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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?"

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    $\begingroup$ 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. $\endgroup$ Sep 22, 2018 at 15:14
  • $\begingroup$ 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. $\endgroup$
    – fleablood
    Sep 22, 2018 at 15:27
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    $\begingroup$ @fleablood It's not pointless to want a category of interest to have an initial object. $\endgroup$ Sep 22, 2018 at 15:50
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    $\begingroup$ @fleablood Look, I'm certainly not saying that the empty semigroup is an interesting structure. But it fits into the entire variety of semigroups in an interesting way, namely as the initial object in the category. And in universal algebra, the main objects of study are entire varieties of algebras, rather than any particular algebra. By analogy: I'm sure you would agree with me that the trivial group $\{e\}$ is a similarly boring structure. Would you be OK with eliminating it from group theory by making the definition that all groups must have at least $2$ elements? $\endgroup$ Sep 22, 2018 at 17:04
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    $\begingroup$ @AlexKruckman To support your point, it's not just the initial object that goes away -- you also lose equalizers of morphisms with disjoint images, and have to check that pullbacks are nonempty before declaring they exist. $\endgroup$ Sep 22, 2018 at 22:02

3 Answers 3

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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 $(A_i)_{i\in I}$ 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 $\prod_{i\in I} A_i$ by the congruence $\theta_U$, defined by $((a_i), (b_i))\in \theta_U$ if and only if $\{i\in I\mid a_i = b_i\}\in U$.

Now we'd really like to have Łoś'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 Łoś'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 $i\in I$ 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: $$\prod_{i\in I} A_i/U = \varinjlim_{X\in U} \prod_{i\in X} A_i.$$ Here we look at each set $X\subseteq I$ in the ultrafilter, and take the $X$-indexed product $\prod_{i\in X} A_i$. Whenever $Y\subseteq X$, 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 Łoś's theorem in this context.

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  • $\begingroup$ So... you're saying that $\forall x(x\neq x)$ is not provably false? Huh. $\endgroup$
    – Asaf Karagila
    Sep 22, 2018 at 17:22
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    $\begingroup$ @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. $\endgroup$ Sep 22, 2018 at 17:25
  • $\begingroup$ 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. $\endgroup$
    – Vera
    Sep 22, 2018 at 18:12
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    $\begingroup$ Another reason some people might favor the "nonempty universe" convention is that they would like $(\exists x)\,P(x)$ to be a logical consequence of $(\forall x)\,P(x)$. I don't find that very convincing because the corresponding principle for restricted quantifiers can fail (when you're restricting to a set that happens to be empty). $\endgroup$ Sep 22, 2018 at 20:23
  • $\begingroup$ @AndreasBlass In Foundation of Mathematics (Kenneth Kunen) I encountered that (page 103, Remark II.8.16). It is explained there in what way non-empty universes can be an obstacle in model-theory. $\endgroup$
    – Vera
    Oct 25, 2018 at 8:34
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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.

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    $\begingroup$ It is a fairly standard convention coming from logic. $\endgroup$ Sep 22, 2018 at 15:30
  • $\begingroup$ @AndrésE.Caicedo, Thanks, looks like I underestimated how common the convention has been; I've edited my answer accordingly. $\endgroup$ Sep 22, 2018 at 16:12
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That the empty set is valid comes from the fact that the laws of associativity and also commutativity are satisfied, such as $$\forall a,b\in A: ab=ba.$$ This means that $$\forall a\forall b[ a\in A\wedge b\in A\Rightarrow ab=ba].$$ This whole assertion is true if the premise is false such as in the case $A=\emptyset$.

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    $\begingroup$ 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). $\endgroup$
    – Vera
    Sep 22, 2018 at 18:16

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