Mixing
Since $emptyset subset A$ where $A$ is any set, does that mean $emptyset in A$?
$begingroup$
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
edited Jan 18 at 8:12
user21820
39.2k543154
asked Jan 17 at 23:37
ZduffZduff
1,671920
$endgroup$
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
edited Jan 18 at 8:12
user21820
39.2k543154
asked Jan 17 at 23:37
ZduffZduff
1,671920
$endgroup$
2
$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
Jan 18 at 0:09
$begingroup$
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
Jan 18 at 0:11
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
edited Jan 18 at 8:12
user21820
39.2k543154
asked Jan 17 at 23:37
ZduffZduff
1,671920
$endgroup$
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 Jan 18 at 8:12
user21820
39.2k543154
asked Jan 17 at 23:37
ZduffZduff
1,671920
edited Jan 18 at 8:12
user21820
39.2k543154
asked Jan 17 at 23:37
ZduffZduff
1,671920
edited Jan 18 at 8:12
user21820
39.2k543154
edited Jan 18 at 8:12
user21820
39.2k543154
edited Jan 18 at 8:12
user21820
39.2k543154
39.2k543154
asked Jan 17 at 23:37
ZduffZduff
1,671920
asked Jan 17 at 23:37
ZduffZduff
1,671920
asked Jan 17 at 23:37
ZduffZduff
1,671920
1,671920
2
$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
Jan 18 at 0:09
$begingroup$
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
Jan 18 at 0:11
add a comment |
2
$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
Jan 18 at 0:09
$begingroup$
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
Jan 18 at 0:11
2
2
$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
Jan 18 at 0:09
$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
Jan 18 at 0:09
$begingroup$
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
Jan 18 at 0:11
$begingroup$
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
Jan 18 at 0:11
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 $0cdot 5=0$.
answered Jan 17 at 23:54
vadim123vadim123
76.2k897191
$endgroup$
1
$begingroup$
upvoted for banana
$endgroup$
– dbx
Jan 18 at 2:33
$begingroup$
$emptyset$ is the natural number $0$ (which can certainly be multiplied by $5$), but to address the question title, it isn't any real number because no Dedekind cut can be an empty set by definition.
$endgroup$
– bjb568
Jan 18 at 6:35
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 $emptyset subset A$ 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 Jan 18 at 0:27
fleabloodfleablood
71.5k22686
$endgroup$
$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
Jan 18 at 2:02
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
Jan 18 at 2:49
$begingroup$
Not the last. I was talking about the concept of $emptysetin A $ for all sets. I don't see how if it were true, it'd have to do with multiplication.
$endgroup$
– fleablood
Jan 18 at 6:13
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.
answered Jan 17 at 23:38
Zubin MukerjeeZubin Mukerjee
15.2k32658
$endgroup$
$begingroup$
Does a relation require members in order to be reflexive?
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– Zduff
Jan 17 at 23:40
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What is required for thing to be a member of a set?
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– Zduff
Jan 17 at 23:44
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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
Jan 17 at 23:46
1
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$R$ 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
Jan 18 at 0:26
add a comment |
$begingroup$
Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,
$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$
answered Jan 17 at 23:46
Yves DaoustYves Daoust
129k675227
$endgroup$
1
$begingroup$
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
Jan 17 at 23:48
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@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
Jan 17 at 23:49
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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.
$endgroup$
– Rob Arthan
Jan 17 at 23:54
$begingroup$
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
Jan 18 at 0:42
1
$begingroup$
@Dair: I said ordinary arithmetic, didn't I ? My answer is for the OP, not for PhD's.
$endgroup$
– Yves Daoust
Jan 18 at 10:35
add a comment |
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4 Answers
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active
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4 Answers
4
active
oldest
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$begingroup$
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 $0cdot 5=0$.
answered Jan 17 at 23:54
vadim123vadim123
76.2k897191
$endgroup$
1
$begingroup$
upvoted for banana
$endgroup$
– dbx
Jan 18 at 2:33
$begingroup$
$emptyset$ is the natural number $0$ (which can certainly be multiplied by $5$), but to address the question title, it isn't any real number because no Dedekind cut can be an empty set by definition.
$endgroup$
– bjb568
Jan 18 at 6:35
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 $0cdot 5=0$.
answered Jan 17 at 23:54
vadim123vadim123
76.2k897191
$endgroup$
1
$begingroup$
upvoted for banana
$endgroup$
– dbx
Jan 18 at 2:33
$begingroup$
$emptyset$ is the natural number $0$ (which can certainly be multiplied by $5$), but to address the question title, it isn't any real number because no Dedekind cut can be an empty set by definition.
