Is the empty set a real number?
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Clearly $emptyset subset A$ where $A$ is any set. But does that mean $emptyset in A$? And if so, would it make sense to try to perform arithmetic operations with it. Like
$$emptyset cdot 5 tag{where $5 in A$}$$
This is inspired by a question that was along the lines of: if a relation is symmetric and transitive, is it reflexive? Where I've seen (and am relatively satisfied by) the answer of: no, consider the empty relation.
elementary-set-theory relations
$endgroup$
add a comment |
$begingroup$
Clearly $emptyset subset A$ where $A$ is any set. But does that mean $emptyset in A$? And if so, would it make sense to try to perform arithmetic operations with it. Like
$$emptyset cdot 5 tag{where $5 in A$}$$
This is inspired by a question that was along the lines of: if a relation is symmetric and transitive, is it reflexive? Where I've seen (and am relatively satisfied by) the answer of: no, consider the empty relation.
elementary-set-theory relations
$endgroup$
$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
2 hours ago
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Why would $x in A$ mean $xcdot 5$ make sense. Ar you think thinking that if $x in A$ for every possible set then $x in mathbb R$ so so $x cdot 5$ makes sense. But we would also have $x in MONKEYS$ but $monkeycdot 5$ does not make any sense for any monkey. Clearly $x in A$ for every set is impossible.
$endgroup$
– fleablood
2 hours ago
add a comment |
$begingroup$
Clearly $emptyset subset A$ where $A$ is any set. But does that mean $emptyset in A$? And if so, would it make sense to try to perform arithmetic operations with it. Like
$$emptyset cdot 5 tag{where $5 in A$}$$
This is inspired by a question that was along the lines of: if a relation is symmetric and transitive, is it reflexive? Where I've seen (and am relatively satisfied by) the answer of: no, consider the empty relation.
elementary-set-theory relations
$endgroup$
Clearly $emptyset subset A$ where $A$ is any set. But does that mean $emptyset in A$? And if so, would it make sense to try to perform arithmetic operations with it. Like
$$emptyset cdot 5 tag{where $5 in A$}$$
This is inspired by a question that was along the lines of: if a relation is symmetric and transitive, is it reflexive? Where I've seen (and am relatively satisfied by) the answer of: no, consider the empty relation.
elementary-set-theory relations
elementary-set-theory relations
edited 1 hour ago
Zduff
asked 3 hours ago
ZduffZduff
1,551820
1,551820
$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
2 hours ago
$begingroup$
Why would $x in A$ mean $xcdot 5$ make sense. Ar you think thinking that if $x in A$ for every possible set then $x in mathbb R$ so so $x cdot 5$ makes sense. But we would also have $x in MONKEYS$ but $monkeycdot 5$ does not make any sense for any monkey. Clearly $x in A$ for every set is impossible.
$endgroup$
– fleablood
2 hours ago
add a comment |
$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
2 hours ago
$begingroup$
Why would $x in A$ mean $xcdot 5$ make sense. Ar you think thinking that if $x in A$ for every possible set then $x in mathbb R$ so so $x cdot 5$ makes sense. But we would also have $x in MONKEYS$ but $monkeycdot 5$ does not make any sense for any monkey. Clearly $x in A$ for every set is impossible.
$endgroup$
– fleablood
2 hours ago
$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
2 hours ago
$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
2 hours ago
$begingroup$
Why would $x in A$ mean $xcdot 5$ make sense. Ar you think thinking that if $x in A$ for every possible set then $x in mathbb R$ so so $x cdot 5$ makes sense. But we would also have $x in MONKEYS$ but $monkeycdot 5$ does not make any sense for any monkey. Clearly $x in A$ for every set is impossible.
$endgroup$
– fleablood
2 hours ago
$begingroup$
Why would $x in A$ mean $xcdot 5$ make sense. Ar you think thinking that if $x in A$ for every possible set then $x in mathbb R$ so so $x cdot 5$ makes sense. But we would also have $x in MONKEYS$ but $monkeycdot 5$ does not make any sense for any monkey. Clearly $x in A$ for every set is impossible.
