Is the empty set a real number?












1












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
    $endgroup$
    – fleablood
    3 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
    3 hours ago
















1












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Subsets are not elements. ${dog, fish} subset {dog, cat, bird, fish}$ but ${dog, fish} not in {dog, cat, bird, fish}$.
    $endgroup$
    – fleablood
    3 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
    3 hours ago














1












1








1





$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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
    3 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
    3 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
    3 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
    3 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
3 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
3 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
3 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
3 hours ago










4 Answers
4






active

oldest

votes


















4












$begingroup$

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 $0cdot 5=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    upvoted for banana
    $endgroup$
    – dbx
    50 mins ago



















2












$begingroup$


"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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    34 mins ago





















0












$begingroup$

Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,



$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$






share|cite|improve this answer









$endgroup$









  • 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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077670%2fis-the-empty-set-a-real-number%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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 $0cdot 5=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    upvoted for banana
    $endgroup$
    – dbx
    50 mins ago
















4












$begingroup$

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 $0cdot 5=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    upvoted for banana
    $endgroup$
    – dbx
    50 mins ago














4












4








4





$begingroup$

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 $0cdot 5=0$.






share|cite|improve this answer









$endgroup$



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 $0cdot 5=0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









vadim123vadim123

75.8k897189




75.8k897189












  • $begingroup$
    upvoted for banana
    $endgroup$
    – dbx
    50 mins ago


















  • $begingroup$
    upvoted for banana
    $endgroup$
    – dbx
    50 mins ago
















$begingroup$
upvoted for banana
$endgroup$
– dbx
50 mins ago




$begingroup$
upvoted for banana
$endgroup$
– dbx
50 mins ago











2












$begingroup$


"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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












$begingroup$


"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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$


"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.






share|cite|improve this answer











$endgroup$




"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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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


















  • $begingroup$
    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











1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    34 mins ago


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    34 mins ago
















1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
    34 mins ago




















  • $begingroup$
    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
    34 mins ago


















$begingroup$
I'm not sure if this counts as an answer.
$endgroup$
– Zduff
1 hour ago




$begingroup$
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
34 mins ago






$begingroup$
Your first instance of $emptyset in A$ should've been $emptyset subseteq A$. Same for the last.
$endgroup$
– Randall
34 mins ago













0












$begingroup$

Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,



$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$






share|cite|improve this answer









$endgroup$









  • 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
















0












$begingroup$

Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,



$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$






share|cite|improve this answer









$endgroup$









  • 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














0












0








0





$begingroup$

Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,



$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$






share|cite|improve this answer









$endgroup$



Ordinary arithmetic is defined on numbers, be them integers, rationals or reals. Obviously,



$$emptysetnotinmathbb Z, mathbb Q, mathbb R.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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














  • 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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077670%2fis-the-empty-set-a-real-number%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Knooppunt Holsloot

Altaar (religie)

Gregoriusmis