In the category $mathbf{Set}$ is “the product of an empty set of sets a one-element set”?











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I was reading these notes on Category Theory and it said (paraphrased to add context):




Exercise 4: Explain why in $textbf{Set}$ (the Category of Sets), the product of an empty set of sets is a one-element set.




which I think is incorrect. The product of two empty sets (or any number) is empty because we are considering:



$$ emptyset times emptyset = { (a,b) : a in emptyset, b in emptyset } = emptyset$$



where $a in emptyset , b in emptyset$ are false, so the above is the $emptyset$ which is NOT a one element set (its a zero element set).



This should be trivial so I am assuming I am somewhere mis reading the natural language of the exercise. Someone help me catch where is it and what the answer should be? i.e. whats being asked and the answer?










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    Your example is similar to asking why in arithmetic $0^0$ is taken to be $1$ rather than $0$
    – Henry
    Nov 21 at 8:22








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    Related : math.stackexchange.com/questions/1991522/…
    – Arnaud D.
    Nov 21 at 13:20















up vote
5
down vote

favorite
2












I was reading these notes on Category Theory and it said (paraphrased to add context):




Exercise 4: Explain why in $textbf{Set}$ (the Category of Sets), the product of an empty set of sets is a one-element set.




which I think is incorrect. The product of two empty sets (or any number) is empty because we are considering:



$$ emptyset times emptyset = { (a,b) : a in emptyset, b in emptyset } = emptyset$$



where $a in emptyset , b in emptyset$ are false, so the above is the $emptyset$ which is NOT a one element set (its a zero element set).



This should be trivial so I am assuming I am somewhere mis reading the natural language of the exercise. Someone help me catch where is it and what the answer should be? i.e. whats being asked and the answer?










share|cite|improve this question




















  • 1




    Your example is similar to asking why in arithmetic $0^0$ is taken to be $1$ rather than $0$
    – Henry
    Nov 21 at 8:22








  • 2




    Related : math.stackexchange.com/questions/1991522/…
    – Arnaud D.
    Nov 21 at 13:20













up vote
5
down vote

favorite
2









up vote
5
down vote

favorite
2






2





I was reading these notes on Category Theory and it said (paraphrased to add context):




Exercise 4: Explain why in $textbf{Set}$ (the Category of Sets), the product of an empty set of sets is a one-element set.




which I think is incorrect. The product of two empty sets (or any number) is empty because we are considering:



$$ emptyset times emptyset = { (a,b) : a in emptyset, b in emptyset } = emptyset$$



where $a in emptyset , b in emptyset$ are false, so the above is the $emptyset$ which is NOT a one element set (its a zero element set).



This should be trivial so I am assuming I am somewhere mis reading the natural language of the exercise. Someone help me catch where is it and what the answer should be? i.e. whats being asked and the answer?










share|cite|improve this question















I was reading these notes on Category Theory and it said (paraphrased to add context):




Exercise 4: Explain why in $textbf{Set}$ (the Category of Sets), the product of an empty set of sets is a one-element set.




which I think is incorrect. The product of two empty sets (or any number) is empty because we are considering:



$$ emptyset times emptyset = { (a,b) : a in emptyset, b in emptyset } = emptyset$$



where $a in emptyset , b in emptyset$ are false, so the above is the $emptyset$ which is NOT a one element set (its a zero element set).



This should be trivial so I am assuming I am somewhere mis reading the natural language of the exercise. Someone help me catch where is it and what the answer should be? i.e. whats being asked and the answer?







category-theory direct-product






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edited Nov 21 at 14:11









Asaf Karagila

301k32422753




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asked Nov 21 at 2:12









Pinocchio

1,86321754




1,86321754








  • 1




    Your example is similar to asking why in arithmetic $0^0$ is taken to be $1$ rather than $0$
    – Henry
    Nov 21 at 8:22








  • 2




    Related : math.stackexchange.com/questions/1991522/…
    – Arnaud D.
    Nov 21 at 13:20














  • 1




    Your example is similar to asking why in arithmetic $0^0$ is taken to be $1$ rather than $0$
    – Henry
    Nov 21 at 8:22








  • 2




    Related : math.stackexchange.com/questions/1991522/…
    – Arnaud D.
    Nov 21 at 13:20








1




1




Your example is similar to asking why in arithmetic $0^0$ is taken to be $1$ rather than $0$
– Henry
Nov 21 at 8:22






Your example is similar to asking why in arithmetic $0^0$ is taken to be $1$ rather than $0$
– Henry
Nov 21 at 8:22






2




2




Related : math.stackexchange.com/questions/1991522/…
– Arnaud D.
Nov 21 at 13:20




Related : math.stackexchange.com/questions/1991522/…
– Arnaud D.
Nov 21 at 13:20










4 Answers
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What the exercise is saying is the following: let $mathcal{C}$ be an empty family of sets (awkward, but $mathcal{C}$ is just isomorphic to $emptyset$). Then, $prod,mathcal{C}=prodlimits_{Cinmathcal{C}},C$ has one element. Simply put, $$prod,emptyset=prod_{Cinemptyset},C$$ has exactly one element. This is the same situation that gives rise to $a^0=1$. The real question is where on heaven this element comes from, and what exactly it is.




You can interpret $prod,emptyset$ as the set of all functions from $emptyset$ to itself. The only function there is the empty function $emptyset$. Therefore, $prod,emptyset={emptyset}$.







share|cite|improve this answer

















  • 3




    So it's not like a product of empty sets, but rather an empty product of sets, right? And thus its having one element is analogous to how a numerical empty product equals 1, and not 0.
    – The_Sympathizer
    Nov 21 at 6:55










  • Yep, that's the idea.
    – Batominovski
    Nov 21 at 7:45


















up vote
5
down vote













A product $Pi_{iin I} A_i$ of objects in a category is an object $B$ and a collection of morphisms $pi_i: Bto A_i$ such that for any object $C$ and collection of morphisms $d_i:Cto A_i,$ there is a unique morphism $f:Cto B$ such that $d_i=pi_icirc f.$ Note such a $(B,pi_i)$ may not exist and is generally not unique, although if $(B,pi_i)$ and $(B',pi_i')$ both satisfy the condition, there is a canonical isomorphism between $B$ and $B'$. If $I$ is empty, then there are no $pi_i$ and the definition reduces to an object $B$ such that for any $C$ there is a unique $f:Cto B$ (i.e. $B$ is a terminal object).



