# Interesting examples of vacuous / void entities

I included this footnote in a paper in which I mentioned that the number of partitions of the empty set is 1 (every member of any partition is a non-empty set, and of course every member of the empty set is a non-empty set):

"Perhaps as a result of studying set theory, I was surprised when I learned that some respectable combinatorialists consider such things as this to be mere convention. One of them even said a case could be made for setting the number of partitions to 0 when $$n=0$$. By stark contrast, Gian-Carlo Rota wrote in \cite{Rota2}, p.~15, that 'the kind of mathematical reasoning that physicists find unbearably pedantic' leads not only to the conclusion that the elementary symmetric function in no variables is 1, but straight from there to the theory of the Euler characteristic, so that 'such reasoning does pay off.' The only other really sexy example I know is from applied statistics: the non-central chi-square distribution with zero degrees of freedom, unlike its 'central' counterpart, is non-trivial."

The cited paper was: G-C.~Rota, Geometric Probability, Mathematical Intelligencer, 20 (4), 1998, pp. 11--16. The paper in which my footnote appears is the first one you see here, doi: 10.37236/1027.

Question: What other really gaudy examples are there?

Some remarks:

• From one point of view, the whole concept of vacuous truth is silly. It is a counterintuitive but true proposition that Minneapolis is at a higher latitude than Toronto. "Ex falso quodlibet" (or whatever the Latin phrase is) and so if you believe Toronto is a more northerly locale than Minneapolis, it will lead you into all sorts of mistakes like $$2 + 2 = 5,$$ etc. But that is nonsense.

• From another point of view, in its proper mathematical context, it makes perfect sense.

• People use examples like propositions about all volcanoes made of pure gold, etc. That's bad pedagogy and bad in other ways. What if I ask whether all cell phones in the classroom have been shut off? If there are no cell phones in the room (that is more realistic than volcanoes made of gold, isn't it??) then the correct answer is "yes". That's a good example, showing, if only in a small way, the utility of the concept when used properly.

• I don't think it's mere convention that the number of partitions of the empty set is 1; it follows logically from some basic things in logic. Those don't make sense in some contexts (see "Minneapolis", "Toronto", etc., above) but in fact the only truth value that can be assigned to $$\text{“}F\Longrightarrow F\text{''}$$ or $$\text{“}F\Longrightarrow T\text{''}$$ that makes it possible to fill in the truth table without knowing the content of the false proposition (and satisfies the other desiderata?) is $$T.$$ That's a fact whose truth doesn't depend on conventions.

• I agree that any standard definition of "partition" will give the empty set one partition. I suspect that the reason some people think of this conclusion as a mere convention is that they are reminded of some superficially similar situations. Is $1$ a prime number? Is $R$ a prime ideal in $R$? Is the empty space connected? Is a trivial module irreducible? Life is easier if you say "no" to all of these, even though this at first seems to call for a special convention. Nov 13, 2010 at 20:06
• @some guy: “the quotient hasn’t any zero-divisors”: true, but nonetheless, it isn't an integral domain! …at least if you define that as a ring where “any product of non-zero elements is non-zero”, since in the trivial ring, the empty product of non-zero elements is zero :-) (I guess most people enjoying this question will agree that this is the “right” definition of integral domain, not the more common version which just considers binary products, but I think plenty of mathematicians might disagree.) Nov 14, 2010 at 0:18
• @Michael Hardy: am I the only person who doesn't actually understand what the question being asked is? Nov 14, 2010 at 0:24
• @Qiaochu Yuan: Maybe you are the only one. But I can't be sure of that. The question asks for other examples of non-trivial and interesting mathematics arising out of seemingly trivial instances of vacuity. Nov 14, 2010 at 1:31
• Oh, let's start flaming about $0^0$ and the degree of the constant zero polynomial while we're here. Nov 14, 2010 at 11:23

How many open covers does the empty topological space have? Not one, not none, but two: the empty cover $$\varnothing$$, since its union is $$\bigcup\varnothing=\varnothing$$, and the cover $$\{\varnothing\}$$, since its union is also $$\bigcup\{\varnothing\} =\varnothing$$.

This comes up when using the Grothendieck plus-construction to sheafify a presheaf. Apply the construction to the (nonseparated) presheaf $$P:\mathcal{O}(X)^\mathrm{op}\to \mathrm{Set}$$ sending every open set to the set $$A$$, with $$|A|\geq 2$$. Then the presheaf $$P^+:\mathcal{O}(X)^\mathrm{op}\to\mathrm{Set}$$ agrees with $$P$$ on every open set except $$\varnothing\subseteq X$$, where $$P^+(\varnothing)$$ is now a one-element set $$\{*\}.$$ This is because the matching families for the cover $$\{\varnothing\}$$ of $$\varnothing$$ (of which there is one for each $$a\in A$$) are all set equal to the unique matching family for the refining cover $$\varnothing\subseteq\{\varnothing\}$$ of $$\varnothing$$.

This elementary example comes from "Sheaves in Geometry and Logic", by Moerdijk and MacLane.

• Here I'm using the notation that if $S$ is a set (in particular, of sets) then $\bigcup S=\bigcup_{A\in S}A$. Nov 15, 2010 at 15:41
• I laughed out loud at this one! At least continuous functions patch easily over both open covers :) Nov 15, 2010 at 21:45
• That is bizarre. Sep 23, 2012 at 6:16
• Well, yeah! Union being an operation $P(P(S)) \to P(S)$, i.e. a function from a 2-element set to a 1-element set if $S$ is empty. Jul 20, 2013 at 19:15
• @LSpice Good point. Likewise, if we allow $\varnothing$ as an element of a partition, then we can say "if $\mathcal P$ and $\mathcal Q$ are partitions of $E$, then so is $\{P\cap Q:P\in\mathcal P,\ Q\in\mathcal Q\}$" instead of having to write $\{P\cap Q:P\in\mathcal P,\ Q\in\mathcal Q,\ P\cap Q\ne\varnothing\}$. Then each equivalence relation would correspond to not one but two partitions. But we can fix that by requiring $\varnothing$ to be an element of every partition.
– bof
May 28, 2021 at 23:50

There is a big difference between statements such as, one the one hand "the empty sum is zero" or "0!=1" and on the other hand "1 is not a prime number". In my opinion, the latter does involve a convention (i.e., a choice) but the former does not.

