Mixing
Can the empty set serve as universe of a semigroup (i.e. set equipped with associative binary operation)?
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The title of my question is more a representative of my more general question.
In A course in universal algebra (nice material!) I encountered in definition 1.3 that the universe of an algebra is not empty.
It is clear to me that this is inevitable if there are nullary operations, but why also demanding this if that is not the case?
Uptil I could not find advantages for that, while I could find disadvantages.
For instance some categories (e.g. the one of sets and the one of semigroups) "loose" their initial object.
So my question is:
"What could wrong if we allow algebras to have an empty universe?"
logic model-theory universal-algebra
edited 3 hours ago
Alex Kruckman
23.9k22454
asked 3 hours ago
Vera
1,956414
 |Â
show 2 more comments
up vote
2
down vote
favorite
The title of my question is more a representative of my more general question.
In A course in universal algebra (nice material!) I encountered in definition 1.3 that the universe of an algebra is not empty.
It is clear to me that this is inevitable if there are nullary operations, but why also demanding this if that is not the case?
Uptil I could not find advantages for that, while I could find disadvantages.
For instance some categories (e.g. the one of sets and the one of semigroups) "loose" their initial object.
So my question is:
"What could wrong if we allow algebras to have an empty universe?"
logic model-theory universal-algebra
edited 3 hours ago
Alex Kruckman
23.9k22454
asked 3 hours ago
Vera
1,956414
2
I suspect that the requirement that algebras are non-empty comes directly from the common assumption in model theory and mathematical logic that all structures are nonempty. (I added the (logic) and (model-theory) tags for this reason.) Of course, it's not really possible for a convention or a definition to be "right" or "wrong" - but some conventions are better than others, and I feel strongly that empty structures should always be allowed, both in universal algebra and in model theory.
– Alex Kruckman
3 hours ago
I don't think anything is wrong. It's just pointless. After all in Def 1.1 we define $A^n$ as all $n$-tuples. Well, that would define $emptyset^n = emptyset$ and there'd only be one vacuus operation. You can't have any groups. You could have a vacous semi-group but its pointless.
– fleablood
2 hours ago
1
@fleablood It's not pointless to want a category of interest to have an initial object.
– Alex Kruckman
2 hours ago
Um... have an initial object is entirely different that having a terminal object. And ending at the empty set is just easy trivial and vaccuus albeit perfectly fine. Just pointless.
– fleablood
1 hour ago
@fleablood I don't understand your comment at all. In a category of algebras, the initial object is the free algebra generated by the constants (which is empty when there are no constants). The terminal object is the trivial algebra with exactly one element. The empty set can't be terminal, because there are no maps from non-empty sets to the empty set. .....Ohhhhh wait, maybe "pointless" is a pun on "empty", and you're trolling me?
– Alex Kruckman
1 hour ago
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The title of my question is more a representative of my more general question.
In A course in universal algebra (nice material!) I encountered in definition 1.3 that the universe of an algebra is not empty.
It is clear to me that this is inevitable if there are nullary operations, but why also demanding this if that is not the case?
Uptil I could not find advantages for that, while I could find disadvantages.
For instance some categories (e.g. the one of sets and the one of semigroups) "loose" their initial object.
So my question is:
"What could wrong if we allow algebras to have an empty universe?"
logic model-theory universal-algebra
edited 3 hours ago
Alex Kruckman
23.9k22454
asked 3 hours ago
Vera
1,956414
The title of my question is more a representative of my more general question.
In A course in universal algebra (nice material!) I encountered in definition 1.3 that the universe of an algebra is not empty.
It is clear to me that this is inevitable if there are nullary operations, but why also demanding this if that is not the case?
Uptil I could not find advantages for that, while I could find disadvantages.
For instance some categories (e.g. the one of sets and the one of semigroups) "loose" their initial object.