Specifying to the category Set, the sets that have a unique map into them from any set are exactly the singletons (if the set had more than one element there would be multiple maps into it from any nonempty set; if it had zero elements, there would be none). So any singleton is a nullary product, and as promised, any two singletons have a canonical bijection between them. In a set theoretical sense, one may define the nullary cartesian product to be ${phi}$ for the sake of having a well-defined operation (just as one defines the binary cartesian product as the set of ordered pairs), but from a category theory perspective these are best thought of as nice representatives of a whole collection of product objects, which are all isomorphic to one another.






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    If the interpretation is $X^0$ then the answer is obvious. If this is not obvious look at the end of this answer.



    It's just the question of what the set $X^0$ is which is the set of functions mapping nothing (the empty set) to a set $X$ which obviously is only the function that does "nothing" (the single element in the set "1"). Formally $X^0 = { f mid f : emptyset to X } $ (the set of functions from the empty set to $X$) now what we want to figure out is what "this set really is". First we recall that a function is simply a relation defined by the graph of $f$ i.e. $f subseteq A times B$ such that $f := { (a,b) in A times B : forall a in A, exists b in B }$. In other words, a function just tells you how to pair the elements of the domain $A$ with the elements in the codomain $B$ via $(a,f(a))$ where $f(a) = b$ for some $b in B$ in the codomain. In this case we have $A = emptyset$ so all this actually become the following:



    $$X^0 = { f mid f : emptyset to X } = [emptyset to X] = { f subseteq A times B mid f: emptyset to X } $$
    which is equal to:



    $$ { f subseteq A times B mid f: emptyset to X } = { { (a,b) in A times B : forall a in emptyset , exists b in B } } = { emptyset } $$



    so,



    $$ X^0 = { emptyset }$$



    the key is realizing that $f = { (a,b) in A times B : forall a in emptyset , exists b in B } = emptyset$ because if we try to form a set of pairs where the first element of the pair comes from the empty set, then of course we won't form any pair becuase $a in emptyset$ is false.



    To make it really clear, recall the graph (i.e. relation set) of a function $f$ is the set of pairs $ f = { (a,b) in A times B : forall a in A, exists b in B } $ which is why we get as a final answer the empty set inside a set (since a function is really just a relation saying how we pair up the first element to the second element and $f$ induces this pairing/relation). In the end $X^0$ is just the set of all function from the empty set to $B$ but a function is a set of "pairs". This set of pairs ends up being empty because we can never select elements for its first location since we are trying to select things form the empty set.





    One comment on what $X^0$ should mean. I realized that perhaps its not entirely obvious what $X^0$ should mean. Recall:



    $$ X^1 = { f mid f : { 1 } to X }$$
    $$ X^2 = { f mid f : { 1,2 } to X }$$
    $$ X^3 = { f mid f: { 1,2,3 } to X }$$
    $$ vdots $$
    $$ X^n = { f mid f : { 1,2,3,dots,n} to X }$$
    $$ vdots $$
    sho what should $X^0$? Well, $X^n$ is really short hand for $X^{ {1,dots,n } }$ so when $n=0$ the set should start at 1 and reach zero. But no set in this context does that, so what we really mean is the empty set. So $X^0$ is short hand for $X^{emptyset}$ i.e.



    $$ X^{0} = X^{emptyset} = { f mid f emptyset to X } $$



    by definition.





    Alternatively with the sequence notation $A^n = { (a_1,dots,a_n) mid a_1 in A , dots, a_n in A }$ we have:



    $$ $A^1 = { (a_1) mid a_1 in A }$ $$
    $$A^2 = { (a_1,a_2) mid a_1 in A , a_2 in A }$$
    $$A^2 = { (a_1,a_2,a_3) mid a_1 in A , a_2 in A, a_3 in A }$$
    $$ vdots $$
    $$A^n = { (a_1,dots,a_n) mid a_1 in A , dots, a_n in A }$$
    $$ vdots $$
    we wonder how to interpret $A^0$? Basically says:




    The set of sequences of size zero from elements from A.




    which is just the empty sequence ${ () }$. Since no element are there (hand waving a little) it's isomorphic to the empty set $() cong emptyset$ (ok this last part would be nice to make more formal).






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    • This is a nice answer; could be better if it were less ambiguous in its notation, therefore slightly clearer. E.g., the second equality is rather unreasonable except to those with sufficient familiarity with set theory to extrapolate the missing pieces. Nonetheless, +1.
      – Musa Al-hassy
      Nov 22 at 10:36










    • @MusaAl-hassy ok I added more details, it should be more accessible to everyone now, right? Thanks for the feedback! :)
      – Pinocchio
      Nov 22 at 18:08








    • 1




      @MusaAl-hassy ok you are right, the notation ${ f : A to B }$ is really confusing. It looks like the set contains 1 element, some function from $A$ to $B$. I decided to change it to something less confusing like ${ f mid f: A to B }$ which says "the set is the set of function with the property that each function $f$ is a function from $A$ to $B$. This is a lot less confusing since it mimics the more standard notations that say "the set is defined by elements $x$ (usually they say $x in X$) such that each element has so property $P(x)$". (continued...)
      – Pinocchio
      Dec 4 at 21:01








    • 1




      To make it more clear I could have done ${ f subseteq A times B mid f: A to B }$ but it felt too close to the formal definition (since I assume not many people think of function as relations or subsets of $A times B$ I intentionally deferred that (so that instead at that point $f$ was thought more like functions from programming, pointers to how to act given a input). Later I explain it clearly so the second time they read that short hand they can infer $f$ is really a subset of $A times B$.
      – Pinocchio
      Dec 4 at 21:04






    • 1




      Way to go :-) Thanks for your efforts!
      – Musa Al-hassy
      Dec 5 at 20:47


















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    Since no-one has said this explicitly:



    Yes, you are misreading the language of the question. The question is asking for



    $$prodlimits_{x in emptyset} x$$



    It's the difference between "the product of an empty set of sets" and "the product of a set of empty sets".