$endgroup$
– fleablood
2 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Normally $emptyset$ is not a number. Multiplication is defined between two numbers. Hence it is equally as meaningful to write $emptyset cdot 5$, as it is to write $banana cdot 5$. They mean nothing on their own, but we can always assign meaning to them.
However, there is an important exception. In a common construction of natural numbers, due to Zermelo and Fraenkel, everything a set: there are no separate "non-set" numbers. The number zero is defined as the empty set, the number one is defined as the set containing the empty set. And so on, as described in the link. In this construction, it is meaningful to write $emptysetcdot 5$, because this translates to just $0cdot 5=0$.
$endgroup$
$begingroup$
upvoted for banana
$endgroup$
– dbx
33 mins ago
add a comment |
$begingroup$
"Clearly $emptyset subset A$ where $A$ is any set. But does that mean $emptyset in A$?"
If $A$ is a subset of $B$, it does not imply that $A$ is an element of $B$.
The empty set is a subset of every set, but it is not an element of every set.
For example, the empty set is not an element of the empty set.
By contrast, the empty set is a subset of the empty set.
$endgroup$
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
3 hours ago
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
3 hours ago
$begingroup$
A relation $R$ is defined over a set $X$. If $R$ is empty and $X$ is nonempty then $R$ is not reflexive. Interestingly, if $R$ is empty and $X$ is empty, then $R$ is reflexive. So, a relation technically does not have to have members to be reflexive.
$endgroup$
– Zubin Mukerjee
3 hours ago
1
$begingroup$
$R$ is reflexive if all elements $x in X$ will have $(x,x)in R$. If $X$ is empty there are no elments in $X$ so there are no $(x,x)$. So they all (all zero of them) are in $R$.
$endgroup$
– fleablood
2 hours ago
add a comment |
$begingroup$
First off $A subset B$ does not mean $A in B$.
Example: $mathbb Q subset mathbb R$. But $mathbb Q not in mathbb R$. If it were so, exactly which number is $mathbb Q$ equal to? It makes no sense.
$subset$ compares two sets as to whether all the elements of a set or also elements of the other. $in$ refers to elements in a set an whether they are in a set.
It is vacuously true that $emptyset in A$ for every set ($emptyset$ has no elements; so every element it has, all zero of them, is in $A$) but it's pretty clear that it is not true that $emptysetin A$ for all sets $A$. After all $emptyset$ is not an elephant so $emptyset not in {Babar, Tantor, Haiti, pink honk-honk}$.
And if so, would it make sense to try to perform arithmetic operations with it. Like ∅⋅5
I have to admit absolute puzzlement as to how the concept of $emptyset in A$ for all sets $A$ could have anything to do with defining arithmetic on $emptyset$ so I'm not sure how to answer this.
Your explanation of empty relationships doesn't seem to make what you are asking clearer.
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$begingroup$
I'm not sure if this counts as an answer.
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– Zduff
1 hour ago
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Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
17 mins ago
add a comment |
$begingroup$
Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,
$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$
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1
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But it is standard in group theory, ring theory etc. to write $XY$ for ${xy | x in X, y in Y}$.
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– Rob Arthan
3 hours ago
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@RobArthan: what is the connection with $emptysetcdot 5$ ?
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– Yves Daoust
3 hours ago
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You said the operations are defined on numbers, but that definition is generally extended to sets of numbers. I was probably being a bit pedantic: if you'd written "The primitive arithmetic operations are" instead of "Ordinary arithmetic is", I wouldn't have raised a comment.
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– Rob Arthan
3 hours ago
$begingroup$
Does this not depend on the foundation you use and the construction of the natural/integers you use? To say "Obviously, $varnothing notin mathbb{Z}$?" begs the question what is in $mathbb{Z}$? Constructing the natural numbers using von Neumann ordinals defines $0 = varnothing$. See Vadim's answer.