So my question is:
"What could wrong if we allow algebras to have an empty universe?"
logic model-theory universal-algebra
logic model-theory universal-algebra
edited 3 hours ago
Alex Kruckman
23.9k22454
asked 3 hours ago
Vera
1,956414
edited 3 hours ago
Alex Kruckman
23.9k22454
asked 3 hours ago
Vera
1,956414
edited 3 hours ago
Alex Kruckman
23.9k22454
edited 3 hours ago
Alex Kruckman
23.9k22454
edited 3 hours ago
Alex Kruckman
23.9k22454
23.9k22454
asked 3 hours ago
Vera
1,956414
asked 3 hours ago
Vera
1,956414
asked 3 hours ago
Vera
1,956414
1,956414
2
I suspect that the requirement that algebras are non-empty comes directly from the common assumption in model theory and mathematical logic that all structures are nonempty. (I added the (logic) and (model-theory) tags for this reason.) Of course, it's not really possible for a convention or a definition to be "right" or "wrong" - but some conventions are better than others, and I feel strongly that empty structures should always be allowed, both in universal algebra and in model theory.
– Alex Kruckman
3 hours ago
I don't think anything is wrong. It's just pointless. After all in Def 1.1 we define $A^n$ as all $n$-tuples. Well, that would define $emptyset^n = emptyset$ and there'd only be one vacuus operation. You can't have any groups. You could have a vacous semi-group but its pointless.
– fleablood
2 hours ago
1
@fleablood It's not pointless to want a category of interest to have an initial object.
– Alex Kruckman
2 hours ago
Um... have an initial object is entirely different that having a terminal object. And ending at the empty set is just easy trivial and vaccuus albeit perfectly fine. Just pointless.
– fleablood
1 hour ago
@fleablood I don't understand your comment at all. In a category of algebras, the initial object is the free algebra generated by the constants (which is empty when there are no constants). The terminal object is the trivial algebra with exactly one element. The empty set can't be terminal, because there are no maps from non-empty sets to the empty set. .....Ohhhhh wait, maybe "pointless" is a pun on "empty", and you're trolling me?
– Alex Kruckman
1 hour ago
 |Â
show 2 more comments
2
I suspect that the requirement that algebras are non-empty comes directly from the common assumption in model theory and mathematical logic that all structures are nonempty. (I added the (logic) and (model-theory) tags for this reason.) Of course, it's not really possible for a convention or a definition to be "right" or "wrong" - but some conventions are better than others, and I feel strongly that empty structures should always be allowed, both in universal algebra and in model theory.
– Alex Kruckman
3 hours ago
I don't think anything is wrong. It's just pointless. After all in Def 1.1 we define $A^n$ as all $n$-tuples. Well, that would define $emptyset^n = emptyset$ and there'd only be one vacuus operation. You can't have any groups. You could have a vacous semi-group but its pointless.
– fleablood
2 hours ago
1
@fleablood It's not pointless to want a category of interest to have an initial object.
– Alex Kruckman
2 hours ago
Um... have an initial object is entirely different that having a terminal object. And ending at the empty set is just easy trivial and vaccuus albeit perfectly fine. Just pointless.
– fleablood
1 hour ago
@fleablood I don't understand your comment at all. In a category of algebras, the initial object is the free algebra generated by the constants (which is empty when there are no constants). The terminal object is the trivial algebra with exactly one element. The empty set can't be terminal, because there are no maps from non-empty sets to the empty set. .....Ohhhhh wait, maybe "pointless" is a pun on "empty", and you're trolling me?
– Alex Kruckman
1 hour ago
2
2
I suspect that the requirement that algebras are non-empty comes directly from the common assumption in model theory and mathematical logic that all structures are nonempty. (I added the (logic) and (model-theory) tags for this reason.) Of course, it's not really possible for a convention or a definition to be "right" or "wrong" - but some conventions are better than others, and I feel strongly that empty structures should always be allowed, both in universal algebra and in model theory.
– Alex Kruckman
3 hours ago
I suspect that the requirement that algebras are non-empty comes directly from the common assumption in model theory and mathematical logic that all structures are nonempty. (I added the (logic) and (model-theory) tags for this reason.) Of course, it's not really possible for a convention or a definition to be "right" or "wrong" - but some conventions are better than others, and I feel strongly that empty structures should always be allowed, both in universal algebra and in model theory.
– Alex Kruckman
3 hours ago
I don't think anything is wrong. It's just pointless. After all in Def 1.1 we define $A^n$ as all $n$-tuples. Well, that would define $emptyset^n = emptyset$ and there'd only be one vacuus operation. You can't have any groups. You could have a vacous semi-group but its pointless.