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      4 Answers
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      up vote
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      accepted










      What the exercise is saying is the following: let $mathcal{C}$ be an empty family of sets (awkward, but $mathcal{C}$ is just isomorphic to $emptyset$). Then, $prod,mathcal{C}=prodlimits_{Cinmathcal{C}},C$ has one element. Simply put, $$prod,emptyset=prod_{Cinemptyset},C$$ has exactly one element. This is the same situation that gives rise to $a^0=1$. The real question is where on heaven this element comes from, and what exactly it is.




      You can interpret $prod,emptyset$ as the set of all functions from $emptyset$ to itself. The only function there is the empty function $emptyset$. Therefore, $prod,emptyset={emptyset}$.







      share|cite|improve this answer

















      • 3




        So it's not like a product of empty sets, but rather an empty product of sets, right? And thus its having one element is analogous to how a numerical empty product equals 1, and not 0.
        – The_Sympathizer
        Nov 21 at 6:55










      • Yep, that's the idea.
        – Batominovski
        Nov 21 at 7:45















      up vote
      7
      down vote



      accepted










      What the exercise is saying is the following: let $mathcal{C}$ be an empty family of sets (awkward, but $mathcal{C}$ is just isomorphic to $emptyset$). Then, $prod,mathcal{C}=prodlimits_{Cinmathcal{C}},C$ has one element. Simply put, $$prod,emptyset=prod_{Cinemptyset},C$$ has exactly one element. This is the same situation that gives rise to $a^0=1$. The real question is where on heaven this element comes from, and what exactly it is.




      You can interpret $prod,emptyset$ as the set of all functions from $emptyset$ to itself. The only function there is the empty function $emptyset$. Therefore, $prod,emptyset={emptyset}$.







      share|cite|improve this answer

















      • 3




        So it's not like a product of empty sets, but rather an empty product of sets, right? And thus its having one element is analogous to how a numerical empty product equals 1, and not 0.
        – The_Sympathizer
        Nov 21 at 6:55










      • Yep, that's the idea.
        – Batominovski
        Nov 21 at 7:45













      up vote
      7
      down vote



      accepted







      up vote
      7
      down vote



      accepted






      What the exercise is saying is the following: let $mathcal{C}$ be an empty family of sets (awkward, but $mathcal{C}$ is just isomorphic to $emptyset$). Then, $prod,mathcal{C}=prodlimits_{Cinmathcal{C}},C$ has one element. Simply put, $$prod,emptyset=prod_{Cinemptyset},C$$ has exactly one element. This is the same situation that gives rise to $a^0=1$. The real question is where on heaven this element comes from, and what exactly it is.




      You can interpret $prod,emptyset$ as the set of all functions from $emptyset$ to itself. The only function there is the empty function $emptyset$. Therefore, $prod,emptyset={emptyset}$.







      share|cite|improve this answer












      What the exercise is saying is the following: let $mathcal{C}$ be an empty family of sets (awkward, but $mathcal{C}$ is just isomorphic to $emptyset$). Then, $prod,mathcal{C}=prodlimits_{Cinmathcal{C}},C$ has one element. Simply put, $$prod,emptyset=prod_{Cinemptyset},C$$ has exactly one element. This is the same situation that gives rise to $a^0=1$. The real question is where on heaven this element comes from, and what exactly it is.




      You can interpret $prod,emptyset$ as the set of all functions from $emptyset$ to itself. The only function there is the empty function $emptyset$. Therefore, $prod,emptyset={emptyset}$.








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      share|cite|improve this answer










      answered Nov 21 at 2:27









      Batominovski

      33.5k33292




      33.5k33292








      • 3




        So it's not like a product of empty sets, but rather an empty product of sets, right? And thus its having one element is analogous to how a numerical empty product equals 1, and not 0.
        – The_Sympathizer
        Nov 21 at 6:55










      • Yep, that's the idea.
        – Batominovski
        Nov 21 at 7:45














      • 3




        So it's not like a product of empty sets, but rather an empty product of sets, right? And thus its having one element is analogous to how a numerical empty product equals 1, and not 0.
        – The_Sympathizer
        Nov 21 at 6:55










      • Yep, that's the idea.
        – Batominovski
        Nov 21 at 7:45








      3




      3




      So it's not like a product of empty sets, but rather an empty product of sets, right? And thus its having one element is analogous to how a numerical empty product equals 1, and not 0.
      – The_Sympathizer
      Nov 21 at 6:55




      So it's not like a product of empty sets, but rather an empty product of sets, right? And thus its having one element is analogous to how a numerical empty product equals 1, and not 0.
      – The_Sympathizer
      Nov 21 at 6:55












      Yep, that's the idea.
      – Batominovski
      Nov 21 at 7:45




      Yep, that's the idea.
      – Batominovski
      Nov 21 at 7:45










      up vote
      5
      down vote













      A product $Pi_{iin I} A_i$ of objects in a category is an object $B$ and a collection of morphisms $pi_i: Bto A_i$ such that for any object $C$ and collection of morphisms $d_i:Cto A_i,$ there is a unique morphism $f:Cto B$ such that $d_i=pi_icirc f.$ Note such a $(B,pi_i)$ may not exist and is generally not unique, although if $(B,pi_i)$ and $(B',pi_i')$ both satisfy the condition, there is a canonical isomorphism between $B$ and $B'$. If $I$ is empty, then there are no $pi_i$ and the definition reduces to an object $B$ such that for any $C$ there is a unique $f:Cto B$ (i.e. $B$ is a terminal object).