The first definition of a prime that comes to mind (and came historically, I guess) is "a natural number with no divisors except 1 and itself". This is a perfectly reasonable notion, but it leads to unpleasant contortions when one tries to state the prime decomposition theorem, including uniqueness. A similar phenomenon explains why an irreducible space is nonempty by definition. In these cases, the definition has been tailored to the need of getting cleaner statements. The question "is the empty space connected?" falls into the same category; I find it strange that the more common convention (which is yes) does not match the other two.

In the case of the empty sum, 0 is the only conceivable value, the other choice being "undefined": a mathematician hostile to the empty set might define finite (nonempty) sums by induction, starting from the one-term case and leaving the empty case meaningless. This would not lead to contradictions, only to lots of traps in proofs because whenever you take the sum of some finite set of numbers you first have to check that it is not empty, or treat the empty case separately.

And of course, if you run the inductive definition "backwards" from 1 term to 0 term you immediately find the right value for the empty sum. This is an efficient way to convince students.

• I once heard it asserted that Euclid and his coevals did not consider 1 a prime number because they did not consider 1 a number. That seems consistent with our own colloquial usage when we say "A number of people commented on this idea." Nov 14, 2010 at 17:36
• @Micheal, in classical greek neither one nor two are properly numbers. Nov 15, 2010 at 1:33
• @Laurent: From my years in Rennes, I remember a coffee-break conversation (back when the room was still on the 4th floor) where people were discussing whether it made sense to talk about the action of the null group on the empty set. (I don't think you were involved in that discussion, but I'm pretty sure Antoine Ducros was.) It stuck in my mind, I could never quite decide if it was very silly or too deep for me. Nov 15, 2010 at 16:25
• @Mariano. So the greeks instinctively knew that 2 is a very odd prime. Nov 15, 2010 at 21:36
• On the other hand, the closely related notion of prime ideal can be very naturally formulated in such a way that the analog of "1 is not prime" comes out as a consequence. E. g. you define a prime ideal of a bounded lattice $L$ to be a subset whose characteristic function $L\to\{\text{false},\text{true}\}$ is a $\textit{bounded}$ lattice homomorphism. Now bounded means in particular that the top of $L$ should not belong to the ideal and the bottom should belong to it, so both "improper" primes $L$ and $\varnothing$ are naturally excluded. Then the only "unnatural" assumption is true$\ne$false Jul 28, 2017 at 6:00

Over the reals, $\sup \emptyset = -\infty$ and $\inf \emptyset = \infty$.

• This is another one which comes from universal properties. One can regard a poset as a category where a \le b iff there is an arrow from a to b. Then the supremum is the colimit and the infimum is the limit. The empty supremum is the empty colimit, which is the initial object (if it exists), and the empty infimum is the empty limit, which is the final object (if it exists). And I guess you mean "extended reals." Nov 15, 2010 at 21:10
• If one defines the distance between two points in a (suitable) space as the length of the shortest path, then this one implies the distance is infinite if there's no path. So that's one way this one is useful. Nov 15, 2010 at 21:30
• This is an example I would have given. In particular, in the extended reals, inf A is only less than or equal sup A if A is nonempty. The same logic explains why the zero polynomial has degree $-\infty$ and why the empty set has dimension $-\infty$. Nov 15, 2010 at 22:03
• Scott, I think you are talking about the empty (n-1)-manifold there, not the empty set :) Nov 16, 2010 at 8:59
• @AkivaWeinberger, I think that that would be a bad convention, since the maximum and minimum of a set, if they exist, should belong to it (whereas there is no such expectation for the supremum and infimum). May 7, 2018 at 17:07

An elementary example, but pedagogically nice: a standard early induction proof example is that you can tile any $2^n \times 2^n$ square with one unit square removed, using L-shaped tiles of three unit squares each.

Surprisingly (to me), many textbooks take the base case as $n=2$. The better ones use $n=1$. But the version in The Book, though, surely starts at $n = 0$!

(Of course, I understand the pedagogy of not starting at 0: it’s usually best to make one point at a time. Trying to use this single example to teach about both induction and vacuity simultaneously would end up confusing most students. But when it’s not needed for the former, it does work nicely for the latter, I think!)

• When inductively calculating homology of spheres I like to take the ${-1}$-sphere (empty space) as base case. Nov 14, 2010 at 1:03
• Better take the unreduced suspension then! You do that to students? Nov 15, 2010 at 8:54
• You wrote "The better ones use $n=1$. But the version in The Book surely starts at $n=0!$. But isn't $0!=1$? And yes, that totally on-topic pun is intended! Nov 15, 2010 at 13:51
• Besides the $(-1)$-sphere, continued fractions are better defined from $n=-1$: $p_n/q_n=[a_0,a_1,\dots,a_n]$ so that $p_0=a_0$, $q_0=1$, but also $p_{-1}=1$, $q_{-1}=0$. Dec 29, 2010 at 14:33
• @StevenStadnicki: note that $n=0$ means a $1 \times 1$ square, not $0 \times 0$. So the first induction step goes from the $1 \times 1$ to $2 \times 2$, and does indeed work uniformly with all the rest: put an L down in the “center”; now each of the $1 \times 1$ subsquares has one square removed, so the case $n = 0$ tells us how to tile those. May 8, 2018 at 2:15

Counting is a special case I think: the number of ways of doing nothing is always 1, because you do exactly that, nothing. The number of ways of doing something impossible is 0, because you can't do it. That's why we have: $$\binom{n}{0}=1 \quad \text{but} \quad \binom{n}{n+1}=0.$$ So I don't think your partition example or the cell phone example are really about vacuous truth the same way the Minneapolis example is. Though if pressed I'm not sure how I would formulate precisely how to make the distinction.

• A zillion times I've had undergraduates ask me why 0! is 1, and that's not hard to answer by saying if you divide 4! by 4 you get 3!, and if you divide that by 3 you get 2!, and keep going and see where the pattern leads. But When they ask why $\binom{6}{0} = 1$, there's a bad way to answer that and a good way. The bad way talks about the empty set, etc. Here's the good way. Toss a coin 6 times; in how many ways can you get "heads" 3 times? The list of 20 includes HHHttt, HHtHtt, HHttHt, etc. Then do the same for 2 out of 6. Then 1 out of 6. Then 0 out of 6. Simple. Nov 13, 2010 at 20:42
• @MichaelHardy why is the toin explanations better than the empty set explanation? As for me, they are equivalent. Nov 23, 2021 at 5:11
• I don't know what you mean by "toin"? But how do you get students to understand why such a thing as the empty set should ever be contemplated? Students "know" that in mathematics one accepts absurdities as infallible dogmas (and it's the fault of mathematicians that they think that) and so they accept the empty set because they're told to. But what if you have a student who actually wants to understand why such a thing as an empty set is worth considering? Nov 23, 2021 at 16:59

(1) The value of any sheaf on the empty set is the terminal object. (Consider the gluing condition for the empty open cover of the empty set.)