– fleablood
2 hours ago
I don't think anything is wrong. It's just pointless. After all in Def 1.1 we define $A^n$ as all $n$-tuples. Well, that would define $emptyset^n = emptyset$ and there'd only be one vacuus operation. You can't have any groups. You could have a vacous semi-group but its pointless.
– fleablood
2 hours ago
1
1
@fleablood It's not pointless to want a category of interest to have an initial object.
– Alex Kruckman
2 hours ago
@fleablood It's not pointless to want a category of interest to have an initial object.
– Alex Kruckman
2 hours ago
Um... have an initial object is entirely different that having a terminal object. And ending at the empty set is just easy trivial and vaccuus albeit perfectly fine. Just pointless.
– fleablood
1 hour ago
Um... have an initial object is entirely different that having a terminal object. And ending at the empty set is just easy trivial and vaccuus albeit perfectly fine. Just pointless.
– fleablood
1 hour ago
@fleablood I don't understand your comment at all. In a category of algebras, the initial object is the free algebra generated by the constants (which is empty when there are no constants). The terminal object is the trivial algebra with exactly one element. The empty set can't be terminal, because there are no maps from non-empty sets to the empty set. .....Ohhhhh wait, maybe "pointless" is a pun on "empty", and you're trolling me?
– Alex Kruckman
1 hour ago
@fleablood I don't understand your comment at all. In a category of algebras, the initial object is the free algebra generated by the constants (which is empty when there are no constants). The terminal object is the trivial algebra with exactly one element. The empty set can't be terminal, because there are no maps from non-empty sets to the empty set. .....Ohhhhh wait, maybe "pointless" is a pun on "empty", and you're trolling me?
– Alex Kruckman
1 hour ago
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
3
down vote
First, I want to reiterate my comment above to the OP - I believe that empty structures should always be allowed, both in universal algebra and in model theory. Nothing goes seriously wrong when you include them.
But let me play the devil's advocate for a moment and point out one thing that could apparently go wrong: ultraproducts (see Sections IV.6 and V.2 in Burris and Sankappanavar). Let $(A_i)_iin I$ be a family of algebras, indexed by the set $I$, and let $U$ be an ultrafilter on $I$. Then the ultraproduct is defined to be the quotient of the product $prod_iin I A_i$ by the congruence $theta_U$, defined by $((a_i), (b_i))in theta_U$ if and only if $iin Imid a_i = b_iin U$.
Now we'd really like to have Ã…Âoś's theorem, which says (in a special case) that an identity holds in the ultraproduct if and only if the set of factors on which it holds is in $U$. But observe that if any single $A_i$ is empty, the product of the $A_i$ is empty, so the ultraproduct is empty. Now Ã…Âoś's theorem can fail: for example, suppse $U$ is a non-principal ultrafilter, one $A_i^*$ is empty, and every other $A_i$ has at least $2$ elements. Then the identity $x = y$ holds in the ultraproduct (vacuously), but the set of $iin I$ such that it holds in $A_i$ is the singleton $i^*$, which is not in $U$.
But this just means we're using the wrong definition of ultraproduct! The correct definition is: $$prod_iin I A_i/U = varinjlim_Xin U prod_iin X A_i.$$
Here we look at each set $Xsubseteq I$ in the ultrafilter, and take the $X$-indexed product $prod_iin X A_i$. Whenever $Ysubseteq X$, we have a projection map $pi^X_Y$ from the $X$-indexed product to the $Y$-indexed product. The resulting system of products and connecting maps is directed (since $U$ is a filter), and we take the directed colimit.
This definition gives an ultraproduct which is isomorphic to the old one in the case when all of the $A_i$ are nonempty. But it gives a non-empty ultraproduct when the set of non-empty factors is in $U$, and it allows you to prove Ã…Âoś's theorem in this context.
answered 2 hours ago
Alex Kruckman
23.9k22454
So... you're saying that $forall x(xneq x)$ is not provably false? Huh.
– Asaf Karagila♦
1 hour ago
@AsafKaragila That depends on your proof system. And of course you should always choose a proof system which is sound and complete for the semantics you have in mind.