      Specifying to the category Set, the sets that have a unique map into them from any set are exactly the singletons (if the set had more than one element there would be multiple maps into it from any nonempty set; if it had zero elements, there would be none). So any singleton is a nullary product, and as promised, any two singletons have a canonical bijection between them. In a set theoretical sense, one may define the nullary cartesian product to be ${phi}$ for the sake of having a well-defined operation (just as one defines the binary cartesian product as the set of ordered pairs), but from a category theory perspective these are best thought of as nice representatives of a whole collection of product objects, which are all isomorphic to one another.






      share|cite|improve this answer



























        up vote
        5
        down vote













        A product $Pi_{iin I} A_i$ of objects in a category is an object $B$ and a collection of morphisms $pi_i: Bto A_i$ such that for any object $C$ and collection of morphisms $d_i:Cto A_i,$ there is a unique morphism $f:Cto B$ such that $d_i=pi_icirc f.$ Note such a $(B,pi_i)$ may not exist and is generally not unique, although if $(B,pi_i)$ and $(B',pi_i')$ both satisfy the condition, there is a canonical isomorphism between $B$ and $B'$. If $I$ is empty, then there are no $pi_i$ and the definition reduces to an object $B$ such that for any $C$ there is a unique $f:Cto B$ (i.e. $B$ is a terminal object).



        Specifying to the category Set, the sets that have a unique map into them from any set are exactly the singletons (if the set had more than one element there would be multiple maps into it from any nonempty set; if it had zero elements, there would be none). So any singleton is a nullary product, and as promised, any two singletons have a canonical bijection between them. In a set theoretical sense, one may define the nullary cartesian product to be ${phi}$ for the sake of having a well-defined operation (just as one defines the binary cartesian product as the set of ordered pairs), but from a category theory perspective these are best thought of as nice representatives of a whole collection of product objects, which are all isomorphic to one another.






        share|cite|improve this answer

























          up vote
          5
          down vote










          up vote
          5
          down vote









          A product $Pi_{iin I} A_i$ of objects in a category is an object $B$ and a collection of morphisms $pi_i: Bto A_i$ such that for any object $C$ and collection of morphisms $d_i:Cto A_i,$ there is a unique morphism $f:Cto B$ such that $d_i=pi_icirc f.$ Note such a $(B,pi_i)$ may not exist and is generally not unique, although if $(B,pi_i)$ and $(B',pi_i')$ both satisfy the condition, there is a canonical isomorphism between $B$ and $B'$. If $I$ is empty, then there are no $pi_i$ and the definition reduces to an object $B$ such that for any $C$ there is a unique $f:Cto B$ (i.e. $B$ is a terminal object).



          Specifying to the category Set, the sets that have a unique map into them from any set are exactly the singletons (if the set had more than one element there would be multiple maps into it from any nonempty set; if it had zero elements, there would be none). So any singleton is a nullary product, and as promised, any two singletons have a canonical bijection between them. In a set theoretical sense, one may define the nullary cartesian product to be ${phi}$ for the sake of having a well-defined operation (just as one defines the binary cartesian product as the set of ordered pairs), but from a category theory perspective these are best thought of as nice representatives of a whole collection of product objects, which are all isomorphic to one another.






          share|cite|improve this answer














          A product $Pi_{iin I} A_i$ of objects in a category is an object $B$ and a collection of morphisms $pi_i: Bto A_i$ such that for any object $C$ and collection of morphisms $d_i:Cto A_i,$ there is a unique morphism $f:Cto B$ such that $d_i=pi_icirc f.$ Note such a $(B,pi_i)$ may not exist and is generally not unique, although if $(B,pi_i)$ and $(B',pi_i')$ both satisfy the condition, there is a canonical isomorphism between $B$ and $B'$. If $I$ is empty, then there are no $pi_i$ and the definition reduces to an object $B$ such that for any $C$ there is a unique $f:Cto B$ (i.e. $B$ is a terminal object).



          Specifying to the category Set, the sets that have a unique map into them from any set are exactly the singletons (if the set had more than one element there would be multiple maps into it from any nonempty set; if it had zero elements, there would be none). So any singleton is a nullary product, and as promised, any two singletons have a canonical bijection between them. In a set theoretical sense, one may define the nullary cartesian product to be ${phi}$ for the sake of having a well-defined operation (just as one defines the binary cartesian product as the set of ordered pairs), but from a category theory perspective these are best thought of as nice representatives of a whole collection of product objects, which are all isomorphic to one another.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 21 at 5:18

























          answered Nov 21 at 5:13









          spaceisdarkgreen

          32k21652




          32k21652






















              up vote
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              down vote













              If the interpretation is $X^0$ then the answer is obvious. If this is not obvious look at the end of this answer.



              It's just the question of what the set $X^0$ is which is the set of functions mapping nothing (the empty set) to a set $X$ which obviously is only the function that does "nothing" (the single element in the set "1"). Formally $X^0 = { f mid f : emptyset to X } $ (the set of functions from the empty set to $X$) now what we want to figure out is what "this set really is". First we recall that a function is simply a relation defined by the graph of $f$ i.e. $f subseteq A times B$ such that $f := { (a,b) in A times B : forall a in A, exists b in B }$. In other words, a function just tells you how to pair the elements of the domain $A$ with the elements in the codomain $B$ via $(a,f(a))$ where $f(a) = b$ for some $b in B$ in the codomain. In this case we have $A = emptyset$ so all this actually become the following:



              $$X^0 = { f mid f : emptyset to X } = [emptyset to X] = { f subseteq A times B mid f: emptyset to X } $$
              which is equal to:



              $$ { f subseteq A times B mid f: emptyset to X } = { { (a,b) in A times B : forall a in emptyset , exists b in B } } = { emptyset } $$



              so,



              $$ X^0 = { emptyset }$$



              the key is realizing that $f = { (a,b) in A times B : forall a in emptyset , exists b in B } = emptyset$ because if we try to form a set of pairs where the first element of the pair comes from the empty set, then of course we won't form any pair becuase $a in emptyset$ is false.



