Talk:Empty set

I removed this entire passage, based on material first submitted by a non-registered Wikipedian:

The empty set is very simple; ironically so simple that many mathematics students (and even professional mathematicians!) have a difficult time applying it correctly.

For example, take the first feature listed above, that the empty set is a subset of any set A. If you look up subset, then you'll see that this claim means that for every element x of {}, x belongs to A. Since "for every" is a strong condition, we intuitively expect that it must be necessary to find many elements of {} that also belong to A. Of course, we can't find any elements of {} that also belong to A. So many people think that {} is not a subset of A after all. But this is a mistake. In fact "for every" may not be a very strong requirement at all if it says "for every element of {}". Since there are no elements of {}, "for every element of {}" is no requirement at all. Every statement satisfies that requirement, which is actually the weakest condition possible.

A statement which claims something about all the elements of {} (i.e. about nothing whatsoever) is an example of a vacuous truth. So it is vacuously true that {} is a subset of A. The concept of vacuous truth can be a difficult one to wrap one's brain around, and this leads to difficulties in applying the empty set correctly.

Here are some of my reasons:

1. "The empty set is very simple" is not NPOV, simplicity being a far from simple concept. I've tried to compensate for deleting this by giving the briefest mention of how intuition can conflict with the formal definition of a set. (This discussion probably belongs in a different article, but...)
2. The second paragraph seems, in context of the first, to be an attempt to describe a common difficulty in "applying" the empty set concept "correctly". However, it is more a justification for why the first property of the empty set listed by the article is reasonable. As such a justification, I think the second paragraph is underdeveloped, and worse than no paragraph at all.
3. The second paragraph has an implicit discussion of vacuous truth and the third has an explicit reference to it. At present, I believe the vacuous truth article does a better job explaining these issues.

-- Ryguasu (Wed Sep 25 01:01:06 EST 2002)

The nonregistered Wikipedian is me.

1. No argument; what you've written is good, better than what I wrote.
2. I don't see why you don't want to have an example of what we're talking about. First, how is this example "undeveloped"? It goes on a lot more than any straightforward application of the definitions; to just prove the statement is trivial! This is a standard example of nonintuitive reasoning that I've seen in several set theory books, and it seems odd to consider it out of place. If anything, we should have more of this sort of thing in here.
3. What you have doesn't even explain how the empty set is related to vacuous truth. That's a major omission! Not every fact must appear on only one page; most should appear on several, since they deal with several things.

— Toby 08:54 Sep 27, 2002 (UTC)

I'm baffled by the bit about everybody, even mathematicians finding it baffling. It is terribly simple. The hard stuff is problems like deciding how many elements a set like { { {}, { {} } }, { } } has, and even that's just a case of keeping a clear head and counting the brackets -- Tarquin

I wrote "and even professional mathematicians!" back in the day because I see such confusion so often. Whenever I read math texts, I'm always on the look out for incorrect handling of degenerate cases in definitions and theorems. (Of course, one can argue about the definitions, but not the theorems!) The empty set pops up here all the time.

The most common example in my personal experience is this statement of the axiom of choice:

Every nonempty collection of nonempty sets has a choice function.

This has one "nonempty" too many (look at our article to see which, because we get it right). Removing the extra "nonempty" results in an equivalent statement, of course, since the empty collection has a choice function (the empty function). But it's presence implies that the word is necessary, giving a false impression; and it wouldn't be there if somebody hadn't had a false impression before.

If you give me a chance to look in some books, I'll find you examples where confusion about the empty set leads not just to misleading inclusion of unncessary words but to actual false statements. Finding and correcting these is almost a hobby of mine.

— Toby 11:25 Sep 28, 2002 (UTC)

Examples of such mistakes would be an interesting addition to the article! (I had a lecturer at university who didn't believe in the axiom of choice... ) -- Tarquin

Hum, now that means that I have to find a selection of standard texts (it's no far just picking things randomly off the shelves) and searching for this particular error. That could take up a lot of time, since I'd naturally get drawn into reading the other parts of these books ^_^. (And I don't believe the axiom of choice either, at least not on weekends.) — Toby 08:18 Sep 29, 2002 (UTC)

Sorry, didn't mean to send you off on a research quest! -- Tarquin

Don't worry, I probably won't do it ^_^. — Toby 10:52 Sep 29, 2002 (UTC)

Question: does the empty set have a well-defined notation? For example, the set S = {1, 2, 3, 4, 5} = {x : 1 <= x <= 5} is well-defined. Does the empty-set have an analagous notation, e.g. ES = {} = {x : x <math>\notin<math> S} for any nonempty set S? The first part of this definition says that the empty set *does* contain elements, otherwise the set is not well-defined. The second part of the definition tells us how to construct the empty set; define a set, then remove all elements until the set is empty. But then we get a contradiction from defining the empty set as a subset of *any* set, because any of these subsets is defined as: {x: x<math>\notin<math>S, x<math>\in<math>S}. So is there a real set-theoretic definition and notation for the empty set or do we just conceptualize it for convenience? — Elijah Gregory

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Regarding the difficulty/simplicity of the concept: the other day I talked to a math Ph.D. who said that he thinks of the empty subset of R² as being different from the empty subset of R. AxelBoldt 23:54 Sep 29, 2002 (UTC)

Does he think of the empty subset of Q as different from the empty subset of R? How about the subset {0}? Or the subset Q?