$endgroup$
– bjb568
Jan 18 at 6:35
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 $0cdot 5=0$.
answered Jan 17 at 23:54
vadim123vadim123
76.2k897191
$endgroup$
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 Jan 17 at 23:54
vadim123vadim123
76.2k897191
answered Jan 17 at 23:54
vadim123vadim123
76.2k897191
answered Jan 17 at 23:54
vadim123vadim123
76.2k897191
answered Jan 17 at 23:54
vadim123vadim123
76.2k897191
76.2k897191
1
$begingroup$
upvoted for banana
$endgroup$
– dbx
Jan 18 at 2:33
$begingroup$
$emptyset$ is the natural number $0$ (which can certainly be multiplied by $5$), but to address the question title, it isn't any real number because no Dedekind cut can be an empty set by definition.
$endgroup$
– bjb568
Jan 18 at 6:35
add a comment |
1
$begingroup$
upvoted for banana
$endgroup$
– dbx
Jan 18 at 2:33
$begingroup$
$emptyset$ is the natural number $0$ (which can certainly be multiplied by $5$), but to address the question title, it isn't any real number because no Dedekind cut can be an empty set by definition.
$endgroup$
– bjb568
Jan 18 at 6:35
1
1
$begingroup$
upvoted for banana
$endgroup$
– dbx
Jan 18 at 2:33
$begingroup$
upvoted for banana
$endgroup$
– dbx
Jan 18 at 2:33
$begingroup$
$emptyset$ is the natural number $0$ (which can certainly be multiplied by $5$), but to address the question title, it isn't any real number because no Dedekind cut can be an empty set by definition.
$endgroup$
– bjb568
Jan 18 at 6:35
$begingroup$
$emptyset$ is the natural number $0$ (which can certainly be multiplied by $5$), but to address the question title, it isn't any real number because no Dedekind cut can be an empty set by definition.
$endgroup$
– bjb568
Jan 18 at 6:35
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 $emptyset subset A$ 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 Jan 18 at 0:27
fleabloodfleablood
71.5k22686
$endgroup$
$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
Jan 18 at 2:02
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
Jan 18 at 2:49
$begingroup$
Not the last. I was talking about the concept of $emptysetin A $ for all sets. I don't see how if it were true, it'd have to do with multiplication.
$endgroup$
– fleablood
Jan 18 at 6:13
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 $emptyset subset A$ 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 Jan 18 at 0:27
fleabloodfleablood
71.5k22686
$endgroup$
$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
Jan 18 at 2:02
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
Jan 18 at 2:49
$begingroup$
Not the last. I was talking about the concept of $emptysetin A $ for all sets. I don't see how if it were true, it'd have to do with multiplication.
$endgroup$
– fleablood
Jan 18 at 6:13
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 $emptyset subset A$ 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 Jan 18 at 0:27
fleabloodfleablood
71.5k22686
$endgroup$
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 $emptyset subset A$ 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 Jan 18 at 0:27
fleabloodfleablood
71.5k22686
edited Jan 18 at 6:11
answered Jan 18 at 0:27
fleabloodfleablood
71.5k22686
answered Jan 18 at 0:27
fleabloodfleablood
71.5k22686
answered Jan 18 at 0:27
fleabloodfleablood
71.5k22686
71.5k22686
$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
Jan 18 at 2:02
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
Jan 18 at 2:49
$begingroup$
Not the last. I was talking about the concept of $emptysetin A $ for all sets. I don't see how if it were true, it'd have to do with multiplication.
$endgroup$
– fleablood
Jan 18 at 6:13
add a comment |
$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
Jan 18 at 2:02
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
Jan 18 at 2:49
$begingroup$
Not the last. I was talking about the concept of $emptysetin A $ for all sets. I don't see how if it were true, it'd have to do with multiplication.
$endgroup$
– fleablood
Jan 18 at 6:13
$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
Jan 18 at 2:02
$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
Jan 18 at 2:02
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
Jan 18 at 2:49
$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
Jan 18 at 2:49
$begingroup$
Not the last. I was talking about the concept of $emptysetin A $ for all sets. I don't see how if it were true, it'd have to do with multiplication.
$endgroup$
– fleablood
Jan 18 at 6:13
$begingroup$
Not the last. I was talking about the concept of $emptysetin A $ for all sets. I don't see how if it were true, it'd have to do with multiplication.