(2) If A→B is a morphism of sets, then we can define the factor set B/A. We have B/∅=B⊔*, where * is a one-element set. (Consider the left adjoint of the forgetful functor from the category of pointed sets to the category of morphisms of sets.)

(3) Sometimes the norm of a morphism of normed spaces f: X→Y is defined as sup_{x∈X: x≠0} ‖f(x)‖/‖x‖ or as sup_{x∈X: ‖x‖=1} ‖f(x)‖. This does not work for X=0. The correct definition is ‖f‖=sup_{x∈X: ‖x‖≤1} ‖f(x)‖. It also works for seminorms.

(4) The zero ring is the terminal object in the category of unital rings. It is not an integral domain, nor a local ring or a field.

(5) The empty manifold is not connected. Its number of connected components is 0. (Think of the following theorem: Every manifold is the coproduct of a unique family of connected manifolds. The cardinality of the family equals the number of connected components.)

(6) Examples in elementary mathematics abound. The zero vector space has an empty basis and a unique endomorphism A. The matrix of A in the unique basis is empty and the determinant of A is 1. There is exactly one function from the empty set to any other set (the empty function). Zero is a natural number, 0^0=1, the sum of the empty family of numbers is 0, the product of the empty family of numbers is 1, the product or the coproduct of an empty family of objects in a category is the terminal or the initial object of this category, the monoidal product of the empty family of objects in a monoidal category is the monoidal unit.

• I’ve long adored the fact that the determinant of the unique 0×0 matrix is one — even though it’s a zero matrix! Nov 14, 2010 at 0:24
• I'd say $0^0 =1$, not $0$ since it's an empty product. And the series expansion $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$ doesn't work when $z=0$ unless $0^0=1$ (consider the first term). (But that's not the whole story, since $0^0$ is an indeterminate form since $g(x)^{f(x)}$ can approach any nonnegative number as $f(x)$ and $g(x)$ approach 0, depending on what $f$ and $g$ are. However, if both are analytic, then the limit is 1. And in order to make $g(x)^{f(x)}$ approach anything but 1, the point $(g(x),f(x)$ must approach 0 along a curve with the $g$ or $f$ axis as a tangent line.) Nov 14, 2010 at 1:48
• @Michael: Oops, I meant 0^0=1, of course. This is a typo. Nov 14, 2010 at 12:07
• I once ran into the issue mentioned in (3) when writing some foundational material about maps between modules for $p$-adic Banach algebras, but I came up with a different fix: because all these sets of norms were clearly living in the non-negative reals, I decreed that all sups should be taken in the set of non-negative reals. This fixed all problems at a stroke because the sup of the empty set was suddenly zero which is the answer one wants. Nov 15, 2010 at 8:00
• If one wants exponentiation to be continuous (as an analyst might) then $0^0$ has to be indeterminate. But I've never been motivated to make $0^x$ continuous at 0, because it becomes undefined just to the left of 0 (i.e., $x<0$) anyway. So I prefer to say $0^0=1$ and tolerate a little discontinuity. Nov 15, 2010 at 13:45

If you've ever written code to convert an integer into a string of decimal digits, you may have come to the conclusion that the integer 0 should map not to the string 0, but to the empty string instead. Most algorithms I've seen need to introduce a kludge to make 0 come out right. After all, when we write 0 we are violating the usual rule of "no leading zeros".

A nice, natural recursive expression of the conversion process is

def itoa(n):
if n==0: return ""
return itoa(n/10) + chr(ord('0') + n%10)


which can be thought of as
The string representation of an integer consists of its leading digits (n/10) followed by its last digit (n%10).

Trying to fix this by returning "0" instead of "" would result in everything getting a superfluous leading zero.

On the other hand writing 0 as the empty string would be rather annoying.

• def itoa(n): (n>9 ? itoa(n/10) : "") + chr(ord('0') + n%10) Nov 16, 2011 at 14:11
• Ancients would write 0 as an empty string. Aug 16, 2015 at 23:41
• In fact "" is one symbol more than needed compared to 0 :D On top of that, this second symbol is a duplication of the first, which makes it even more superfluous May 10, 2016 at 9:14
• In the same way, the 0 polynomial should really be of degree -1. May 7, 2018 at 22:02
• Isn't the 0 polynomial of degree $-\infty$? May 6, 2020 at 8:06

The usual axiomatizations of set theory (without urelements) mean that every set in the entire set-theoretic universe is ultimately built from copies of the emptyset, in complex empty-box-in-a-box-in-a-box constructions.

• Amazing. I gave this answer (in terms of ordinals) four hours before and ot one vote only ! Nov 14, 2010 at 8:10
• Oh, sorry, I had't noticed the second part of your answer! Since the observation applies in ZFC not only to ordinals, but to all sets, I'll leave this answer up. Nov 14, 2010 at 11:23
• In a similar vein, Conway's theory of Games (which include the Surreal Numbers) gets quite a lot of mileage out of considering the "empty game" in which neither player has any move at all. Every game can be considered as a built from it, just as in Set theory. Nov 15, 2010 at 8:43

What about the two orientations of a point? $$*$$

(Pro trivialogia). I'd like to add some general remarks about the question you raised in the comment below, which seems to me a question of general interest. I see at least three general good reasons why it is worth dealing with trivial cases of mathematical notions.

• Sometimes we simply do not know whether $x$ is a trivial object. Even if our main interest is in non-trivial cases, in the course of a proof or a computation we deal with unknown objects $x,\, y\dots,$ that may possibly degenerate. Therefore, we would like theorems, methods, rules, to hold with the minimum of assumptions, avoiding special separate treatments for degenerate cases (think to some classifications into unnecessary special cases, for algebraic equations, used at the beginning of algebra).

• Abstraction. It is a great feature of modern mathematics the ability of translating a complicated notion belonging to a simple setting, into a simple notion belonging to a possibly more complicated setting (in many abstract contexts the cost of this operation is zero). Example: a limit or a colimit in a category, and in fact any universal construction, is just a zero-object in a suitable category (as an application, Freyd's theorem about existence of adjoint functors, &c.)