              To make it really clear, recall the graph (i.e. relation set) of a function $f$ is the set of pairs $ f = { (a,b) in A times B : forall a in A, exists b in B } $ which is why we get as a final answer the empty set inside a set (since a function is really just a relation saying how we pair up the first element to the second element and $f$ induces this pairing/relation). In the end $X^0$ is just the set of all function from the empty set to $B$ but a function is a set of "pairs". This set of pairs ends up being empty because we can never select elements for its first location since we are trying to select things form the empty set.





              One comment on what $X^0$ should mean. I realized that perhaps its not entirely obvious what $X^0$ should mean. Recall:



              $$ X^1 = { f mid f : { 1 } to X }$$
              $$ X^2 = { f mid f : { 1,2 } to X }$$
              $$ X^3 = { f mid f: { 1,2,3 } to X }$$
              $$ vdots $$
              $$ X^n = { f mid f : { 1,2,3,dots,n} to X }$$
              $$ vdots $$
              sho what should $X^0$? Well, $X^n$ is really short hand for $X^{ {1,dots,n } }$ so when $n=0$ the set should start at 1 and reach zero. But no set in this context does that, so what we really mean is the empty set. So $X^0$ is short hand for $X^{emptyset}$ i.e.



              $$ X^{0} = X^{emptyset} = { f mid f emptyset to X } $$



              by definition.





              Alternatively with the sequence notation $A^n = { (a_1,dots,a_n) mid a_1 in A , dots, a_n in A }$ we have:



              $$ $A^1 = { (a_1) mid a_1 in A }$ $$
              $$A^2 = { (a_1,a_2) mid a_1 in A , a_2 in A }$$
              $$A^2 = { (a_1,a_2,a_3) mid a_1 in A , a_2 in A, a_3 in A }$$
              $$ vdots $$
              $$A^n = { (a_1,dots,a_n) mid a_1 in A , dots, a_n in A }$$
              $$ vdots $$
              we wonder how to interpret $A^0$? Basically says:




              The set of sequences of size zero from elements from A.




              which is just the empty sequence ${ () }$. Since no element are there (hand waving a little) it's isomorphic to the empty set $() cong emptyset$ (ok this last part would be nice to make more formal).






              share|cite|improve this answer























              • This is a nice answer; could be better if it were less ambiguous in its notation, therefore slightly clearer. E.g., the second equality is rather unreasonable except to those with sufficient familiarity with set theory to extrapolate the missing pieces. Nonetheless, +1.
                – Musa Al-hassy
                Nov 22 at 10:36










              • @MusaAl-hassy ok I added more details, it should be more accessible to everyone now, right? Thanks for the feedback! :)
                – Pinocchio
                Nov 22 at 18:08








              • 1




                @MusaAl-hassy ok you are right, the notation ${ f : A to B }$ is really confusing. It looks like the set contains 1 element, some function from $A$ to $B$. I decided to change it to something less confusing like ${ f mid f: A to B }$ which says "the set is the set of function with the property that each function $f$ is a function from $A$ to $B$. This is a lot less confusing since it mimics the more standard notations that say "the set is defined by elements $x$ (usually they say $x in X$) such that each element has so property $P(x)$". (continued...)
                – Pinocchio
                Dec 4 at 21:01








              • 1




                To make it more clear I could have done ${ f subseteq A times B mid f: A to B }$ but it felt too close to the formal definition (since I assume not many people think of function as relations or subsets of $A times B$ I intentionally deferred that (so that instead at that point $f$ was thought more like functions from programming, pointers to how to act given a input). Later I explain it clearly so the second time they read that short hand they can infer $f$ is really a subset of $A times B$.
                – Pinocchio
                Dec 4 at 21:04






              • 1




                Way to go :-) Thanks for your efforts!
                – Musa Al-hassy
                Dec 5 at 20:47















              up vote
              4
              down vote













              If the interpretation is $X^0$ then the answer is obvious. If this is not obvious look at the end of this answer.



              It's just the question of what the set $X^0$ is which is the set of functions mapping nothing (the empty set) to a set $X$ which obviously is only the function that does "nothing" (the single element in the set "1"). Formally $X^0 = { f mid f : emptyset to X } $ (the set of functions from the empty set to $X$) now what we want to figure out is what "this set really is". First we recall that a function is simply a relation defined by the graph of $f$ i.e. $f subseteq A times B$ such that $f := { (a,b) in A times B : forall a in A, exists b in B }$. In other words, a function just tells you how to pair the elements of the domain $A$ with the elements in the codomain $B$ via $(a,f(a))$ where $f(a) = b$ for some $b in B$ in the codomain. In this case we have $A = emptyset$ so all this actually become the following:



              $$X^0 = { f mid f : emptyset to X } = [emptyset to X] = { f subseteq A times B mid f: emptyset to X } $$
              which is equal to:



              $$ { f subseteq A times B mid f: emptyset to X } = { { (a,b) in A times B : forall a in emptyset , exists b in B } } = { emptyset } $$



              so,



              $$ X^0 = { emptyset }$$



              the key is realizing that $f = { (a,b) in A times B : forall a in emptyset , exists b in B } = emptyset$ because if we try to form a set of pairs where the first element of the pair comes from the empty set, then of course we won't form any pair becuase $a in emptyset$ is false.



              To make it really clear, recall the graph (i.e. relation set) of a function $f$ is the set of pairs $ f = { (a,b) in A times B : forall a in A, exists b in B } $ which is why we get as a final answer the empty set inside a set (since a function is really just a relation saying how we pair up the first element to the second element and $f$ induces this pairing/relation). In the end $X^0$ is just the set of all function from the empty set to $B$ but a function is a set of "pairs". This set of pairs ends up being empty because we can never select elements for its first location since we are trying to select things form the empty set.