$endgroup$
– Dair
2 hours ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Normally $emptyset$ is not a number. Multiplication is defined between two numbers. Hence it is equally as meaningful to write $emptyset cdot 5$, as it is to write $banana cdot 5$. They mean nothing on their own, but we can always assign meaning to them.
However, there is an important exception. In a common construction of natural numbers, due to Zermelo and Fraenkel, everything a set: there are no separate "non-set" numbers. The number zero is defined as the empty set, the number one is defined as the set containing the empty set. And so on, as described in the link. In this construction, it is meaningful to write $emptysetcdot 5$, because this translates to just $0cdot 5=0$.
$endgroup$
$begingroup$
upvoted for banana
$endgroup$
– dbx
33 mins ago
add a comment |
$begingroup$
Normally $emptyset$ is not a number. Multiplication is defined between two numbers. Hence it is equally as meaningful to write $emptyset cdot 5$, as it is to write $banana cdot 5$. They mean nothing on their own, but we can always assign meaning to them.
However, there is an important exception. In a common construction of natural numbers, due to Zermelo and Fraenkel, everything a set: there are no separate "non-set" numbers. The number zero is defined as the empty set, the number one is defined as the set containing the empty set. And so on, as described in the link. In this construction, it is meaningful to write $emptysetcdot 5$, because this translates to just $0cdot 5=0$.
$endgroup$
$begingroup$
upvoted for banana
$endgroup$
– dbx
33 mins ago
add a comment |
$begingroup$
Normally $emptyset$ is not a number. Multiplication is defined between two numbers. Hence it is equally as meaningful to write $emptyset cdot 5$, as it is to write $banana cdot 5$. They mean nothing on their own, but we can always assign meaning to them.
However, there is an important exception. In a common construction of natural numbers, due to Zermelo and Fraenkel, everything a set: there are no separate "non-set" numbers. The number zero is defined as the empty set, the number one is defined as the set containing the empty set. And so on, as described in the link. In this construction, it is meaningful to write $emptysetcdot 5$, because this translates to just $0cdot 5=0$.
$endgroup$
Normally $emptyset$ is not a number. Multiplication is defined between two numbers. Hence it is equally as meaningful to write $emptyset cdot 5$, as it is to write $banana cdot 5$. They mean nothing on their own, but we can always assign meaning to them.
However, there is an important exception. In a common construction of natural numbers, due to Zermelo and Fraenkel, everything a set: there are no separate "non-set" numbers. The number zero is defined as the empty set, the number one is defined as the set containing the empty set. And so on, as described in the link. In this construction, it is meaningful to write $emptysetcdot 5$, because this translates to just $0cdot 5=0$.
answered 3 hours ago
vadim123vadim123
75.8k897189
75.8k897189
$begingroup$
upvoted for banana
$endgroup$
– dbx
33 mins ago
add a comment |
$begingroup$
upvoted for banana
$endgroup$
– dbx
33 mins ago
$begingroup$
upvoted for banana
$endgroup$
– dbx
33 mins ago
$begingroup$
upvoted for banana
$endgroup$
– dbx
33 mins ago
add a comment |
$begingroup$
"Clearly $emptyset subset A$ where $A$ is any set. But does that mean $emptyset in A$?"
If $A$ is a subset of $B$, it does not imply that $A$ is an element of $B$.
The empty set is a subset of every set, but it is not an element of every set.
For example, the empty set is not an element of the empty set.
By contrast, the empty set is a subset of the empty set.
$endgroup$
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
3 hours ago
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
3 hours ago
$begingroup$
A relation $R$ is defined over a set $X$. If $R$ is empty and $X$ is nonempty then $R$ is not reflexive. Interestingly, if $R$ is empty and $X$ is empty, then $R$ is reflexive. So, a relation technically does not have to have members to be reflexive.