– Alex Kruckman
58 mins ago
Thank you. Ultraproducts are not in my scope (yet), but in this situation that has a positive side: the argumentation of the devil must come from far. Posing the question I was actually hoping for answers that would (mainly) tell me that an empty carrier set does no harm in general.
– Vera
11 mins ago
add a comment |Â
up vote
3
down vote
There is nothing wrong with allowing the carrier set of an algebra to be empty. Arguably it is a more natural and modern approach to allow for structures to be empty in general. The Wikipedia definitions of algebra and semigroup, for instance, do not exclude this possibility. It's not uncommon though, especially in older texts, to see a requirement that structures be nonempty, which does create some blemishes in the theory, as you've observed.
answered 2 hours ago
Brent Kerby
4,609414
1
It is a fairly standard convention coming from logic.
– Andrés E. Caicedo
2 hours ago
@AndrésE.Caicedo, Thanks, looks like I underestimated how common the convention has been; I've edited my answer accordingly.
– Brent Kerby
2 hours ago
add a comment |Â
up vote
-1
down vote
That the empty set is valid comes from the fact that the laws of associativity and also commutativity are satisfied, such as
$$forall a,bin A: ab=ba.$$
This means that
$$forall aforall b[ ain Awedge bin ARightarrow ab=ba].$$
This whole assertion is true if the premise is false such as in the case $A=emptyset$.
answered 2 hours ago
Wuestenfux
1,048128
Thank you for answering, but I did not have any doubts concerning this topic. So actually this is not really an answer to my question. Sorry (I did not downvote, though).
– Vera
7 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
First, I want to reiterate my comment above to the OP - I believe that empty structures should always be allowed, both in universal algebra and in model theory. Nothing goes seriously wrong when you include them.
But let me play the devil's advocate for a moment and point out one thing that could apparently go wrong: ultraproducts (see Sections IV.6 and V.2 in Burris and Sankappanavar). Let $(A_i)_iin I$ be a family of algebras, indexed by the set $I$, and let $U$ be an ultrafilter on $I$. Then the ultraproduct is defined to be the quotient of the product $prod_iin I A_i$ by the congruence $theta_U$, defined by $((a_i), (b_i))in theta_U$ if and only if $iin Imid a_i = b_iin U$.
Now we'd really like to have Ã…Âoś's theorem, which says (in a special case) that an identity holds in the ultraproduct if and only if the set of factors on which it holds is in $U$. But observe that if any single $A_i$ is empty, the product of the $A_i$ is empty, so the ultraproduct is empty. Now Ã…Âoś's theorem can fail: for example, suppse $U$ is a non-principal ultrafilter, one $A_i^*$ is empty, and every other $A_i$ has at least $2$ elements. Then the identity $x = y$ holds in the ultraproduct (vacuously), but the set of $iin I$ such that it holds in $A_i$ is the singleton $i^*$, which is not in $U$.
But this just means we're using the wrong definition of ultraproduct! The correct definition is: $$prod_iin I A_i/U = varinjlim_Xin U prod_iin X A_i.$$
Here we look at each set $Xsubseteq I$ in the ultrafilter, and take the $X$-indexed product $prod_iin X A_i$. Whenever $Ysubseteq X$, we have a projection map $pi^X_Y$ from the $X$-indexed product to the $Y$-indexed product. The resulting system of products and connecting maps is directed (since $U$ is a filter), and we take the directed colimit.
This definition gives an ultraproduct which is isomorphic to the old one in the case when all of the $A_i$ are nonempty. But it gives a non-empty ultraproduct when the set of non-empty factors is in $U$, and it allows you to prove Ã…Âoś's theorem in this context.
answered 2 hours ago
Alex Kruckman
23.9k22454
So... you're saying that $forall x(xneq x)$ is not provably false? Huh.
– Asaf Karagila♦
1 hour ago
@AsafKaragila That depends on your proof system. And of course you should always choose a proof system which is sound and complete for the semantics you have in mind.
– Alex Kruckman
58 mins ago
Thank you. Ultraproducts are not in my scope (yet), but in this situation that has a positive side: the argumentation of the devil must come from far. Posing the question I was actually hoping for answers that would (mainly) tell me that an empty carrier set does no harm in general.
– Vera
11 mins ago
add a comment |Â
up vote
3
down vote
First, I want to reiterate my comment above to the OP - I believe that empty structures should always be allowed, both in universal algebra and in model theory. Nothing goes seriously wrong when you include them.