              One comment on what $X^0$ should mean. I realized that perhaps its not entirely obvious what $X^0$ should mean. Recall:



              $$ X^1 = { f mid f : { 1 } to X }$$
              $$ X^2 = { f mid f : { 1,2 } to X }$$
              $$ X^3 = { f mid f: { 1,2,3 } to X }$$
              $$ vdots $$
              $$ X^n = { f mid f : { 1,2,3,dots,n} to X }$$
              $$ vdots $$
              sho what should $X^0$? Well, $X^n$ is really short hand for $X^{ {1,dots,n } }$ so when $n=0$ the set should start at 1 and reach zero. But no set in this context does that, so what we really mean is the empty set. So $X^0$ is short hand for $X^{emptyset}$ i.e.



              $$ X^{0} = X^{emptyset} = { f mid f emptyset to X } $$



              by definition.





              Alternatively with the sequence notation $A^n = { (a_1,dots,a_n) mid a_1 in A , dots, a_n in A }$ we have:



              $$ $A^1 = { (a_1) mid a_1 in A }$ $$
              $$A^2 = { (a_1,a_2) mid a_1 in A , a_2 in A }$$
              $$A^2 = { (a_1,a_2,a_3) mid a_1 in A , a_2 in A, a_3 in A }$$
              $$ vdots $$
              $$A^n = { (a_1,dots,a_n) mid a_1 in A , dots, a_n in A }$$
              $$ vdots $$
              we wonder how to interpret $A^0$? Basically says:




              The set of sequences of size zero from elements from A.




              which is just the empty sequence ${ () }$. Since no element are there (hand waving a little) it's isomorphic to the empty set $() cong emptyset$ (ok this last part would be nice to make more formal).






              share|cite|improve this answer























              • This is a nice answer; could be better if it were less ambiguous in its notation, therefore slightly clearer. E.g., the second equality is rather unreasonable except to those with sufficient familiarity with set theory to extrapolate the missing pieces. Nonetheless, +1.
                – Musa Al-hassy
                Nov 22 at 10:36










              • @MusaAl-hassy ok I added more details, it should be more accessible to everyone now, right? Thanks for the feedback! :)
                – Pinocchio
                Nov 22 at 18:08








              • 1




                @MusaAl-hassy ok you are right, the notation ${ f : A to B }$ is really confusing. It looks like the set contains 1 element, some function from $A$ to $B$. I decided to change it to something less confusing like ${ f mid f: A to B }$ which says "the set is the set of function with the property that each function $f$ is a function from $A$ to $B$. This is a lot less confusing since it mimics the more standard notations that say "the set is defined by elements $x$ (usually they say $x in X$) such that each element has so property $P(x)$". (continued...)
                – Pinocchio
                Dec 4 at 21:01








              • 1




                To make it more clear I could have done ${ f subseteq A times B mid f: A to B }$ but it felt too close to the formal definition (since I assume not many people think of function as relations or subsets of $A times B$ I intentionally deferred that (so that instead at that point $f$ was thought more like functions from programming, pointers to how to act given a input). Later I explain it clearly so the second time they read that short hand they can infer $f$ is really a subset of $A times B$.
                – Pinocchio
                Dec 4 at 21:04






              • 1




                Way to go :-) Thanks for your efforts!
                – Musa Al-hassy
                Dec 5 at 20:47













              up vote
              4
              down vote










              up vote
              4
              down vote









              If the interpretation is $X^0$ then the answer is obvious. If this is not obvious look at the end of this answer.



              It's just the question of what the set $X^0$ is which is the set of functions mapping nothing (the empty set) to a set $X$ which obviously is only the function that does "nothing" (the single element in the set "1"). Formally $X^0 = { f mid f : emptyset to X } $ (the set of functions from the empty set to $X$) now what we want to figure out is what "this set really is". First we recall that a function is simply a relation defined by the graph of $f$ i.e. $f subseteq A times B$ such that $f := { (a,b) in A times B : forall a in A, exists b in B }$. In other words, a function just tells you how to pair the elements of the domain $A$ with the elements in the codomain $B$ via $(a,f(a))$ where $f(a) = b$ for some $b in B$ in the codomain. In this case we have $A = emptyset$ so all this actually become the following:



              $$X^0 = { f mid f : emptyset to X } = [emptyset to X] = { f subseteq A times B mid f: emptyset to X } $$
              which is equal to:



              $$ { f subseteq A times B mid f: emptyset to X } = { { (a,b) in A times B : forall a in emptyset , exists b in B } } = { emptyset } $$



              so,



              $$ X^0 = { emptyset }$$



              the key is realizing that $f = { (a,b) in A times B : forall a in emptyset , exists b in B } = emptyset$ because if we try to form a set of pairs where the first element of the pair comes from the empty set, then of course we won't form any pair becuase $a in emptyset$ is false.



              To make it really clear, recall the graph (i.e. relation set) of a function $f$ is the set of pairs $ f = { (a,b) in A times B : forall a in A, exists b in B } $ which is why we get as a final answer the empty set inside a set (since a function is really just a relation saying how we pair up the first element to the second element and $f$ induces this pairing/relation). In the end $X^0$ is just the set of all function from the empty set to $B$ but a function is a set of "pairs". This set of pairs ends up being empty because we can never select elements for its first location since we are trying to select things form the empty set.





              One comment on what $X^0$ should mean. I realized that perhaps its not entirely obvious what $X^0$ should mean. Recall:



              $$ X^1 = { f mid f : { 1 } to X }$$
              $$ X^2 = { f mid f : { 1,2 } to X }$$
              $$ X^3 = { f mid f: { 1,2,3 } to X }$$
              $$ vdots $$
              $$ X^n = { f mid f : { 1,2,3,dots,n} to X }$$
              $$ vdots $$
              sho what should $X^0$? Well, $X^n$ is really short hand for $X^{ {1,dots,n } }$ so when $n=0$ the set should start at 1 and reach zero. But no set in this context does that, so what we really mean is the empty set. So $X^0$ is short hand for $X^{emptyset}$ i.e.



              $$ X^{0} = X^{emptyset} = { f mid f emptyset to X } $$



              by definition.





