseteq S$, will continue to hold with $A$ empty:
Suppose $A subseteq S$ is finite and $b in S setminus A$. Then
$$op limits_x in A cup b F(x) = left[ op limits_x in A F(x) right] ast F(b).$$
That justifies the definition of the product over the empty set to be $1$. As for $min$ and $max$ as operations on $mathbbR$, they don't have neutral elements, so they are not defined on the empty set. However, we can extend the order $(mathbbR, leqslant)$ to $overlinemathbbR = mathbbR cup -infty, +infty $ so that $-infty < a < +infty$ for every $a in mathbbR$. Then the extended order $overlinemathbbR$ has the greatest element $+infty$, which becomes the neutral element of $min$, so it makes sense to define $min varnothing = +infty$ and similarly, $max varnothing = -infty$.
answered Jul 15 at 11:18
Adayah
6,575822
edited Jul 26 at 11:30
answered Jul 15 at 11:18
Adayah
6,575822
answered Jul 15 at 11:18
Adayah
6,575822
answered Jul 15 at 11:18
Adayah
6,575822
6,575822
What about other operations over an empty set? For example, $max$, $min$, $prod$.
– Dave
Jul 15 at 11:44
1
@Dave If $s=max S$ where $S$ is a set then $sin S$. This is not possible for $S=varnothing$ so $maxvarnothing$ is not defined. Same story for $min$. Further $sum_xinvarnothingx$ and $prod_xinvarnothingx$ both correspond with neutral elements wrt certain operations. The first neutral wrt what commonly is called addition (hence is often denoted as $0$), the second neutral wrt what is commonly called multiplication (hence is often denoted as $1$).
– drhab
Jul 15 at 12:29
@Dave I've edited my post to answer your question.
– Adayah
Jul 15 at 12:54
If $max S$ is defined then always $max Sin S$. So I disagree with $minvarnothing=+infty$ and $maxvarnothing=-infty$. We have situations where $infvarnothing=+infty$ and $supvarnothing=-infty$.
– drhab
Jul 15 at 12:55
@drhab I wrote that $max$ and $min$ of $varnothing$ formally don't exist and in certain situations we abuse the notation and write $min varnothing = infty$ and $max varnothing = -infty$. After all, it's only a convention, and if you would like me to, I can think of some realistic situation where it would be useful. Also, by definition $inf varnothing = max a in mathbbR : (forall b in varnothing) , a leqslant b = max mathbbR$ so $max mathbbR notin mathbbR$ if we assume (which you agreed with in some situations :p) that $inf varnothing = +infty$.
– Adayah
Jul 15 at 13:15
 |Â
show 1 more comment
What about other operations over an empty set? For example, $max$, $min$, $prod$.
– Dave
Jul 15 at 11:44
1
@Dave If $s=max S$ where $S$ is a set then $sin S$. This is not possible for $S=varnothing$ so $maxvarnothing$ is not defined. Same story for $min$. Further $sum_xinvarnothingx$ and $prod_xinvarnothingx$ both correspond with neutral elements wrt certain operations. The first neutral wrt what commonly is called addition (hence is often denoted as $0$), the second neutral wrt what is commonly called multiplication (hence is often denoted as $1$).
– drhab
Jul 15 at 12:29
@Dave I've edited my post to answer your question.
– Adayah
Jul 15 at 12:54
If $max S$ is defined then always $max Sin S$. So I disagree with $minvarnothing=+infty$ and $maxvarnothing=-infty$. We have situations where $infvarnothing=+infty$ and $supvarnothing=-infty$.
– drhab
Jul 15 at 12:55
@drhab I wrote that $max$ and $min$ of $varnothing$ formally don't exist and in certain situations we abuse the notation and write $min varnothing = infty$ and $max varnothing = -infty$. After all, it's only a convention, and if you would like me to, I can think of some realistic situation where it would be useful. Also, by definition $inf varnothing = max a in mathbbR : (forall b in varnothing) , a leqslant b = max mathbbR$ so $max mathbbR notin mathbbR$ if we assume (which you agreed with in some situations :p) that $inf varnothing = +infty$.
– Adayah
Jul 15 at 13:15
What about other operations over an empty set? For example, $max$, $min$, $prod$.
– Dave
Jul 15 at 11:44
What about other operations over an empty set? For example, $max$, $min$, $prod$.
– Dave
Jul 15 at 11:44
1
1
@Dave If $s=max S$ where $S$ is a set then $sin S$. This is not possible for $S=varnothing$ so $maxvarnothing$ is not defined. Same story for $min$. Further $sum_xinvarnothingx$ and $prod_xinvarnothingx$ both correspond with neutral elements wrt certain operations. The first neutral wrt what commonly is called addition (hence is often denoted as $0$), the second neutral wrt what is commonly called multiplication (hence is often denoted as $1$).
– drhab
Jul 15 at 12:29
@Dave If $s=max S$ where $S$ is a set then $sin S$. This is not possible for $S=varnothing$ so $maxvarnothing$ is not defined. Same story for $min$. Further $sum_xinvarnothingx$ and $prod_xinvarnothingx$ both correspond with neutral elements wrt certain operations. The first neutral wrt what commonly is called addition (hence is often denoted as $0$), the second neutral wrt what is commonly called multiplication (hence is often denoted as $1$).
– drhab
Jul 15 at 12:29
@Dave I've edited my post to answer your question.
– Adayah
Jul 15 at 12:54
@Dave I've edited my post to answer your question.
– Adayah
Jul 15 at 12:54
If $max S$ is defined then always $max Sin S$. So I disagree with $minvarnothing=+infty$ and $maxvarnothing=-infty$. We have situations where $infvarnothing=+infty$ and $supvarnothing=-infty$.
– drhab
Jul 15 at 12:55
If $max S$ is defined then always $max Sin S$. So I disagree with $minvarnothing=+infty$ and $maxvarnothing=-infty$. We have situations where $infvarnothing=+infty$ and $supvarnothing=-infty$.
– drhab
Jul 15 at 12:55
@drhab I wrote that $max$ and $min$ of $varnothing$ formally don't exist and in certain situations we abuse the notation and write $min varnothing = infty$ and $max varnothing = -infty$. After all, it's only a convention, and if you would like me to, I can think of some realistic situation where it would be useful. Also, by definition $inf varnothing = max a in mathbbR : (forall b in varnothing) , a leqslant b = max mathbbR$ so $max mathbbR notin mathbbR$ if we assume (which you agreed with in some situations :p) that $inf varnothing = +infty$.
– Adayah
Jul 15 at 13:15
@drhab I wrote that $max$ and $min$ of $varnothing$ formally don't exist and in certain situations we abuse the notation and write $min varnothing = infty$ and $max varnothing = -infty$. After all, it's only a convention, and if you would like me to, I can think of some realistic situation where it would be useful. Also, by definition $inf varnothing = max a in mathbbR : (forall b in varnothing) , a leqslant b = max mathbbR$ so $max mathbbR notin mathbbR$ if we assume (which you agreed with in some situations :p) that $inf varnothing = +infty$.
– Adayah
Jul 15 at 13:15
 |Â
show 1 more comment
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