Sexy vacuity .... - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:08:13Z http://mathoverflow.net/feeds/question/45951 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45951/sexy-vacuity Sexy vacuity .... Michael Hardy 2010-11-13T19:47:02Z 2010-11-19T22:34:53Z <p>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):</p> <p>"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."</p> <p>The cited paper was: G-C.~Rota, Geometric Probability, <em>Mathematical Intelligencer</em>, 20 (4), 1998, pp. 11--16. The paper in which my footnote appears is the first one you see <a href="http://www.combinatorics.org/Volume_13/v13i1toc.html" rel="nofollow">here</a>.</p> <p><b>Question:</b> What other really gaudy examples are there?</p> <p>Some remarks:</p> <ul> <li><p>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.</p></li> <li><p>From another point of view, in its proper mathematical context, it makes perfect sense.</p></li> <li><p>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 <em>is</em> 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.</p></li> <li><p>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 <em>only</em> truth value that can be assigned to "F--->F" or "F--->T" 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.</p></li> </ul> http://mathoverflow.net/questions/45951/sexy-vacuity/45955#45955 Answer by Thierry Zell for Sexy vacuity .... Thierry Zell 2010-11-13T19:59:32Z 2010-11-13T19:59:32Z <p>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.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/45956#45956 Answer by Eivind Dahl for Sexy vacuity .... Eivind Dahl 2010-11-13T20:00:38Z 2010-11-13T20:00:38Z <p>The terminal object of a category is the product over the empty set of objects.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/45961#45961 Answer by Peter Krautzberger for Sexy vacuity .... Peter Krautzberger 2010-11-13T20:18:22Z 2010-11-13T20:18:22Z <p>$\bigcap \emptyset = V$</p> <p>Unfortunately, I have read more than one philosophical comment on the "set theoretic depth" of this logical triviality.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/45967#45967 Answer by Denis Serre for Sexy vacuity .... Denis Serre 2010-11-13T20:43:00Z 2010-11-13T20:43:00Z <p>The empty product in a group $G$ is the unit of $G$. This is the only way to avoid mistake in calculations.</p> <p>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 asled to fund mathematics) might find this pedantic, but it is actually powerful.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/45969#45969 Answer by Dmitri Pavlov for Sexy vacuity .... Dmitri Pavlov 2010-11-13T20:57:16Z 2010-11-14T17:32:22Z <p>(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.)</p> <p>(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.)</p> <p>(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.</p> <p>(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.</p> <p>(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.)</p> <p>(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.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/45986#45986 Answer by Martin Brandenburg for Sexy vacuity .... Martin Brandenburg 2010-11-13T22:56:22Z 2010-11-14T01:18:48Z <p>Let $\otimes_{i \in I} M_i$ denote the tensor product of $R$-modules $M_i$. Then $\otimes_{i \in \emptyset} M_i$ is $R$.</p> <p>(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 $R \to N$.)</p> <p>Another one: I know this is really silly and already contained somehow in the other answers, but anyway:</p> <p>$\prod_{i \in \emptyset} 0 = 1$</p> http://mathoverflow.net/questions/45951/sexy-vacuity/45989#45989 Answer by Pietro Majer for Sexy vacuity .... Pietro Majer 2010-11-13T23:35:16Z 2010-11-15T14:25:54Z <p>What about the two orientations of a point? $$*$$</p> <p><em>(Pro trivialogia)</em>. 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.</p> <ul> <li><p><em>Sometimes we simply do not know whether $x$ is a trivial object</em>. 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).</p></li> <li><p><em>Abstraction</em>. 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, &amp;c.) </p></li> <li><p><em>Constructions and proofs by induction.</em> 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.</p></li> </ul> http://mathoverflow.net/questions/45951/sexy-vacuity/45993#45993 Answer by Joel David Hamkins for Sexy vacuity .... Joel David Hamkins 2010-11-14T00:17:42Z 2010-11-14T00:17:42Z <p>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.