• Constructions and proofs by induction. As soon as the induction step from $n$ to $n+1$ is suitably clarified, the validity of the general fact is reduced to that of the trivial case, that becomes the heart of the whole story. So for instance, the very reason of some facts about spheres is some (trivial, but important) fact about $\mathbb{S}^0$. This is of course also the case of constructions and operations with orientations.

• Talking about a point $p$ as a connected real manifold $M=${p} of dimension $0$. It has two canonical orientations, $+1$ and $-1$, the generators of the top-degree exterior power $\Lambda^{top}(T_p M)=\mathbb{R}^1$. It seems to me another instance of a trivial case of a mathematical notion, as interesting and useful as the other ones (and slightly paradoxical as well). Nov 14, 2010 at 7:07
• And this is essential when considering 1-dimensional framed bordism. Nov 15, 2010 at 4:35
• I don't know what a 1-dimensional framed bordism is, but I'm glad someone mentioned that there's a (nontrivial?) reason to think about the concept. Nov 15, 2010 at 4:48
• I wasn't convinced about orienting points -- as $GL(0)/SL(0)$ has one component, not two -- until I hit the statement "If $A = B\oplus C$, orienting any two uniquely determines an orientation on the third." Nov 10, 2011 at 4:44
• I had similar thoughts to Allen, but I resolved them like this. More natural than SL(n) is "BSL(n)", the homotopy fiber of the determinant map BGL(n) --> BGL(1). For n>0 this is connected and indeed is homotopy equivalent with BSL(n), but for n=0 it's homotopically two points. Feb 21, 2019 at 21:38

$\bigcap \emptyset = V$

Unfortunately, I have read more than one philosophical comment on the "set theoretic depth" of this logical triviality.

• In some topology book I read that it's redundant to say the whole space is an open set after you've said the intersection of any finite set of open sets is open. Nov 13, 2010 at 20:45
• Formally there's nothing wrong. The way $\bigcap$ is usually defined (in the language of set theory) is $(\forall x) x \in \bigcap A \Leftrightarrow ((\forall y) y \in A \Rightarrow x \in y)$ And, of course vacuously $(\forall x)(\forall y) y \in \emptyset \Rightarrow x \in y$. Nov 14, 2010 at 16:28
• @Harry: the basic POV of traditional set-theory based foundations considers all sets as once-and-for-all embedded in the class $V$. So then we really can consider intersections among them, and we do get $\bigcap \empty = V$. We're taking the limit of $\empty$ not in the category $\mathbf{Sets}$ (which gives $*$, as you say), but in partially-ordered class of subsets of $V$ (or equivalently, in $V$ itself ordered by $\subseteq$). Several authors in Algebraic Set Theory have considered the connection between these two structures. Nov 15, 2010 at 1:10
• Harry, for a moment ignore my ZFC "shorthand" of writing $\bigcap$ and $V$ (which I blatantly used to make it more, well, sexy). Instead, just look at the last formula in my previous comment. For any set theory (first order logic, signature $\{ \in \}$ etc) which proves an empty set exists, this theorem is (vacuously) true. That's all I'm saying (and as I said, I do not believe it is deep, it's just classical logic). Feel free to drop by my office (4826) if you have further questions. Nov 15, 2010 at 16:37
• To return to Michael's first comment, the non-philosophical (and useful) statement is that the intersection of an empty family of subsets of a set X is X itself. This is why we define X to be open in X for a topological space X, because it is a finite intersection. Nov 15, 2010 at 21:51

I regard "negative thinking" in category theory as an example of cool vacuity: see e.g. https://ncatlab.org/nlab/show/negative+thinking. As category theory is not set theory, such vacuity does not necessarily involve the empty set directly, but the same principle of backwards generalization is used.

The fact that a set is uniquely determined by its elements (i.e., has no additional structure beyond the equality relation between its elements) is summarized by saying that a (-1)-category is a truth value: a morphism between two elements in a set is either true (the elements are the same) or false (they are not). So the morphisms in a 0-category (a set) are (-1)-categories (either true or false), just as the morphisms in a (1-)category are 0-categories (sets). This admits generalizations to situations where "truth" is a more subtle concept (e.g. parameter dependent).

The terminal object of a category is the product over the empty set of objects.

• That's exactly the answer I would have given :-P Nov 13, 2010 at 22:04
• Indeed, probably every answer here is this one in disguise. Aug 16, 2015 at 23:49

The determinant of the $0\times 0$ matrix is $1$. First of all, it's the only way to make determinant of a direct sum of matrices be a product of determinants. Second, it is the sum of $0!$ terms each of which is a product of $0$ factors.

• Third, with this convention, Dogson's condensation formula for the determinant of a $2\times 2$ matrix boils down to the usual formula. Feb 10, 2014 at 15:52
• I guess I have to add that the sign of the permutation of $0$ elements is $1$, because the number of inversions is even. Feb 10, 2014 at 17:54
• @LevBorisov, also because it's the identity element of $S_0$, and the sign map is a homomorphism. Feb 15, 2017 at 15:54
• It's also the (empty) product of the eigenvalues of the $0 \times 0$ matrix. Apr 15, 2018 at 18:37
• And the volume of the 0-dimensional space consisting of a single point. :) May 6, 2020 at 8:12

'Silly' but I like them anyways:

$$\prod_{y\in\emptyset}\left(\sum_{x\in \emptyset}x\right)=1,$$

and

$$\sum_{y\in\emptyset}\left(\prod_{x\in \emptyset}x\right)=0.$$

When defining a topology.

Do not say

the intersection of any two open sets is open

any finite intersection of open sets is open

That way, you assert in particular that the empty intersection of open sets (i.e. the whole space) is open.

Of course we also postulate

an arbitrary union of open sets is open

telling us that the empty union (i.e. $$\varnothing$$) is open.

similar
Caratheodory's "Method I" for constructing an outer measure $$m^*$$ in a set $$X$$ starting from a set-function $$E : \mathcal E \to [0,+\infty]$$ is done like this:

$$m^*(A) = \inf\sum_{i=1}^\infty E(A_i)$$ where the inf is over all sequences $$(A_i)_{i=1}^\infty \subseteq \mathcal E$$ such that $$\bigcup_{i=1}^\infty A_i \supseteq A$$.