But let me play the devil's advocate for a moment and point out one thing that could apparently go wrong: ultraproducts (see Sections IV.6 and V.2 in Burris and Sankappanavar). Let $(A_i)_iin I$ be a family of algebras, indexed by the set $I$, and let $U$ be an ultrafilter on $I$. Then the ultraproduct is defined to be the quotient of the product $prod_iin I A_i$ by the congruence $theta_U$, defined by $((a_i), (b_i))in theta_U$ if and only if $iin Imid a_i = b_iin U$.
Now we'd really like to have Ã…Âoś's theorem, which says (in a special case) that an identity holds in the ultraproduct if and only if the set of factors on which it holds is in $U$. But observe that if any single $A_i$ is empty, the product of the $A_i$ is empty, so the ultraproduct is empty. Now Ã…Âoś's theorem can fail: for example, suppse $U$ is a non-principal ultrafilter, one $A_i^*$ is empty, and every other $A_i$ has at least $2$ elements. Then the identity $x = y$ holds in the ultraproduct (vacuously), but the set of $iin I$ such that it holds in $A_i$ is the singleton $i^*$, which is not in $U$.
But this just means we're using the wrong definition of ultraproduct! The correct definition is: $$prod_iin I A_i/U = varinjlim_Xin U prod_iin X A_i.$$
Here we look at each set $Xsubseteq I$ in the ultrafilter, and take the $X$-indexed product $prod_iin X A_i$. Whenever $Ysubseteq X$, we have a projection map $pi^X_Y$ from the $X$-indexed product to the $Y$-indexed product. The resulting system of products and connecting maps is directed (since $U$ is a filter), and we take the directed colimit.
This definition gives an ultraproduct which is isomorphic to the old one in the case when all of the $A_i$ are nonempty. But it gives a non-empty ultraproduct when the set of non-empty factors is in $U$, and it allows you to prove Ã…Âoś's theorem in this context.
answered 2 hours ago
Alex Kruckman
23.9k22454
So... you're saying that $forall x(xneq x)$ is not provably false? Huh.
– Asaf Karagila♦
1 hour ago
@AsafKaragila That depends on your proof system. And of course you should always choose a proof system which is sound and complete for the semantics you have in mind.
– Alex Kruckman
58 mins ago
Thank you. Ultraproducts are not in my scope (yet), but in this situation that has a positive side: the argumentation of the devil must come from far. Posing the question I was actually hoping for answers that would (mainly) tell me that an empty carrier set does no harm in general.
– Vera
11 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
First, I want to reiterate my comment above to the OP - I believe that empty structures should always be allowed, both in universal algebra and in model theory. Nothing goes seriously wrong when you include them.
But let me play the devil's advocate for a moment and point out one thing that could apparently go wrong: ultraproducts (see Sections IV.6 and V.2 in Burris and Sankappanavar). Let $(A_i)_iin I$ be a family of algebras, indexed by the set $I$, and let $U$ be an ultrafilter on $I$. Then the ultraproduct is defined to be the quotient of the product $prod_iin I A_i$ by the congruence $theta_U$, defined by $((a_i), (b_i))in theta_U$ if and only if $iin Imid a_i = b_iin U$.
Now we'd really like to have Ã…Âoś's theorem, which says (in a special case) that an identity holds in the ultraproduct if and only if the set of factors on which it holds is in $U$. But observe that if any single $A_i$ is empty, the product of the $A_i$ is empty, so the ultraproduct is empty. Now Ã…Âoś's theorem can fail: for example, suppse $U$ is a non-principal ultrafilter, one $A_i^*$ is empty, and every other $A_i$ has at least $2$ elements. Then the identity $x = y$ holds in the ultraproduct (vacuously), but the set of $iin I$ such that it holds in $A_i$ is the singleton $i^*$, which is not in $U$.
But this just means we're using the wrong definition of ultraproduct! The correct definition is: $$prod_iin I A_i/U = varinjlim_Xin U prod_iin X A_i.$$
Here we look at each set $Xsubseteq I$ in the ultrafilter, and take the $X$-indexed product $prod_iin X A_i$. Whenever $Ysubseteq X$, we have a projection map $pi^X_Y$ from the $X$-indexed product to the $Y$-indexed product. The resulting system of products and connecting maps is directed (since $U$ is a filter), and we take the directed colimit.