$endgroup$
– Zubin Mukerjee
3 hours ago
1
$begingroup$
$R$ is reflexive if all elements $x in X$ will have $(x,x)in R$. If $X$ is empty there are no elments in $X$ so there are no $(x,x)$. So they all (all zero of them) are in $R$.
$endgroup$
– fleablood
2 hours ago
add a comment |
$begingroup$
"Clearly $emptyset subset A$ where $A$ is any set. But does that mean $emptyset in A$?"
If $A$ is a subset of $B$, it does not imply that $A$ is an element of $B$.
The empty set is a subset of every set, but it is not an element of every set.
For example, the empty set is not an element of the empty set.
By contrast, the empty set is a subset of the empty set.
$endgroup$
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
3 hours ago
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
3 hours ago
$begingroup$
A relation $R$ is defined over a set $X$. If $R$ is empty and $X$ is nonempty then $R$ is not reflexive. Interestingly, if $R$ is empty and $X$ is empty, then $R$ is reflexive. So, a relation technically does not have to have members to be reflexive.
$endgroup$
– Zubin Mukerjee
3 hours ago
1
$begingroup$
$R$ is reflexive if all elements $x in X$ will have $(x,x)in R$. If $X$ is empty there are no elments in $X$ so there are no $(x,x)$. So they all (all zero of them) are in $R$.
$endgroup$
– fleablood
2 hours ago
add a comment |
$begingroup$
"Clearly $emptyset subset A$ where $A$ is any set. But does that mean $emptyset in A$?"
If $A$ is a subset of $B$, it does not imply that $A$ is an element of $B$.
The empty set is a subset of every set, but it is not an element of every set.
For example, the empty set is not an element of the empty set.
By contrast, the empty set is a subset of the empty set.
$endgroup$
"Clearly $emptyset subset A$ where $A$ is any set. But does that mean $emptyset in A$?"
If $A$ is a subset of $B$, it does not imply that $A$ is an element of $B$.
The empty set is a subset of every set, but it is not an element of every set.
For example, the empty set is not an element of the empty set.
By contrast, the empty set is a subset of the empty set.
edited 3 hours ago
answered 3 hours ago
Zubin MukerjeeZubin Mukerjee
14.8k32657
14.8k32657
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
3 hours ago
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
3 hours ago
$begingroup$
A relation $R$ is defined over a set $X$. If $R$ is empty and $X$ is nonempty then $R$ is not reflexive. Interestingly, if $R$ is empty and $X$ is empty, then $R$ is reflexive. So, a relation technically does not have to have members to be reflexive.
$endgroup$
– Zubin Mukerjee
3 hours ago
1
$begingroup$
$R$ is reflexive if all elements $x in X$ will have $(x,x)in R$. If $X$ is empty there are no elments in $X$ so there are no $(x,x)$. So they all (all zero of them) are in $R$.
$endgroup$
– fleablood
2 hours ago
add a comment |
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
3 hours ago
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
3 hours ago
$begingroup$
A relation $R$ is defined over a set $X$. If $R$ is empty and $X$ is nonempty then $R$ is not reflexive. Interestingly, if $R$ is empty and $X$ is empty, then $R$ is reflexive. So, a relation technically does not have to have members to be reflexive.
$endgroup$
– Zubin Mukerjee
3 hours ago
1
$begingroup$
$R$ is reflexive if all elements $x in X$ will have $(x,x)in R$. If $X$ is empty there are no elments in $X$ so there are no $(x,x)$. So they all (all zero of them) are in $R$.
$endgroup$
– fleablood
2 hours ago
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
3 hours ago
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
3 hours ago
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
3 hours ago
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
3 hours ago
$begingroup$
A relation $R$ is defined over a set $X$. If $R$ is empty and $X$ is nonempty then $R$ is not reflexive. Interestingly, if $R$ is empty and $X$ is empty, then $R$ is reflexive. So, a relation technically does not have to have members to be reflexive.