But then we have to include artificial hypotheses like "$$\mathcal E$$ covers the whole space" and "$$\varnothing \in \mathcal E$$ and $$E(\varnothing) = 0$$".
But instead we should do it like this

$$m^*(A) = \inf\sum_{F \in \mathcal F}E(F)$$ where the inf is over all countable collections $$\mathcal F \subseteq \mathcal E$$ such that $$\bigcup_{F \in \mathcal F} F \supseteq A$$

Then: we need not postulate that $$\mathcal E$$ covers $$X$$, since if there is no countable $$\mathcal F$$ with $$\bigcup_{F \in \mathcal F} F \supseteq A$$, then we simply get $$m^*(A) = \inf\varnothing = +\infty$$. And for the empty set, we have (possibly among others) the empty cover for $$\varnothing$$, so that $$m^*(\varnothing) \le 0$$, the empty sum. So we get $$m^*(\varnothing) = 0$$ even if (for example) $$E$$ is identically $$+\infty$$.

• Both Carathéodory-type definitions agree on the meaning of $m^*(A)$ if $\mathcal E$ does not contain a countable cover of $A$. May 7, 2018 at 17:36
• The empty cover is not countable though Oct 29, 2021 at 15:37
• @JohannesHahn ... I consider $\varnothing$ to be a countable set (not uncountable). Oct 29, 2021 at 15:42

Zero is a limit ordinal, because it is the union of its elements.

Transfinite induction has two canonical statements. The "strong" statement, $$(\forall \alpha)((\forall \beta)((\beta<\alpha) \rightarrow P(\beta)) \rightarrow P(\alpha))\rightarrow (\forall \alpha)P(\alpha),$$ doesn't split anything into cases. The version used most frequently in proofs says that any property preserved under unions and successors holds for all ordinals. Zero should rarely be a special case.

Also, "limit ordinals" should totally be called "colimit ordinals". The term "limit ordinal" refers to limit points in the order topology, thus excluding zero, but this is silly.

• Category theory does not have a monopoly on the word "limit". Nov 19, 2010 at 23:46
• True :) I just think things should have names and definitions that reflect the way they're most often used. Dec 3, 2010 at 11:15

The Generalized Continuum Hypothesis is the assertion that $$2^\kappa=\kappa^+$$ for all infinite cardinals $$\kappa$$, or in other words that the power set of a set of size $$\kappa$$ has the next larger cardinal size above $$\kappa$$.

If we consider all cardinals, rather than only the infinite cardinals, then the two provable instances of this equation occur in the following vacuous and near-vacuous facts:

• The power set of a set with $$0$$ members has $$1$$ member.

• The power set of a set with $$1$$ member has $$2$$ members.

All other instances of $$2^\kappa=\kappa^+$$, finite or infinite, are either false or independent of ZFC.

• I heard once that at a Logic Colloquium talk in Berkeley when Foreman or Woodin was speaking on their theorem concerning the consistency that the GCH fails everywhere, the speaker wrote "$2^\kappa\neq\kappa^+$ for all $\kappa$" on the blackboard. Leo Harrington reportedly went silently up to the blackboard, adding the words "except $\kappa=0$ or $1$." Nov 14, 2010 at 0:54

The empty product in a group $$G$$ is the unit of $$G$$. This is the only way to avoid mistake in calculations.

Set theory begins by the construction of finite ordinals. The first one is $$\emptyset$$ and is denoted $$0$$. The next one is $$\{\emptyset\}$$, which is not empty ! It is denoted $$1$$. More generally, every finite ordinal is defined only in terms of the empty set recursively: $$n+1:=n\cup\{ n \}$$. Physicists (or administrators, politicians, whoever is asked to fund mathematics) might find this pedantic, but it is actually powerful.

• The idea of empty products and empty sums seems to make sense only if the operation involved is associative, and then whenever there's an identity element. When asked why the product of no numbers is 1, one can answer that multiplying something by no numbers just leaves it fixed, so it's the same as multiplying it by 1. But you need associativity for that to make sense. Nov 13, 2010 at 22:35
• ....and so that's why it works in all groups. Nov 13, 2010 at 22:35
• This was discussed in a question on math.stackexchange: math.stackexchange.com/questions/6832/… Nov 14, 2010 at 0:14
• Related to this, one can combine the closure, associativity, and identity axioms of a group by requiring that a group be closed under all finite ordered products (including the empty product), together with a unified associativity axiom $\prod_{\alpha \in A} \prod_{\beta \in B_\alpha} g_\beta = \prod_{\beta \in \biguplus_{\alpha \in A} B_\alpha} g_\beta$ (maintaining the ordering of the index sets in the obvious manner), as well as an axiom that a unary product of a group element $g$ is equal to $g$. Dec 26, 2020 at 18:56

Zsbán Ambrus brings up an interesting example in the comments: the degree of the zero polynomial. The first time I was told about this issue I was told that it is largely a matter of convention. Well, maybe. Here is some evidence suggesting that $$\deg 0 = \infty$$:

• The most basic one: the zero polynomial has infinitely many roots in an algebraic closure.

• If one wants the degree of a polynomial to be a valuation, then we must define $$\deg 0 = \infty$$. This is the unique choice consistent with the requirements that $$\deg fg = \deg f + \deg g$$ and $$\deg (f+g) \ge \text{min}(\deg f, \deg g)$$, and it is necessary in order to make the corresponding absolute value nondegenerate.

• One way to say the above geometrically when $$F = \mathbb{C}$$ is that the degree should describe the order of the pole of $$f$$ at infinity on the Riemann sphere. The function $$0$$ decays faster than the reciprocal of any polynomial in the neighborhood of infinity. In fact, the sequence of functions $$x^n$$ converges uniformly to it in a neighborhood of infinity as $$n \to \infty$$.

• Another way to say the above is that, in the natural topology on $$F[[x]]$$, we have $$x^n \to 0$$. This can be appreciated even if you are, for example, a combinatorialist, because it allows you to say natural things about generating functions like $$\frac{1 - x^n}{1 - x} \to \frac{1}{1 - x}$$ as $$n \to \infty$$.

• One can also define the degree of $$f$$ as $$[F[x]/(f(x)) : F]$$, in which case again we find that $$\deg 0 = \infty$$. This is just a fancier version of the first reason.