              Alternatively with the sequence notation $A^n = { (a_1,dots,a_n) mid a_1 in A , dots, a_n in A }$ we have:



              $$ $A^1 = { (a_1) mid a_1 in A }$ $$
              $$A^2 = { (a_1,a_2) mid a_1 in A , a_2 in A }$$
              $$A^2 = { (a_1,a_2,a_3) mid a_1 in A , a_2 in A, a_3 in A }$$
              $$ vdots $$
              $$A^n = { (a_1,dots,a_n) mid a_1 in A , dots, a_n in A }$$
              $$ vdots $$
              we wonder how to interpret $A^0$? Basically says:




              The set of sequences of size zero from elements from A.




              which is just the empty sequence ${ () }$. Since no element are there (hand waving a little) it's isomorphic to the empty set $() cong emptyset$ (ok this last part would be nice to make more formal).






              share|cite|improve this answer














              If the interpretation is $X^0$ then the answer is obvious. If this is not obvious look at the end of this answer.



              It's just the question of what the set $X^0$ is which is the set of functions mapping nothing (the empty set) to a set $X$ which obviously is only the function that does "nothing" (the single element in the set "1"). Formally $X^0 = { f mid f : emptyset to X } $ (the set of functions from the empty set to $X$) now what we want to figure out is what "this set really is". First we recall that a function is simply a relation defined by the graph of $f$ i.e. $f subseteq A times B$ such that $f := { (a,b) in A times B : forall a in A, exists b in B }$. In other words, a function just tells you how to pair the elements of the domain $A$ with the elements in the codomain $B$ via $(a,f(a))$ where $f(a) = b$ for some $b in B$ in the codomain. In this case we have $A = emptyset$ so all this actually become the following:



              $$X^0 = { f mid f : emptyset to X } = [emptyset to X] = { f subseteq A times B mid f: emptyset to X } $$
              which is equal to:



              $$ { f subseteq A times B mid f: emptyset to X } = { { (a,b) in A times B : forall a in emptyset , exists b in B } } = { emptyset } $$



              so,



              $$ X^0 = { emptyset }$$



              the key is realizing that $f = { (a,b) in A times B : forall a in emptyset , exists b in B } = emptyset$ because if we try to form a set of pairs where the first element of the pair comes from the empty set, then of course we won't form any pair becuase $a in emptyset$ is false.



              To make it really clear, recall the graph (i.e. relation set) of a function $f$ is the set of pairs $ f = { (a,b) in A times B : forall a in A, exists b in B } $ which is why we get as a final answer the empty set inside a set (since a function is really just a relation saying how we pair up the first element to the second element and $f$ induces this pairing/relation). In the end $X^0$ is just the set of all function from the empty set to $B$ but a function is a set of "pairs". This set of pairs ends up being empty because we can never select elements for its first location since we are trying to select things form the empty set.





              One comment on what $X^0$ should mean. I realized that perhaps its not entirely obvious what $X^0$ should mean. Recall:



              $$ X^1 = { f mid f : { 1 } to X }$$
              $$ X^2 = { f mid f : { 1,2 } to X }$$
              $$ X^3 = { f mid f: { 1,2,3 } to X }$$
              $$ vdots $$
              $$ X^n = { f mid f : { 1,2,3,dots,n} to X }$$
              $$ vdots $$
              sho what should $X^0$? Well, $X^n$ is really short hand for $X^{ {1,dots,n } }$ so when $n=0$ the set should start at 1 and reach zero. But no set in this context does that, so what we really mean is the empty set. So $X^0$ is short hand for $X^{emptyset}$ i.e.



              $$ X^{0} = X^{emptyset} = { f mid f emptyset to X } $$



              by definition.





              Alternatively with the sequence notation $A^n = { (a_1,dots,a_n) mid a_1 in A , dots, a_n in A }$ we have:



              $$ $A^1 = { (a_1) mid a_1 in A }$ $$
              $$A^2 = { (a_1,a_2) mid a_1 in A , a_2 in A }$$
              $$A^2 = { (a_1,a_2,a_3) mid a_1 in A , a_2 in A, a_3 in A }$$
              $$ vdots $$
              $$A^n = { (a_1,dots,a_n) mid a_1 in A , dots, a_n in A }$$
              $$ vdots $$
              we wonder how to interpret $A^0$? Basically says:




              The set of sequences of size zero from elements from A.




              which is just the empty sequence ${ () }$. Since no element are there (hand waving a little) it's isomorphic to the empty set $() cong emptyset$ (ok this last part would be nice to make more formal).







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 4 at 21:55

























              answered Nov 21 at 2:38









              Pinocchio

              1,86321754




              1,86321754












              • This is a nice answer; could be better if it were less ambiguous in its notation, therefore slightly clearer. E.g., the second equality is rather unreasonable except to those with sufficient familiarity with set theory to extrapolate the missing pieces. Nonetheless, +1.
                – Musa Al-hassy
                Nov 22 at 10:36










              • @MusaAl-hassy ok I added more details, it should be more accessible to everyone now, right? Thanks for the feedback! :)
                – Pinocchio
                Nov 22 at 18:08








              • 1




                @MusaAl-hassy ok you are right, the notation ${ f : A to B }$ is really confusing. It looks like the set contains 1 element, some function from $A$ to $B$. I decided to change it to something less confusing like ${ f mid f: A to B }$ which says "the set is the set of function with the property that each function $f$ is a function from $A$ to $B$. This is a lot less confusing since it mimics the more standard notations that say "the set is defined by elements $x$ (usually they say $x in X$) such that each element has so property $P(x)$". (continued...)
                – Pinocchio
                Dec 4 at 21:01








              • 1




                To make it more clear I could have done ${ f subseteq A times B mid f: A to B }$ but it felt too close to the formal definition (since I assume not many people think of function as relations or subsets of $A times B$ I intentionally deferred that (so that instead at that point $f$ was thought more like functions from programming, pointers to how to act given a input). Later I explain it clearly so the second time they read that short hand they can infer $f$ is really a subset of $A times B$.
                – Pinocchio
                Dec 4 at 21:04