This definition gives an ultraproduct which is isomorphic to the old one in the case when all of the $A_i$ are nonempty. But it gives a non-empty ultraproduct when the set of non-empty factors is in $U$, and it allows you to prove Ã…Âoś's theorem in this context.
answered 2 hours ago
Alex Kruckman
23.9k22454
First, I want to reiterate my comment above to the OP - I believe that empty structures should always be allowed, both in universal algebra and in model theory. Nothing goes seriously wrong when you include them.
But let me play the devil's advocate for a moment and point out one thing that could apparently go wrong: ultraproducts (see Sections IV.6 and V.2 in Burris and Sankappanavar). Let $(A_i)_iin I$ be a family of algebras, indexed by the set $I$, and let $U$ be an ultrafilter on $I$. Then the ultraproduct is defined to be the quotient of the product $prod_iin I A_i$ by the congruence $theta_U$, defined by $((a_i), (b_i))in theta_U$ if and only if $iin Imid a_i = b_iin U$.
Now we'd really like to have Ã…Âoś's theorem, which says (in a special case) that an identity holds in the ultraproduct if and only if the set of factors on which it holds is in $U$. But observe that if any single $A_i$ is empty, the product of the $A_i$ is empty, so the ultraproduct is empty. Now Ã…Âoś's theorem can fail: for example, suppse $U$ is a non-principal ultrafilter, one $A_i^*$ is empty, and every other $A_i$ has at least $2$ elements. Then the identity $x = y$ holds in the ultraproduct (vacuously), but the set of $iin I$ such that it holds in $A_i$ is the singleton $i^*$, which is not in $U$.
But this just means we're using the wrong definition of ultraproduct! The correct definition is: $$prod_iin I A_i/U = varinjlim_Xin U prod_iin X A_i.$$
Here we look at each set $Xsubseteq I$ in the ultrafilter, and take the $X$-indexed product $prod_iin X A_i$. Whenever $Ysubseteq X$, we have a projection map $pi^X_Y$ from the $X$-indexed product to the $Y$-indexed product. The resulting system of products and connecting maps is directed (since $U$ is a filter), and we take the directed colimit.
This definition gives an ultraproduct which is isomorphic to the old one in the case when all of the $A_i$ are nonempty. But it gives a non-empty ultraproduct when the set of non-empty factors is in $U$, and it allows you to prove Ã…Âoś's theorem in this context.
answered 2 hours ago
Alex Kruckman
23.9k22454
edited 2 hours ago
answered 2 hours ago
Alex Kruckman
23.9k22454
answered 2 hours ago
Alex Kruckman
23.9k22454
answered 2 hours ago
Alex Kruckman
23.9k22454
23.9k22454
So... you're saying that $forall x(xneq x)$ is not provably false? Huh.
– Asaf Karagila♦
1 hour ago
@AsafKaragila That depends on your proof system. And of course you should always choose a proof system which is sound and complete for the semantics you have in mind.
– Alex Kruckman
58 mins ago
Thank you. Ultraproducts are not in my scope (yet), but in this situation that has a positive side: the argumentation of the devil must come from far. Posing the question I was actually hoping for answers that would (mainly) tell me that an empty carrier set does no harm in general.
– Vera
11 mins ago
add a comment |Â
So... you're saying that $forall x(xneq x)$ is not provably false? Huh.
– Asaf Karagila♦
1 hour ago
@AsafKaragila That depends on your proof system. And of course you should always choose a proof system which is sound and complete for the semantics you have in mind.
– Alex Kruckman
58 mins ago
Thank you. Ultraproducts are not in my scope (yet), but in this situation that has a positive side: the argumentation of the devil must come from far. Posing the question I was actually hoping for answers that would (mainly) tell me that an empty carrier set does no harm in general.
– Vera
11 mins ago
So... you're saying that $forall x(xneq x)$ is not provably false? Huh.
– Asaf Karagila♦
1 hour ago
So... you're saying that $forall x(xneq x)$ is not provably false? Huh.