Edit: As James Borger points out, the middle ideas are mistaken. Corrected, they actually suggest that $$\deg 0 = -\infty$$:

• $$\deg 0 = -\infty$$ is the unique choice consistent with the requirements that $$\deg fg = \deg f + \deg g$$ and $$\deg (f + g) \le \max(\deg f, \deg g)$$. With this definition, $$|f| = 2^{\deg f}$$ is now an absolute value.

• Geometrically, when $$F = \mathbb{C}$$ the function $$0$$ has a zero of infinite order at infinity, hence a pole of order $$-\infty$$.

• The relevant local ring here is really $$F[[ \frac{1}{x} ]]$$, and in the natural topology on this ring we have $$\frac{1}{x^n} \to 0$$.

• Another reason to like this definition is that it gives a uniform statement of the division algorithm on $$F[x]$$.

• Dear Qiaochu, I'm not sure I agree with all this. First, it's not true that $\deg(f+g)\geq \min(\deg f,\deg g)$. Take $f=x^{100}$, $g=1-x^{100}$. It is true that $\deg(f+g)\leq \max(\deg f, \deg g)$. Second, 0 has a zero of infinite order at $x=\infty$, and hence a pole of order $-\infty$ there. Third, it seems beside the point to take convergence in $F[[x]]$, which is the local ring at $x=0$. I think the local ring at infinity $F[[x^{-1}]]$ is more relevant, and then $x^n \to 0$ as $n\to-\infty$. So I'd prefer to set $\deg 0 = -\infty$. Nov 15, 2010 at 10:20
• $-\infty$ crops up here (and in other places such as the dimension of the empty set) because it is the supremum of the empty set: in this case it is the supremum of the set of powers with nonzero coefficients. Nov 15, 2010 at 21:56
• So it's not true that if you differentiate a polynomial then its degree goes down by 1? (I suppose that even if you set deg(0)=-1 that rule is violated, so this isn't such a strong argument.) Nov 20, 2010 at 15:03
• @gowers: that rule is also false in positive characteristic, so I don't think it's universal enough to count as evidence either way. Nov 20, 2010 at 15:46
• @TomGoodwillie: The fight wasn't about differentiating zero itself, but about passing from a non-zero constant to zero. Mar 9, 2017 at 8:17

One is not a prime number, but zero is! It's different than the other prime numbers in $\mathbb{Z}$, though, because it has height zero rather than height one.

Zero is prime in any integral domain. Remember that the trivial ring is not an integral domain.

• I don't find this convincing. Yes, zero generates a prime ideal, which is an element of Spec, etc. But "prime number" has a specific definition that excludes $0$. Feb 9, 2014 at 19:20

I'm always amused (and unpopular for it) when mathematicians claim that some terminology or notation is "true" rather than "convention", as if there is some God of Mathematics out there who hands us definitions that us mere humans are forbidden to tamper with. Actually all mathematical notation is convention.

We could, if we agreed as a body to do so, define "$m+n$" to mean what it used to mean, unless $m=n=1$ in which case it means 3. It isn't forbidden, because we invented "$+$"; it belongs to us and we can choose what it means. We won't do that, though, not because it is wrong according to some objective source of truth, but because it would cause us a whole heap of trouble. For example we would have to rewrite a huge number of theorems to add exceptional cases, and we hate exceptional cases. It isn't a question of "truth" since all of modern mathematics could be correctly stated using the new definition.

In the same way, we choose to set the empty product equal to 1, and the empty sum equal to 0, because those are the conventions that make our symbol manipulations so much easier and simpler than any alternative conventions would make them. It isn't really different, except in degree, to excluding 1 from being a prime. We do that because otherwise we would have to spend all day writing "let $p$ be a prime other than 1". Qualitatively the reason is the same: we like our theorems to be cleaner, with fewer preconditions and fewer subcases. Simplicity and generality is a fundamental aesthetic of mathematics.

• We mathematicians also seem to be hung up on repeatability and logical as well as otber forms of consistency. Why are we so hidebound, I wonder? Feb 10, 2014 at 16:59
• How about grue primes?
– cody
May 10, 2016 at 15:22
• While it's certainly true that definitions cannot be true (or false), I think that it is meaningful to distinguish among different types of conventions. For example, I think that there is a difference between the convention "the empty product equals 1" and the convention "$\mathbb N$ contains $0$" (or the opposite of the latter convention, if that's the one you use). Perhaps "forced by consistency" (or, even more explicitly, "forced by consistency with …") is better than "true"? May 7, 2018 at 17:30
• I don't think it's that we want our theorems to be cleaner. I think it's that the theorems (and the proof ideas) are what we really care about. We want them to be nice. The axioms are largely just a way to support the theorems. We develop axiom systems and theories (in the sense of a body of work) to support the theorems we care about. If the axioms don't give us the theorems we want, then we change the axioms. Of course, one person's axioms are another's theorems. Example: ring theorists have studied "general rings", but some (many?) algebraic geometers (ok, me) stick to rings with $1$... May 13, 2020 at 15:05

Recall that an abstract simplicial complex consists of a family of finite sets $$K\subseteq 2^V$$ such that $$\sigma\in K$$ and $$\tau\subseteq\sigma$$ implies $$\tau\in K$$. (Sometimes it is also assumed that $$\{v\}\in K$$ for all $$v\in V$$.) Elements of $$K$$ are called faces of $$K$$. Dimension of a face is its cardinality minus 1.

According to the above definition, if $$S\not=\emptyset$$, then the empty set $$\emptyset$$ is a face of the complex (of dimension $$-1$$, since it has $$0$$ elements). Applying the usual definition of simplicial homology gives us what is called reduced homology, which is often much better-behaved than the nonreduced one (obtained when we forget about the empty face).

In this context it is important to distinguish between the empty simplicial complex $$K_e=\{\emptyset\}$$ and the void simplicial complex $$K_v=\{\}$$, which may be understood as the $$(-1)$$-sphere and the $$(-1)$$-disk. (In particular $$H_{-1}(K_e)=\mathbb{Z}$$, $$H_n(K_e)=0$$ for all $$n\not=-1$$, and $$H_m(K_v)=0$$ for all $$m$$.)