$endgroup$
– fleablood
Jan 18 at 6:13
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.
answered Jan 17 at 23:38
Zubin MukerjeeZubin Mukerjee
15.2k32658
$endgroup$
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
Jan 17 at 23:40
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
Jan 17 at 23:44
$begingroup$
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
Jan 17 at 23:46
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
Jan 18 at 0:26
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.
answered Jan 17 at 23:38
Zubin MukerjeeZubin Mukerjee
15.2k32658
$endgroup$
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
Jan 17 at 23:40
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
Jan 17 at 23:44
$begingroup$
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
Jan 17 at 23:46
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
Jan 18 at 0:26
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.
answered Jan 17 at 23:38
Zubin MukerjeeZubin Mukerjee
15.2k32658
$endgroup$
"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.
answered Jan 17 at 23:38
Zubin MukerjeeZubin Mukerjee
15.2k32658
edited Jan 17 at 23:51
answered Jan 17 at 23:38
Zubin MukerjeeZubin Mukerjee
15.2k32658
answered Jan 17 at 23:38
Zubin MukerjeeZubin Mukerjee
15.2k32658
answered Jan 17 at 23:38
Zubin MukerjeeZubin Mukerjee
15.2k32658
15.2k32658
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
Jan 17 at 23:40
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
Jan 17 at 23:44
$begingroup$
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
Jan 17 at 23:46
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
Jan 18 at 0:26
add a comment |
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
Jan 17 at 23:40
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
Jan 17 at 23:44
$begingroup$
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
Jan 17 at 23:46
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
Jan 18 at 0:26
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
Jan 17 at 23:40
$begingroup$
Does a relation require members in order to be reflexive?
$endgroup$
– Zduff
Jan 17 at 23:40
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
Jan 17 at 23:44
$begingroup$
What is required for thing to be a member of a set?
$endgroup$
– Zduff
Jan 17 at 23:44
$begingroup$
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
Jan 17 at 23:46
$begingroup$
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
Jan 17 at 23:46
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
Jan 18 at 0:26
$begingroup$
$R$ 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
Jan 18 at 0:26
add a comment |
$begingroup$
Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,
$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$
answered Jan 17 at 23:46
Yves DaoustYves Daoust
129k675227
$endgroup$
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
Jan 17 at 23:48
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
Jan 17 at 23:49
$begingroup$
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
Jan 17 at 23:54
$begingroup$
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
Jan 18 at 0:42
1
$begingroup$
@Dair: I said ordinary arithmetic, didn't I ? My answer is for the OP, not for PhD's.
$endgroup$
– Yves Daoust
Jan 18 at 10:35
add a comment |
$begingroup$
Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,
$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$
answered Jan 17 at 23:46
Yves DaoustYves Daoust
129k675227
$endgroup$
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
Jan 17 at 23:48
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
Jan 17 at 23:49
$begingroup$
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
Jan 17 at 23:54
$begingroup$
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
Jan 18 at 0:42
1
$begingroup$
@Dair: I said ordinary arithmetic, didn't I ? My answer is for the OP, not for PhD's.
$endgroup$
– Yves Daoust
Jan 18 at 10:35
add a comment |
$begingroup$
Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,
$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$
answered Jan 17 at 23:46
Yves DaoustYves Daoust
129k675227
$endgroup$
Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,
$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$
answered Jan 17 at 23:46
Yves DaoustYves Daoust
129k675227
answered Jan 17 at 23:46
Yves DaoustYves Daoust
129k675227
answered Jan 17 at 23:46
Yves DaoustYves Daoust
129k675227
answered Jan 17 at 23:46
Yves DaoustYves Daoust
129k675227
129k675227
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
Jan 17 at 23:48
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
Jan 17 at 23:49
$begingroup$
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
Jan 17 at 23:54
$begingroup$
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
Jan 18 at 0:42
1
$begingroup$
@Dair: I said ordinary arithmetic, didn't I ? My answer is for the OP, not for PhD's.
$endgroup$
– Yves Daoust
Jan 18 at 10:35
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
Jan 17 at 23:48
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
Jan 17 at 23:49
$begingroup$
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
Jan 17 at 23:54
$begingroup$
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
Jan 18 at 0:42
1
$begingroup$
@Dair: I said ordinary arithmetic, didn't I ? My answer is for the OP, not for PhD's.
$endgroup$
– Yves Daoust
Jan 18 at 10:35
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
Jan 17 at 23:48
$begingroup$
But it is standard in group theory, ring theory etc. to write $XY$ for ${xy | x in X, y in Y}$.
$endgroup$
– Rob Arthan
Jan 17 at 23:48
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
Jan 17 at 23:49
$begingroup$
@RobArthan: what is the connection with $emptysetcdot 5$ ?
$endgroup$
– Yves Daoust
Jan 17 at 23:49
$begingroup$
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
Jan 17 at 23:54
$begingroup$
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
Jan 17 at 23:54
$begingroup$
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
Jan 18 at 0:42
$begingroup$
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
Jan 18 at 0:42
1
1
$begingroup$
@Dair: I said ordinary arithmetic, didn't I ? My answer is for the OP, not for PhD's.
$endgroup$
– Yves Daoust
Jan 18 at 10:35
$begingroup$
@Dair: I said ordinary arithmetic, didn't I ? My answer is for the OP, not for PhD's.
$endgroup$
– Yves Daoust
Jan 18 at 10:35
add a comment |
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$begingroup$
Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
$endgroup$
– fleablood
Jan 18 at 0:09
$begingroup$
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
Jan 18 at 0:11