$endgroup$
– Zubin Mukerjee
3 hours ago
$begingroup$
A relation $R$ is defined over a set $X$. If $R$ is empty and $X$ is nonempty then $R$ is not reflexive. Interestingly, if $R$ is empty and $X$ is empty, then $R$ is reflexive. So, a relation technically does not have to have members to be reflexive.
$endgroup$
– Zubin Mukerjee
3 hours ago
1
1
$begingroup$
$R$ is reflexive if all elements $x in X$ will have $(x,x)in R$. If $X$ is empty there are no elments in $X$ so there are no $(x,x)$. So they all (all zero of them) are in $R$.
$endgroup$
– fleablood
2 hours ago
$begingroup$
$R$ is reflexive if all elements $x in X$ will have $(x,x)in R$. If $X$ is empty there are no elments in $X$ so there are no $(x,x)$. So they all (all zero of them) are in $R$.
$endgroup$
– fleablood
2 hours ago
add a comment |
$begingroup$
First off $A subset B$ does not mean $A in B$.
Example: $mathbb Q subset mathbb R$. But $mathbb Q not in mathbb R$. If it were so, exactly which number is $mathbb Q$ equal to? It makes no sense.
$subset$ compares two sets as to whether all the elements of a set or also elements of the other. $in$ refers to elements in a set an whether they are in a set.
It is vacuously true that $emptyset in A$ for every set ($emptyset$ has no elements; so every element it has, all zero of them, is in $A$) but it's pretty clear that it is not true that $emptysetin A$ for all sets $A$. After all $emptyset$ is not an elephant so $emptyset not in {Babar, Tantor, Haiti, pink honk-honk}$.
And if so, would it make sense to try to perform arithmetic operations with it. Like ∅⋅5
I have to admit absolute puzzlement as to how the concept of $emptyset in A$ for all sets $A$ could have anything to do with defining arithmetic on $emptyset$ so I'm not sure how to answer this.
Your explanation of empty relationships doesn't seem to make what you are asking clearer.
$endgroup$
$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
1 hour ago
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
17 mins ago
add a comment |
$begingroup$
First off $A subset B$ does not mean $A in B$.
Example: $mathbb Q subset mathbb R$. But $mathbb Q not in mathbb R$. If it were so, exactly which number is $mathbb Q$ equal to? It makes no sense.
$subset$ compares two sets as to whether all the elements of a set or also elements of the other. $in$ refers to elements in a set an whether they are in a set.
It is vacuously true that $emptyset in A$ for every set ($emptyset$ has no elements; so every element it has, all zero of them, is in $A$) but it's pretty clear that it is not true that $emptysetin A$ for all sets $A$. After all $emptyset$ is not an elephant so $emptyset not in {Babar, Tantor, Haiti, pink honk-honk}$.
And if so, would it make sense to try to perform arithmetic operations with it. Like ∅⋅5
I have to admit absolute puzzlement as to how the concept of $emptyset in A$ for all sets $A$ could have anything to do with defining arithmetic on $emptyset$ so I'm not sure how to answer this.
Your explanation of empty relationships doesn't seem to make what you are asking clearer.
$endgroup$
$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
1 hour ago
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
17 mins ago
add a comment |
$begingroup$
First off $A subset B$ does not mean $A in B$.
Example: $mathbb Q subset mathbb R$. But $mathbb Q not in mathbb R$. If it were so, exactly which number is $mathbb Q$ equal to? It makes no sense.
$subset$ compares two sets as to whether all the elements of a set or also elements of the other. $in$ refers to elements in a set an whether they are in a set.
It is vacuously true that $emptyset in A$ for every set ($emptyset$ has no elements; so every element it has, all zero of them, is in $A$) but it's pretty clear that it is not true that $emptysetin A$ for all sets $A$. After all $emptyset$ is not an elephant so $emptyset not in {Babar, Tantor, Haiti, pink honk-honk}$.