              • 1




                Way to go :-) Thanks for your efforts!
                – Musa Al-hassy
                Dec 5 at 20:47


















              • This is a nice answer; could be better if it were less ambiguous in its notation, therefore slightly clearer. E.g., the second equality is rather unreasonable except to those with sufficient familiarity with set theory to extrapolate the missing pieces. Nonetheless, +1.
                – Musa Al-hassy
                Nov 22 at 10:36










              • @MusaAl-hassy ok I added more details, it should be more accessible to everyone now, right? Thanks for the feedback! :)
                – Pinocchio
                Nov 22 at 18:08








              • 1




                @MusaAl-hassy ok you are right, the notation ${ f : A to B }$ is really confusing. It looks like the set contains 1 element, some function from $A$ to $B$. I decided to change it to something less confusing like ${ f mid f: A to B }$ which says "the set is the set of function with the property that each function $f$ is a function from $A$ to $B$. This is a lot less confusing since it mimics the more standard notations that say "the set is defined by elements $x$ (usually they say $x in X$) such that each element has so property $P(x)$". (continued...)
                – Pinocchio
                Dec 4 at 21:01








              • 1




                To make it more clear I could have done ${ f subseteq A times B mid f: A to B }$ but it felt too close to the formal definition (since I assume not many people think of function as relations or subsets of $A times B$ I intentionally deferred that (so that instead at that point $f$ was thought more like functions from programming, pointers to how to act given a input). Later I explain it clearly so the second time they read that short hand they can infer $f$ is really a subset of $A times B$.
                – Pinocchio
                Dec 4 at 21:04






              • 1




                Way to go :-) Thanks for your efforts!
                – Musa Al-hassy
                Dec 5 at 20:47
















              This is a nice answer; could be better if it were less ambiguous in its notation, therefore slightly clearer. E.g., the second equality is rather unreasonable except to those with sufficient familiarity with set theory to extrapolate the missing pieces. Nonetheless, +1.
              – Musa Al-hassy
              Nov 22 at 10:36




              This is a nice answer; could be better if it were less ambiguous in its notation, therefore slightly clearer. E.g., the second equality is rather unreasonable except to those with sufficient familiarity with set theory to extrapolate the missing pieces. Nonetheless, +1.
              – Musa Al-hassy
              Nov 22 at 10:36












              @MusaAl-hassy ok I added more details, it should be more accessible to everyone now, right? Thanks for the feedback! :)
              – Pinocchio
              Nov 22 at 18:08






              @MusaAl-hassy ok I added more details, it should be more accessible to everyone now, right? Thanks for the feedback! :)
              – Pinocchio
              Nov 22 at 18:08






              1




              1




              @MusaAl-hassy ok you are right, the notation ${ f : A to B }$ is really confusing. It looks like the set contains 1 element, some function from $A$ to $B$. I decided to change it to something less confusing like ${ f mid f: A to B }$ which says "the set is the set of function with the property that each function $f$ is a function from $A$ to $B$. This is a lot less confusing since it mimics the more standard notations that say "the set is defined by elements $x$ (usually they say $x in X$) such that each element has so property $P(x)$". (continued...)
              – Pinocchio
              Dec 4 at 21:01






              @MusaAl-hassy ok you are right, the notation ${ f : A to B }$ is really confusing. It looks like the set contains 1 element, some function from $A$ to $B$. I decided to change it to something less confusing like ${ f mid f: A to B }$ which says "the set is the set of function with the property that each function $f$ is a function from $A$ to $B$. This is a lot less confusing since it mimics the more standard notations that say "the set is defined by elements $x$ (usually they say $x in X$) such that each element has so property $P(x)$". (continued...)
              – Pinocchio
              Dec 4 at 21:01






              1




              1




              To make it more clear I could have done ${ f subseteq A times B mid f: A to B }$ but it felt too close to the formal definition (since I assume not many people think of function as relations or subsets of $A times B$ I intentionally deferred that (so that instead at that point $f$ was thought more like functions from programming, pointers to how to act given a input). Later I explain it clearly so the second time they read that short hand they can infer $f$ is really a subset of $A times B$.
              – Pinocchio
              Dec 4 at 21:04




              To make it more clear I could have done ${ f subseteq A times B mid f: A to B }$ but it felt too close to the formal definition (since I assume not many people think of function as relations or subsets of $A times B$ I intentionally deferred that (so that instead at that point $f$ was thought more like functions from programming, pointers to how to act given a input). Later I explain it clearly so the second time they read that short hand they can infer $f$ is really a subset of $A times B$.
              – Pinocchio
              Dec 4 at 21:04




              1




              1




              Way to go :-) Thanks for your efforts!
              – Musa Al-hassy
              Dec 5 at 20:47




              Way to go :-) Thanks for your efforts!
              – Musa Al-hassy
              Dec 5 at 20:47










              up vote
              3
              down vote













              Since no-one has said this explicitly:



              Yes, you are misreading the language of the question. The question is asking for



              $$prodlimits_{x in emptyset} x$$



              It's the difference between "the product of an empty set of sets" and "the product of a set of empty sets".






              share|cite|improve this answer

























                up vote
                3
                down vote













                Since no-one has said this explicitly:



                Yes, you are misreading the language of the question. The question is asking for



                $$prodlimits_{x in emptyset} x$$



                It's the difference between "the product of an empty set of sets" and "the product of a set of empty sets".






                share|cite|improve this answer























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  Since no-one has said this explicitly:



                  Yes, you are misreading the language of the question. The question is asking for



                  $$prodlimits_{x in emptyset} x$$



                  It's the difference between "the product of an empty set of sets" and "the product of a set of empty sets".






                  share|cite|improve this answer












                  Since no-one has said this explicitly:



                  Yes, you are misreading the language of the question. The question is asking for



                  $$prodlimits_{x in emptyset} x$$



                  It's the difference between "the product of an empty set of sets" and "the product of a set of empty sets".







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 21 at 9:52









                  Christopher

                  6,42711628




                  6,42711628






























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