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/45997#45997 Answer by Peter LeFanu Lumsdaine for Sexy vacuity .... Peter LeFanu Lumsdaine 2010-11-14T00:34:20Z 2010-11-14T00:34:20Z <p>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.</p> <p>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 $n = 0$!</p> <p>(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!)</p> http://mathoverflow.net/questions/45951/sexy-vacuity/45999#45999 Answer by Joel David Hamkins for Sexy vacuity .... Joel David Hamkins 2010-11-14T00:52:50Z 2010-11-14T00:52:50Z <p>The <a href="http://en.wikipedia.org/wiki/Continuum_hypothesis#The_generalized_continuum_hypothesis" rel="nofollow">Generalized Continuum Hypothesis</a> 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 $\kappa$.</p> <p>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:</p> <ul> <li><p>The power set of a set with $0$ members has $1$ member.</p></li> <li><p>The power set of a set with $1$ member has $2$ members.</p></li> </ul> <p>All other instances of $2^\kappa=\kappa^+$, finite or infinite, are either false or independent of ZFC.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46001#46001 Answer by Michael Hardy for Sexy vacuity .... Michael Hardy 2010-11-14T01:39:48Z 2010-11-14T01:39:48Z <p>I suppose another example would be those proofs by induction in which you don't need a basis, although no nice examples come to mind instantly. Here I mean proofs in which you show that if $P(m)$ for all $m &lt; n$, then $P(n)$. You don't need to show $P(n)$ holds for the smallest value of $n$, since it is vacuously true that $P(m)$ holds for all smaller values, and therefore the thing proved in the inductive step entails the smallest instance as a special case.</p> <p>This works not only for integers, but for infinite ordinals.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46009#46009 Answer by Richard Stanley for Sexy vacuity .... Richard Stanley 2010-11-14T03:18:07Z 2010-11-14T23:21:56Z <p>Is the span of the empty set in a vector space equal to $\lbrace 0\rbrace$, or does it have no span? The "correct" answer in my opinion is the latter. See Example 3.10.3 of <a href="http://math.mit.edu/~rstan/ec/ec1.pdf" rel="nofollow">http://math.mit.edu/~rstan/ec/ec1.pdf</a> for a reason. On the other hand, a reason (which I find unconvincing) for the span to be $\lbrace 0\rbrace$ is given by PBRMEASAP at <a href="http://www.physicsforums.com/archive/index.php/t-84017.html" rel="nofollow">http://www.physicsforums.com/archive/index.php/t-84017.html</a>. This site has a discussion of whether the empty set is a vector space. The correct answer is that it isn't, because one of the axioms is the existence of an additive identity 0.</p> <p><strong>Update.</strong> I agree with the comments that the span of the empty set is $\lbrace 0\rbrace$. What I said above was foolish.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46010#46010 Answer by Georges Elencwajg for Sexy vacuity .... Georges Elencwajg 2010-11-14T03:28:16Z 2010-11-14T03:28:16Z <p><strong>A</strong> 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.</p> <p><strong>B</strong> Dually, given a field $K$, the zero algebra over $K$ is diagonal and in particular étale: the morphism of affine schemes $\varnothing \to Spec(K)$ 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.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46031#46031 Answer by Laurent Moret-Bailly for Sexy vacuity .... Laurent Moret-Bailly 2010-11-14T11:04:29Z 2010-11-14T11:04:29Z <p>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.</p> <p>The fist 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. </p> <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46073#46073 Answer by Owen Biesel for Sexy vacuity .... Owen Biesel 2010-11-14T21:37:52Z 2010-11-14T21:37:52Z <p>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 $\bigcup${$\varnothing$}$\ =\varnothing$.</p> <p>This comes up when using the Grothendieck plus-construction to sheafify a presheaf. Apply the construction to the (nonseparated) presheaf $P:\mathcal{O}(X)^{op}\to \mathrm{Set}$ sending every open set to the set $A$, with $|A|\geq 2$. Then the presheaf $P^+:\mathcal{O}(X)^{op}\to\mathrm{Set}$ 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 $\varnothing$.