– Asaf Karagila♦
1 hour ago
@AsafKaragila That depends on your proof system. And of course you should always choose a proof system which is sound and complete for the semantics you have in mind.
– Alex Kruckman
58 mins ago
@AsafKaragila That depends on your proof system. And of course you should always choose a proof system which is sound and complete for the semantics you have in mind.
– Alex Kruckman
58 mins ago
Thank you. Ultraproducts are not in my scope (yet), but in this situation that has a positive side: the argumentation of the devil must come from far. Posing the question I was actually hoping for answers that would (mainly) tell me that an empty carrier set does no harm in general.
– Vera
11 mins ago
Thank you. Ultraproducts are not in my scope (yet), but in this situation that has a positive side: the argumentation of the devil must come from far. Posing the question I was actually hoping for answers that would (mainly) tell me that an empty carrier set does no harm in general.
– Vera
11 mins ago
add a comment |Â
up vote
3
down vote
There is nothing wrong with allowing the carrier set of an algebra to be empty. Arguably it is a more natural and modern approach to allow for structures to be empty in general. The Wikipedia definitions of algebra and semigroup, for instance, do not exclude this possibility. It's not uncommon though, especially in older texts, to see a requirement that structures be nonempty, which does create some blemishes in the theory, as you've observed.
answered 2 hours ago
Brent Kerby
4,609414
1
It is a fairly standard convention coming from logic.
– Andrés E. Caicedo
2 hours ago
@AndrésE.Caicedo, Thanks, looks like I underestimated how common the convention has been; I've edited my answer accordingly.
– Brent Kerby
2 hours ago
add a comment |Â
up vote
3
down vote
There is nothing wrong with allowing the carrier set of an algebra to be empty. Arguably it is a more natural and modern approach to allow for structures to be empty in general. The Wikipedia definitions of algebra and semigroup, for instance, do not exclude this possibility. It's not uncommon though, especially in older texts, to see a requirement that structures be nonempty, which does create some blemishes in the theory, as you've observed.
answered 2 hours ago
Brent Kerby
4,609414
1
It is a fairly standard convention coming from logic.
– Andrés E. Caicedo
2 hours ago
@AndrésE.Caicedo, Thanks, looks like I underestimated how common the convention has been; I've edited my answer accordingly.
– Brent Kerby
2 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
There is nothing wrong with allowing the carrier set of an algebra to be empty. Arguably it is a more natural and modern approach to allow for structures to be empty in general. The Wikipedia definitions of algebra and semigroup, for instance, do not exclude this possibility. It's not uncommon though, especially in older texts, to see a requirement that structures be nonempty, which does create some blemishes in the theory, as you've observed.
answered 2 hours ago
Brent Kerby
4,609414
There is nothing wrong with allowing the carrier set of an algebra to be empty. Arguably it is a more natural and modern approach to allow for structures to be empty in general. The Wikipedia definitions of algebra and semigroup, for instance, do not exclude this possibility. It's not uncommon though, especially in older texts, to see a requirement that structures be nonempty, which does create some blemishes in the theory, as you've observed.
answered 2 hours ago
Brent Kerby
4,609414
edited 2 hours ago
answered 2 hours ago
Brent Kerby
4,609414
answered 2 hours ago
Brent Kerby
4,609414
answered 2 hours ago
Brent Kerby
4,609414
4,609414
1
It is a fairly standard convention coming from logic.
– Andrés E. Caicedo
2 hours ago
@AndrésE.Caicedo, Thanks, looks like I underestimated how common the convention has been; I've edited my answer accordingly.
– Brent Kerby
2 hours ago
add a comment |Â
1
It is a fairly standard convention coming from logic.
– Andrés E. Caicedo
2 hours ago
@AndrésE.Caicedo, Thanks, looks like I underestimated how common the convention has been; I've edited my answer accordingly.
– Brent Kerby
2 hours ago
1
1
It is a fairly standard convention coming from logic.
– Andrés E. Caicedo
2 hours ago
It is a fairly standard convention coming from logic.
– Andrés E. Caicedo
2 hours ago
@AndrésE.Caicedo, Thanks, looks like I underestimated how common the convention has been; I've edited my answer accordingly.
– Brent Kerby
2 hours ago
@AndrésE.Caicedo, Thanks, looks like I underestimated how common the convention has been; I've edited my answer accordingly.