And if so, would it make sense to try to perform arithmetic operations with it. Like ∅⋅5
I have to admit absolute puzzlement as to how the concept of $emptyset in A$ for all sets $A$ could have anything to do with defining arithmetic on $emptyset$ so I'm not sure how to answer this.
Your explanation of empty relationships doesn't seem to make what you are asking clearer.
$endgroup$
First off $A subset B$ does not mean $A in B$.
Example: $mathbb Q subset mathbb R$. But $mathbb Q not in mathbb R$. If it were so, exactly which number is $mathbb Q$ equal to? It makes no sense.
$subset$ compares two sets as to whether all the elements of a set or also elements of the other. $in$ refers to elements in a set an whether they are in a set.
It is vacuously true that $emptyset in A$ for every set ($emptyset$ has no elements; so every element it has, all zero of them, is in $A$) but it's pretty clear that it is not true that $emptysetin A$ for all sets $A$. After all $emptyset$ is not an elephant so $emptyset not in {Babar, Tantor, Haiti, pink honk-honk}$.
And if so, would it make sense to try to perform arithmetic operations with it. Like ∅⋅5
I have to admit absolute puzzlement as to how the concept of $emptyset in A$ for all sets $A$ could have anything to do with defining arithmetic on $emptyset$ so I'm not sure how to answer this.
Your explanation of empty relationships doesn't seem to make what you are asking clearer.
answered 2 hours ago
fleabloodfleablood
68.9k22685
68.9k22685
$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
1 hour ago
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
17 mins ago
add a comment |
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I'm not sure if this counts as an answer.
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– Zduff
1 hour ago
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Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
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– Randall
17 mins ago
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I'm not sure if this counts as an answer.
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– Zduff
1 hour ago
$begingroup$
I'm not sure if this counts as an answer.
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– Zduff
1 hour ago
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
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– Randall
17 mins ago
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
17 mins ago
add a comment |
$begingroup$
Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,
$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$
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1
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But it is standard in group theory, ring theory etc. to write $XY$ for ${xy | x in X, y in Y}$.
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– Rob Arthan
3 hours ago
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@RobArthan: what is the connection with $emptysetcdot 5$ ?
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– Yves Daoust
3 hours ago
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You said the operations are defined on numbers, but that definition is generally extended to sets of numbers. I was probably being a bit pedantic: if you'd written "The primitive arithmetic operations are" instead of "Ordinary arithmetic is", I wouldn't have raised a comment.
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– Rob Arthan
3 hours ago
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Does this not depend on the foundation you use and the construction of the natural/integers you use? To say "Obviously, $varnothing notin mathbb{Z}$?" begs the question what is in $mathbb{Z}$? Constructing the natural numbers using von Neumann ordinals defines $0 = varnothing$. See Vadim's answer.
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– Dair
2 hours ago
add a comment |
$begingroup$
Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,
$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$
$endgroup$
1
$begingroup$
But it is standard in group theory, ring theory etc. to write $XY$ for ${xy | x in X, y in Y}$.
$endgroup$
– Rob Arthan
3 hours ago
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
3 hours ago
$begingroup$
You said the operations are defined on numbers, but that definition is generally extended to sets of numbers. I was probably being a bit pedantic: if you'd written "The primitive arithmetic operations are" instead of "Ordinary arithmetic is", I wouldn't have raised a comment.
$endgroup$
– Rob Arthan
3 hours ago
$begingroup$
Does this not depend on the foundation you use and the construction of the natural/integers you use? To say "Obviously, $varnothing notin mathbb{Z}$?" begs the question what is in $mathbb{Z}$? Constructing the natural numbers using von Neumann ordinals defines $0 = varnothing$. See Vadim's answer.
$endgroup$
– Dair
2 hours ago
add a comment |
$begingroup$
Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,
$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$
$endgroup$
Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,
$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$
answered 3 hours ago
Yves DaoustYves Daoust
125k671222
125k671222
1
$begingroup$
But it is standard in group theory, ring theory etc. to write $XY$ for ${xy | x in X, y in Y}$.