I often think of the same set as different subsets, depending on what set I'm thinking of it as being a subset of, much as I think of the same set of ordered pairs as different functions, depending on what set I'm think of as the codomain. (And this is more than just an analogy, since subsets of X can be identified with equivalence classes of injections into X, and indeed it's that concept that is used to define subobjects in category theory.) This is a matter of keeping track of context, and is useful when answering questions such as "What is the complement of {}?". You can formalise it as an ordered pair (A,X), where A is a subset of X (just as you can formalise a function as an ordered pair (G,Y), where G is the graph and Y is the codomain). Then the question "What is the complement of A?", normally ambiguous, is unambiguous, because "A" really means (A,X) and you're just abusing notation when you call it A.

But if your colleague treats {} specially in this regard, then I'm afraid that I can't help him. — Toby 07:56 Sep 30, 2002 (UTC)

The question "what is the complement of X" is meaningless. It's an acceptable shorthand when we all know what the superset is, but "complement" is a 3-way relation (or a binary operation): "A is the complement of X in Y". I wonder, does Axel's friend think of the zero in R as different from the zero in the Q? -- Tarquin 12:25 Sep 30, 2002 (UTC)

First, how can you say that "What is the complement of A?" is meaningless if you agree that it's acceptable shorthand in certain contexts? In those contexts, it has a perfectly good meaning! That is why you'll often see sentences just like it in reasonable math books. All meaning depends on context, even in math.

Second, the 0 element of R is different from the 0 element of Q in much the same sense as their empty subsets are different. Even more so, because people that stick to pure set-theoretic reductionism in their math will still see them as different (unless they go out of their way to avoid this), which you can't say in the subset case. (Say, one 0 is an ordered pair of sets of rational numbers, while another 0 is a set of ordered pairs of integers.)

My point is that when you ask if two things in math are equal or distinct, the question only has meaning when it is relevant, that is in a context where the two things are already of a certain type (such as subsets of R² or elements of Q). Yes, the question "Are they equal?" has a meaning in a reductionist set-theoretic sense (using the axiom of extension), but since the answer to the question varies depending on how one carries out the reduction, it is ultimately meaningless. Thus, it's best to regard such things as so different that they are "not even distinct" (as Wolfgang Pauli might have put it). Or to put it another way, you're asking if they're the same species, when they're not even the same genus. — Toby 10:23 Oct 7, 2002 (UTC)

"meaningless" in that it's not clearly defined.

But it is perfectly well defined if you specify ahead of time that you're dealing with subsets of X. If you like algebra, then you might want to ask yourself what is the cokernel of the group homomorphism sgn defined on the symmetric group S₃, where sgn σ is 1 or −1 depending on whether σ is odd or even. Well, the answer to my question is not clearly defined.

For most purposes in algebra, however, you can work in a context where you assume that signs belong to the group {1,−1}, since these are the only values that signs can ever take in any permutation situation. Then the cokernel is trivial, since the image of f is the entire group {1,−1}. On the other hand, you could work in a context where you take signs to belong to the group U(1) of unit complex numbers (which is just what you want in some applications to quantum physics), and then the cokernel is an infinite group (in fact isomorphic to U(1) again).

So the answer depends on the context. Absolutely, the problem is ill defined if the context is not specified (or not specified enough). But by the same token, the problem is quite well defined if the context is sufficiently specified.

Indeed, it's become fashionable (thanks to Bourbaki) to define group homomorphism in such a way that the necessary context for the above problem must be specified. Wikipedia's own definition requires a homomorphism to be a function (obvious enough) and then defines functions to be differnt if they have the same domain and graph but different codomains. And the codomain is exactly the context that we needed here; it was (in my examples) either {1,−1} or U(1).

We don't do this with the term subset; if you say "A is a subset", then you don't need to specify what it's a subset of. A subset of Q and a subset of R² could well be equal under this definition. Yet I hope that you realise how unsatisfactory it is to say simply "A is a subset". You want to say "A is a subset of X", or else you're not really finished with what you wanted to say. So we could (and doubtless would, had Bourbaki seen fit to so influence us) define a subset to be an ordered pair (A,X) such that ∀x∈A, x ∈ X, much as we define a function be an ordered pair (gr,Y), where gr is a set (of ordered pairs) satisfying a certain condition involving Y. Back in the day, they just said that gr was the function, but ultimately that's missing something relevant. Similarly, to say that A is a subset is not really enough to do any good; to get anywhere, you have to say what A is a subset of.