• I would change the definition to avoid having to view the void complex as a simplicial complex. The $-1$-sphere on the other hand is a perfectly decent space, defined as the points in $\mathbb{R}^0$ at distance 1 from the origin, a sort of boundary of the 0-disk.
– IJL
Nov 22, 2021 at 10:55

The ‘divides’ relation $$\mid$$ should properly be called a ‘has-a-multiple-of’ relation and defined without any reference to division as

$$a \mid b \iff \exists c: ac = b$$

This definition implies $$\forall a: a \mid 0$$ (including $$0 \mid 0$$) and $$\forall a: 0 \mid a \Longrightarrow a = 0$$. And over the non-negative integers in particular, the relation becomes a complete lattice, where the infimum is the GCD and the supremum is the LCM.

This comes up as the inductive base case when computing the GCD/LCM of arbitrary sets of numbers. Sometimes people say that the GCD of no numbers doesn’t exist, but when the GCD is defined as the infimum of the $$\mid$$ relation, it becomes perfectly natural to set it to 0 at the empty set.

Somewhat related, though perhaps too silly to mention: division by zero is undefined. However, modulo by zero is not: it is simply the identity function.

One way to see this is to observe that no matter how integer division by zero $$(a \mapsto a \div 0)$$ is defined, there is only one way to define $$a \bmod 0$$ such that the identity

$$(a \div b) \cdot b + (a \bmod b) = a$$

is maintained. (Of course, if $$a \div 0$$ is defined, it will necessarily break the usual property that $$|a - (a \div b) \cdot b| < |b|$$ for every $$a$$ and $$b$$ where it is defined.)

Another is to notice that $$(a \mapsto a \bmod b) : \mathbb{Z} \to \mathbb{Z}/b\mathbb {Z}$$ is a unity-preserving homomorphism of rings. For $$b = 0$$, the codomain becomes $$\mathbb{Z}/0\mathbb {Z} = \mathbb{Z}$$, and the requirement that $$1 \bmod 0 = 1$$ forces the function to be the identity function.

(Setting $$a \bmod 0 = a$$ also agrees with the property that $$b \mid a$$ if and only if $$a \bmod b = 0$$.)

• Another reason to say that the gcd of the empty set is 0 is that in Euclid's algorithm, the gcd always remains the same from one step to the next ($\gcd(a,b) = \gcd(a,b-a)$) and it stops at $0$ and you then have $\gcd(c,0)=c,$ so that $0$ is an identity element for this binary operation. $\qquad$ May 6, 2020 at 22:25
• I'm not totally sure I understand "GCD is defined as the infimum of the $\mid$ relation", but I think it should be supremum: $\gcd(S) = \sup_\mid \operatorname{CD}_S$, where $\operatorname{CD}_S = \{n \in \mathbb Z_{\ge 0} \mathrel| \forall s \in S,\,n \mid s\}$. Indeed, since $\operatorname{CD}_\emptyset = \mathbb Z_{\ge 0}$, the infimum would be $1$, not $0$. Or maybe you meant to take the infimum of something else? Jul 9, 2020 at 13:50
• $\operatorname{gcd}(S) := \inf_\mid(S)$. Your restatement relies on the fact that the infimum of a set is the supremum of the set of its lower bounds. Jul 9, 2020 at 15:19
• Oh! Somehow that way of thinking had never occurred to me. Jul 9, 2020 at 22:02

Let $\bigotimes_{i \in I} M_i$ denote the tensor product of $R$-modules $M_i$. Then $\bigotimes_{i \in \emptyset} M_i$ is $R$.

(Reason: $\prod_{i \in \emptyset} M_i$ is the terminal object in the category of sets (see the answer of Eivind Dahl), i.e. a point, and multilinear maps on this to $N$ are just elements of $N$, i.e. homomorphisms $R \to N$.)

Another one: I know this is really silly and already contained somehow in the other answers, but anyway:

$$\prod_{i \in \emptyset} 0 = 1$$

• It makes if you consider what it would mean for a function be "0-linear". The empty product is as a module isomorphic to 0, but if we want to consider multilinear functions, we're not considering it as a module; as a set it's $\{()\}$. So we have functions from a one-element set, i.e. just any function, with the requirement that they be "0-linear", i.e. linear in all of the 0 arguments, which is a trivial requirement! So it does work after all. Nov 13, 2010 at 23:10
• This can be motivated either by the fact that tensor products represent multilinear maps, or (as per another answer) the fact that \otimes is an associative, unital operation, so “$k$-ary tensor product” with $k=0$ must give the unit object for $\otimes$. (All up to coherent natural isomorphism, of course!) Nov 14, 2010 at 0:11
• Or, to make it even more vacuous: it is a particular case of the fact that the composition of the zero amount of endofunctors is the identity functor... May 8, 2018 at 4:52

The full rank factorization of the 3x2 zero matrix is the product of a 3x0 matrix times a 0x2 matrix. There exist empty matrices with n>0 rows and 0 column. The 0x0 empty matrix is the only non nonsingular zero matrix.

A The empty set is a covering map of any topological space. More generally, a covering map needn't be surjective (although many books claim just that). For example the inclusion of a closed and open subset of a space is a covering. Strangely, I would argue that this entails that the empty topological space, although connected, is not simply connected.

B Dually, given a field $$K$$, the zero algebra over $$K$$ is diagonal and in particular étale: the morphism of affine schemes $$\varnothing \to \operatorname{Spec}(K)$$ is étale. In the same vein, a nonzero constant polynomial over $$K$$ is separable (its nonexistent roots in an algebraic closure of $$K$$ are certainly distinct) . We may then say without any exception that the $$K$$-algebra $$K[X]/(f(X))$$ is étale iff $$f(X)$$ is a separable polynomial.

• Agreed, except that I strongly prefer not to call the empty space connected. How many components does it have? Nov 14, 2010 at 3:46
• If we assume disjoint union is additive on the number of connected components, then the empty space should have zero connected components. I think this should correspond to the usual distinction between primes (connected spaces) and units (empty spaces) in situations where we have unique factorization. Nov 14, 2010 at 10:22
• I would define a space X to be connected if, whenever it is expressed as a topological disjoint sum (i.e., coproduct) of spaces, one of the summands is X itself. This leads to the conclusion that the empty space is not connected, because t is the topological disjoint sum of zero spaces. (Analogously, I would define a positive integer to be prime if, whenever it is expressed as a product, it equals one of the factors. That makes the empty product, 1, not prime.) Nov 14, 2010 at 22:49
• @Georges: In the category of spaces, there can't be a universal covering. The theory of universal covering is a theory of pointed spaces: A pointed cover (Y,y) of a pointed space (X,x) is a universal covering if, for any pointed cover (Z,z) of (X,x), there is a unique X-morphism $f:(Y,y)\rightarrow (Z,z)$ of pointed covers.
– ACL
Nov 14, 2010 at 23:48
• Dear ACL: yes, I also like the idea of restricting the notion of (universal) covering space to pointed spaces. I vaguely remember reading that our Overlords (Deligne et al.) are quite categorical on this point. Nov 15, 2010 at 12:48

A "convention" floating around in (co)homology theory is that given a space $$X$$ the quotient with the empty set is $$X/\emptyset = X_+ = X \sqcup *.$$ This seems rather arbitrary if one defines $$X/A$$ to be given by coequalizing the inclusion $$A \subseteq X$$ and a constant map $$\smash{A \rightarrow * \underset{a}{\rightarrow} X}$$ to some point $$a \in A$$.