– Brent Kerby
2 hours ago
add a comment |Â
up vote
-1
down vote
That the empty set is valid comes from the fact that the laws of associativity and also commutativity are satisfied, such as
$$forall a,bin A: ab=ba.$$
This means that
$$forall aforall b[ ain Awedge bin ARightarrow ab=ba].$$
This whole assertion is true if the premise is false such as in the case $A=emptyset$.
answered 2 hours ago
Wuestenfux
1,048128
Thank you for answering, but I did not have any doubts concerning this topic. So actually this is not really an answer to my question. Sorry (I did not downvote, though).
– Vera
7 mins ago
add a comment |Â
up vote
-1
down vote
That the empty set is valid comes from the fact that the laws of associativity and also commutativity are satisfied, such as
$$forall a,bin A: ab=ba.$$
This means that
$$forall aforall b[ ain Awedge bin ARightarrow ab=ba].$$
This whole assertion is true if the premise is false such as in the case $A=emptyset$.
answered 2 hours ago
Wuestenfux
1,048128
Thank you for answering, but I did not have any doubts concerning this topic. So actually this is not really an answer to my question. Sorry (I did not downvote, though).
– Vera
7 mins ago
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
That the empty set is valid comes from the fact that the laws of associativity and also commutativity are satisfied, such as
$$forall a,bin A: ab=ba.$$
This means that
$$forall aforall b[ ain Awedge bin ARightarrow ab=ba].$$
This whole assertion is true if the premise is false such as in the case $A=emptyset$.
answered 2 hours ago
Wuestenfux
1,048128
That the empty set is valid comes from the fact that the laws of associativity and also commutativity are satisfied, such as
$$forall a,bin A: ab=ba.$$
This means that
$$forall aforall b[ ain Awedge bin ARightarrow ab=ba].$$
This whole assertion is true if the premise is false such as in the case $A=emptyset$.
answered 2 hours ago
Wuestenfux
1,048128
answered 2 hours ago
Wuestenfux
1,048128
answered 2 hours ago
Wuestenfux
1,048128
answered 2 hours ago
Wuestenfux
1,048128
1,048128
Thank you for answering, but I did not have any doubts concerning this topic. So actually this is not really an answer to my question. Sorry (I did not downvote, though).
– Vera
7 mins ago
add a comment |Â
Thank you for answering, but I did not have any doubts concerning this topic. So actually this is not really an answer to my question. Sorry (I did not downvote, though).
– Vera
7 mins ago
Thank you for answering, but I did not have any doubts concerning this topic. So actually this is not really an answer to my question. Sorry (I did not downvote, though).
– Vera
7 mins ago
Thank you for answering, but I did not have any doubts concerning this topic. So actually this is not really an answer to my question. Sorry (I did not downvote, though).
– Vera
7 mins ago
add a comment |Â
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2
I suspect that the requirement that algebras are non-empty comes directly from the common assumption in model theory and mathematical logic that all structures are nonempty. (I added the (logic) and (model-theory) tags for this reason.) Of course, it's not really possible for a convention or a definition to be "right" or "wrong" - but some conventions are better than others, and I feel strongly that empty structures should always be allowed, both in universal algebra and in model theory.
– Alex Kruckman
3 hours ago
I don't think anything is wrong. It's just pointless. After all in Def 1.1 we define $A^n$ as all $n$-tuples. Well, that would define $emptyset^n = emptyset$ and there'd only be one vacuus operation. You can't have any groups. You could have a vacous semi-group but its pointless.
– fleablood
2 hours ago
1
@fleablood It's not pointless to want a category of interest to have an initial object.
– Alex Kruckman
2 hours ago
Um... have an initial object is entirely different that having a terminal object. And ending at the empty set is just easy trivial and vaccuus albeit perfectly fine. Just pointless.
– fleablood
1 hour ago
@fleablood I don't understand your comment at all. In a category of algebras, the initial object is the free algebra generated by the constants (which is empty when there are no constants). The terminal object is the trivial algebra with exactly one element. The empty set can't be terminal, because there are no maps from non-empty sets to the empty set. .....Ohhhhh wait, maybe "pointless" is a pun on "empty", and you're trolling me?
– Alex Kruckman
1 hour ago