"one 0 is an ordered pair of sets of rational numbers, while another 0 is a set of ordered pairs of integers" -- yes, but once R is built, do we still consider Q to be the old construction of pairs of integers, or do we consider Q to now be a subset of R? I suppose there is an "original" Q and a "meta"-Q. Ouch! -- Tarquin

Yes, and you'll sometimes even see that written up in textbooks. The alternative method, which you will also see written up in textbooks, is to take the newly constructed set R, remove all of the metarationals from it, and replace these by the original rationals, thus getting a proR ("προ" being the opposite of "μετα").

The important thing to realise, however, is that all that you actually need to do mathematics is a field Q with certain properties (prime, charcteristic 0), a field R with certain properties (ordered, Dedekind complete), and a monomorphism (that is, both a homomorphism and an injection) from Q to R (there is only one). Exactly what the sets are is irrelevant, and it's even unnecessary to assume that Q is literally a subset of R; you can consider it as just an abuse of notation (we could use an article on that) when you pretend that it is.

This all makes perfect sense from the viewpoint of category theory, where the notion of subset doesn't generalise precisely to arbitrary categories but there is a notion of subobject — according to which, any monomorphism is in exactly the same position as an actual subset inclusion. That is, when you generalise to arbitrary categories, you can no longer tell whether you really have a subset or just a monomorphism, and it doesn't matter. — Toby 10:46 Oct 10, 2002 (UTC)

I suppose that is similar to the question: "is there only one group C3 or are there many groups isomorphic to it?" -- Tarquin

Right, although there is a subtle difference between the case of C3 and the cases of R and Q.

What is really odd is that there is a sense in which it's OK to speak of "the" cyclic group with 2 elements but not OK to speak of "the" cyclic group with 3 elements. The reason is that not only are any two cyclic groups G and H each with 2 elements isomorphic, there is also only one isomorphism between them. Thus, if you and I (discussing "the" such group with one another) come up with different groups in our respective imaginations, we can not only rest assured that there is an ismorphism between our groups but also that we know what that isomorphism is. For some purposes, we need that level of certainty to communicate with one another, a level of certainty that just isn't available with cyclic groups each with 3 elements.

To see this in a more familiar place, consider "the" Dedekind complete ordered field. It's reasonable to use this as the definition of the set R of real numbers, because if you have one set R (say a set of Dedekind cuts) and I have another set R' (say a set of equivalence classes of Cauchy sequences), then they are isomorphic. But more than that, they are uniquely isomorphic. Thus I not only know that when I say "√2" to you that there is something in your R that corresponds to my √2 ∈ R', but I also know that there's only one thing that it could be.

In contrast, it's less safe to define the set C of complex numbers as "the" algebraic completion of R. The reason is that there is no way to distinguish i from −i; they have precisely identical algebraic (and topological) properties. The result is that we have to set conventions, for every representation of the complex numbers, for which square root of −1 is the official i. For example, in the Argand plane, i is counterclockwise from 1, while −i is clockwise. But it could just as easily have been the other way around. In Schroedinger's equation, momentum is −ihd/dx, not ihd/dx; but it could just as easily have been otherwise. Every time complex numbers appear, this convention must be set, or conflicting implementations will result. No such problems arise with the real numbers.

— Toby 10:06 Oct 18, 2002 (UTC)

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I removed:

In fact, in set theory, all objects which don't contain themselves can be constructed from the empty set and the operation of making a set containing some available elements. The empty set acts as a starting point.

First, no set can contain itself, a consequence of the axiom of foundation. The claim that all sets can somehow be constructed from the empty set should probably be formalized as "all sets are constructible", a statement which can neither be proved nor disproved from the ZFC axioms. AxelBoldt 13:09, 29 Nov 2003 (UTC)

The empty set symbol

The letters Φ (obtained by typing &Phi;), φ (&phi;) or <math>{}{}\phi<math> (<math>\phi<<\math>) are not the symbol for the empty set in mathematics, and should not be used as such in Wikipedia.

In Unicode, the empty set symbol ∅ (&#8709;) occupies code point U+2205. But many fonts in use today don't include this character and render it as a small rectangle.

The TeX symbol <math>\emptyset<math> (<math>\emptyset</math>) looks funny and seems to dance above the baseline.

Therefore, I recommend using either Ø (&Oslash;) or {} ({}) to indicate the empty set.

—Herbee 2004-02-18

I just found <math>\varnothing<math> (TeX \varnothing) which now looks best to me.
—Herbee 23:18, 2004 Mar 18 (UTC)

I agree varnothing looks best. If we really want to avoid tex-png than Oslash seems the best choice. It also looks the most similar to varnothing. MarSch 13:46, 20 Apr 2005 (UTC)

The article Ø talks about ∅ (&#8709;). I think that this article should as well—or at least present the same symbol as a graphic. Perhaps there should be a new section at the end about representing the empty set in Unicode. BlankVerse ∅ 05:32, 3 May 2005 (UTC)

Mention &#8709; by all means. However, to be frank, I rather dislike all the sections in various articles about how to represent all kind of symbols in Unicode. I feel that this is a technical issue that has little to do with the concept empty set and is of little use to our readers. -- Jitse Niesen 10:01, 3 May 2005 (UTC)