But there is another possible definition: $$X/A$$ is the pushout $$\begin{array}{cc} A & \rightarrow & X\\ \downarrow & & \downarrow\\ * & \rightarrow & X/A \end{array}$$ Setting $$A=\emptyset$$ we can recover $$X/\emptyset = X \sqcup * = X_+$$. In particular this specializes to the counterintuitive identity $$\emptyset / \emptyset = *$$.

• Should "a constant map to some point of $A$" be understood as "a constant map, on $A$, to some point of the image of $A$ in $X$"? Otherwise I have trouble understanding how to co-equalise two such maps. Jul 8, 2021 at 13:32
• Yes this is what I meant. I hope my edit makes this clearer. Thanks for asking! Jul 8, 2021 at 13:36

Let $$p,q$$ be Hilbert space projections. If $$pq+qp$$ is a non-zero projection then, where $$\varphi$$ is the golden ratio, $$\lVert pq\rVert=\frac{1}{\varphi}.$$

It follows very quickly from Walters - Anticommutator Norm Formula for Projection Operators… but is it vacuous? This answer to Is $$p q + q p$$ ever a projection? on MSE seemed to resolve the issue but has an issue.

Pfister's local-global theorem has an interesting "vacuous" instance:

Remember that quadratic forms over a field $$F$$ of characteristic $$\neq 2$$ are diagonalizable, and hence a nondegenerate quadratic form can be represented by a sequence $$\langle a_1, \ldots, a_m \rangle$$ of elements $$a_i \in F^*$$.

One can "add" diagonal quadratic forms by declaring $$\langle a_1, \ldots, a_m \rangle \oplus \langle b_1, \ldots, b_k \rangle :=\langle a_1, \ldots, a_m, b_1, \ldots, b_k \rangle$$ and one can multiply quadratic forms with natural numbers by declaring $$n\cdot\langle a_1, \ldots, a_m \rangle:=\underbrace{\langle a_1, \ldots, a_m \rangle\oplus \ldots \oplus \langle a_1, \ldots, a_m \rangle}_{n \text{ times}}$$

If "$$<$$" is an ordering on $$F$$ (compatible with $$+$$ and $$\cdot$$), then to a quadratic form $$f=\langle a_1, \ldots, a_m\rangle$$ one can assign its signature $$\sigma(f)(<):=\#\{ a_i \mid 0 < a_i\} - \#\{a_i \mid a_i < 0\}$$. That is, the signature is the number of positive minus the number of negative diagonal entries, according to the ordering "$$<$$". One can show that this signature is independent of the choice of diagonalization.

Thus, if we denote by $$X(F)$$ the set of all orderings of $$F$$, from a fixed quadratic form $$f$$ we obtain a well-defined map $$X(F)\to \mathbb{Z}, \ < \,\mapsto \sigma(f)(<)$$.

Pfister's local-global theorem says something about how much the map $$\sigma(f)$$ determines $$f$$:

Let $$f,g$$ be quadratic forms of the same rank over a field $$F$$. If $$\sigma(f)=\sigma(g)$$, then there exists $$\ell \in \mathbb{N}$$ such that $$2^\ell\cdot f \cong 2^\ell \cdot g$$ (i.e. the multiplied forms are isometric). [The theorem says more, but this is the relevant part.]

When $$F$$ is not orderable at all (e.g. if $$F$$ is algebraically closed or of characteristic $$>0$$, or $$F=\mathbb{Q}_p$$), then $$X(F)=\emptyset$$. Therefore there exists only one map $$S\colon X(F) \to \mathbb{Z}$$, i.e. any two quadratic forms $$f,g$$ satisfy $$\sigma(f)=\sigma(g)$$. Thus, for a non-orderable field, any two quadratic forms have isometric multiples: $$2^\ell\cdot f \cong 2^\ell \cdot g$$ for some $$\ell$$.

I learned this from a wonderful expository note by Pete L. Clark.

If we need to define the value of the Euler's function $\varphi$ at infinity, the best choice will be $\varphi(\infty)=2$ This is because $\varphi(n)$ is the number of generating elements in the cyclic group of order $n$, and so if $n\to \infty$, the the cyclic group tends to $\mathbb{Z}$ which has just two generating elements.

$0^0=1$ is another example, which is easy to prove for the natural zero, but it is not true for the real zero!

• If the base is the real or complex zero and the exponent is the natural zero, then $0^0=1$. For example: $\displaystyle e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$. If $z=0$, then the first term in this expansion is $0^0/0!=1$. Feb 10, 2014 at 16:58
• @YemonChoi, I think that "the number of generating elements" means "the number of elements that form singleton generating sets", rather than "the minimum cardinality of a generating set". May 7, 2018 at 17:27
• I think that's an argument for setting the value at zero — as it is the number of generating elements of $\mathbb Z/0\mathbb Z$. Jun 30, 2020 at 12:53
• -1 for “It is not true for real zero.” I’d like to see an argument for this. Real exponentiation is an extension of natural exponentiation, i.e. it ought to have the same value wherever the latter is defined. Jul 10, 2020 at 18:23
• The conflaton of $x \uparrow y := \prod_{k\in(y, 0]\cap\mathbb{Z}} \frac 1 x \cdot \prod_{k\in[0, y)\cap\mathbb{Z}} x$ with $\exp(y\ln(x))$ by the notation $x^y$ is a mistake, and the question of how to define $0^0$ is just one piece of evidence against it: some situations demand one convention, others demand another. Another is that $(-1) \uparrow k$ is perfectly straightforward, but $\exp(k\ln(-1))$ is meaningless. Dec 20, 2020 at 12:48