</p> <p>This elementary example comes from "Sheaves in Geometry and Logic", by Moerdijk and MacLane.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46074#46074 Answer by Qiaochu Yuan for Sexy vacuity .... Qiaochu Yuan 2010-11-14T22:12:14Z 2010-11-15T11:34:49Z <p>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 $\deg 0 = \infty$:</p> <ul> <li><p>The most basic one: the zero polynomial has infinitely many roots in an algebraic closure.</p></li> <li><p><strike>If one wants the degree of a polynomial to be a <a href="http://en.wikipedia.org/wiki/Valuation_(algebra)" rel="nofollow">valuation</a>, 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 <a href="http://en.wikipedia.org/wiki/Absolute_value#Fields" rel="nofollow">absolute value</a> nondegenerate.</strike> </p></li> <li><p><strike>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 $n \to \infty$.</strike></p></li> <li><p><strike>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 $n \to \infty$.</strike></p></li> <li><p>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.</p></li> </ul> <p><strong>Edit:</strong> As James Borger points out, the middle ideas are mistaken. Corrected, they actually suggest that $\deg 0 = -\infty$:</p> <ul> <li><p>$\deg 0 = -\infty$ is the unique choice consistent with the requirements that $\deg fg = \deg f + \deg g$ and $\deg (f + g) \ge \text{min}(\deg f, \deg g)$. With this definition, $|f| = 2^{\deg f}$ is now an absolute value.</p></li> <li><p>Geometrically, when $F = \mathbb{C}$ the function $0$ has a zero of infinite order at infinity, hence a pole of order $-\infty$. </p></li> <li><p>The relevant local ring here is really $F[[ \frac{1}{x} ]]$, and in the natural topology on this ring we have $\frac{1}{x^n} \to 0$. </p></li> <li><p>Another reason to like this definition is that it gives a uniform statement of the division algorithm on $F[x]$.</p></li> </ul> http://mathoverflow.net/questions/45951/sexy-vacuity/46076#46076 Answer by Gene S. Kopp for Sexy vacuity .... Gene S. Kopp 2010-11-14T22:24:17Z 2010-11-14T22:24:17Z <p>Zero is a limit ordinal, because it is the union of its elements.</p> <p>Transfinite induction has two canonical statements. The "strong" statement, $$(\forall \alpha,\beta)((\beta&lt;\alpha) \wedge P(\beta) \rightarrow P(\alpha))\rightarrow (\forall \alpha)P(\alpha),$$ 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.</p> <p>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.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46077#46077 Answer by Gene S. Kopp for Sexy vacuity .... Gene S. Kopp 2010-11-14T22:34:07Z 2010-11-14T22:34:07Z <p>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.</p> <p>Zero is prime in any integral domain. Remember that the trivial ring is <em>not</em> an integral domain.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46096#46096 Answer by I. J. Kennedy for Sexy vacuity .... I. J. Kennedy 2010-11-15T01:21:36Z 2010-11-17T04:26:10Z <p>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 <em>0</em>, 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".</p> <p>A nice, natural recursive expression of the conversion process is</p> <pre><code>def itoa(n): if n==0: return "" return itoa(n/10) + chr(ord('0') + n%10) </code></pre> <p>which can be thought of as <br /> <em>The string representation of an integer consists of its leading digits (n/10) followed by its last digit (n%10).</em></p> <p>Trying to fix this by returning "0" instead of "" would result in everything getting a superfluous leading zero. </p> <p>On the other hand writing <em>0</em> as the empty string would be rather annoying.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46150#46150 Answer by Gabe Cunningham for Sexy vacuity .... Gabe Cunningham 2010-11-15T20:22:29Z 2010-11-15T20:22:29Z <p>Over the reals, $\sup \emptyset = -\infty$ and $\inf \emptyset = \infty$.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46688#46688 Answer by David MJC for Sexy vacuity .... David MJC 2010-11-19T22:34:53Z 2010-11-19T22:34:53Z <p>I regard "negative thinking" in category theory as an example of cool vacuity: see e.g. <a href="http://ncatlab.org/nlab/show/negative+thinking" rel="nofollow">http://ncatlab.org/nlab/show/negative+thinking</a>. 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.</p> <p>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).</p>