$endgroup$
– Rob Arthan
3 hours ago
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
3 hours ago
$begingroup$
You said the operations are defined on numbers, but that definition is generally extended to sets of numbers. I was probably being a bit pedantic: if you'd written "The primitive arithmetic operations are" instead of "Ordinary arithmetic is", I wouldn't have raised a comment.
$endgroup$
– Rob Arthan
3 hours ago
$begingroup$
Does this not depend on the foundation you use and the construction of the natural/integers you use? To say "Obviously, $varnothing notin mathbb{Z}$?" begs the question what is in $mathbb{Z}$? Constructing the natural numbers using von Neumann ordinals defines $0 = varnothing$. See Vadim's answer.
$endgroup$
– Dair
2 hours ago
add a comment |
1
$begingroup$
But it is standard in group theory, ring theory etc. to write $XY$ for ${xy | x in X, y in Y}$.
$endgroup$
– Rob Arthan
3 hours ago
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
3 hours ago
$begingroup$
You said the operations are defined on numbers, but that definition is generally extended to sets of numbers. I was probably being a bit pedantic: if you'd written "The primitive arithmetic operations are" instead of "Ordinary arithmetic is", I wouldn't have raised a comment.
$endgroup$
– Rob Arthan
3 hours ago
$begingroup$
Does this not depend on the foundation you use and the construction of the natural/integers you use? To say "Obviously, $varnothing notin mathbb{Z}$?" begs the question what is in $mathbb{Z}$? Constructing the natural numbers using von Neumann ordinals defines $0 = varnothing$. See Vadim's answer.
$endgroup$
– Dair
2 hours ago
1
1
$begingroup$
But it is standard in group theory, ring theory etc. to write $XY$ for ${xy | x in X, y in Y}$.
$endgroup$
– Rob Arthan
3 hours ago
$begingroup$
But it is standard in group theory, ring theory etc. to write $XY$ for ${xy | x in X, y in Y}$.
$endgroup$
– Rob Arthan
3 hours ago
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
3 hours ago
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
3 hours ago
$begingroup$
You said the operations are defined on numbers, but that definition is generally extended to sets of numbers. I was probably being a bit pedantic: if you'd written "The primitive arithmetic operations are" instead of "Ordinary arithmetic is", I wouldn't have raised a comment.
$endgroup$
– Rob Arthan
3 hours ago
$begingroup$
You said the operations are defined on numbers, but that definition is generally extended to sets of numbers. I was probably being a bit pedantic: if you'd written "The primitive arithmetic operations are" instead of "Ordinary arithmetic is", I wouldn't have raised a comment.
$endgroup$
– Rob Arthan
3 hours ago
$begingroup$
Does this not depend on the foundation you use and the construction of the natural/integers you use? To say "Obviously, $varnothing notin mathbb{Z}$?" begs the question what is in $mathbb{Z}$? Constructing the natural numbers using von Neumann ordinals defines $0 = varnothing$. See Vadim's answer.
$endgroup$
– Dair
2 hours ago
$begingroup$
Does this not depend on the foundation you use and the construction of the natural/integers you use? To say "Obviously, $varnothing notin mathbb{Z}$?" begs the question what is in $mathbb{Z}$? Constructing the natural numbers using von Neumann ordinals defines $0 = varnothing$. See Vadim's answer.
$endgroup$
– Dair
2 hours ago
add a comment |
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$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
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– fleablood
2 hours ago
$begingroup$
Why would $x in A$ mean $xcdot 5$ make sense. Ar you think thinking that if $x in A$ for every possible set then $x in mathbb R$ so so $x cdot 5$ makes sense. But we would also have $x in MONKEYS$ but $monkeycdot 5$ does not make any sense for any monkey. Clearly $x in A$ for every set is impossible.
$endgroup$
– fleablood
2 hours ago