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This is a classical question, that has led to many a heated argument:

Should the symbol $\mathbb{N}$ stand for $0,1,2,3,\dots$ or $1,2,3,\dots$?

It is immediately obvious that the question is not quite well posed. This convention, as many others, are not carved in stone, and there is nothing to prevent mathematician $A$ define $\mathbb{N}$ to be the positive integers including $0$, and mathematician $B$ to define $\mathbb{N}$ to be the positive integers excluding $0$. It does not seem that one definition is accepted widely enough for it to be "the right definition", and even if this was the case, the fashion might change in the future.

I am, however, hoping that there might be a semi-mathematical reason to prefer one notion over the other. For example, I have spent much of my mathematical life believing that $0 \in \mathbb{N}$ because: 1) morally, $\mathbb{N}$ is the cardinalities of finite sets 2) the empty set is a set with $0$ elements. However, recently I realised that this reasoning applies to $\omega$ rather than $\mathbb{N}$, and - much to my horror - I saw $\omega$ and $\mathbb{N}$ used side by side with the only distinction being that $0 \in \omega$ while $0 \not \in \mathbb{N}$. For another example, $\mathbb{N}$ seems to be a much nicer semigroup if $0 \not \in \mathbb{N}$ (and in any case, adding $0$ to a semigroup is a more natural operation than removing it), which would an argument for taking $0 \not \in \mathbb{N}$.

The arguments mentioned above are, of course, rather weak, but perhaps just enough to tip the scale. In any case, this is the general type of argument I am looking for.

Question: Does there exists a convincing argument for deciding if $0 \in \mathbb{N}$ ?

(I consider it quite possible that the answer is negative because in some context one convention is preferable, and in other context the other one. The problem could be dismissed by using $\mathbb{Z}_+$ (or even $\mathbb{Z}_{>0}$ and $\mathbb{Z}_{\geq 0}$) to avoid confusion, but note that in some context one definitely does not want to do this. I would be interested in an argument that is universal in the sense that it makes overall mathematical landscape more elegant, and does not spoil any detail too much. I do not hope that the argument would be convincing to every mathematician, especially one working in a very specific and narrow area.)

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closed as not constructive by fgp, Amzoti, Keenan Kidwell, tomasz, vadim123 May 15 '13 at 1:27

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

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I’ve voted to close this as not being constructive. Although the Question is formally not an invitation to argument, it really does just solicit opinions. Worse, those opinions mean virtually nothing, since people will in general use their preferred notation regardless of the views of others. –  Brian M. Scott May 14 '13 at 21:28
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I think this question can invite worthwhile discussion, if we strip away the distractions and focus solely on the central question: Does there exist a convincing argument for deciding if 0 ∈ N? –  xisk May 14 '13 at 21:31
    
The most common notations I've seen are $\mathbb N_0$ ($\mathbb N$ with $0$) and $\mathbb N^*$ ($\mathbb N$ without the additive identity). If it doesn't matter which to use, then one can use $\mathbb N$. –  Samuel May 14 '13 at 21:33
    
@BrianM.Scott: Apologies if I did ask an unconstructive question. I do agree that it is close to the verge between constructive and not, but I think it (barely) fits on the right side. I do believe that "existence of a convincing argument" is something that can be established and is not strongly subject to opinion. If I had asked if $0 \in \mathbb{N}$, period, then I would fully agree with the close. –  Feanor May 14 '13 at 21:34
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I agree with Brian, but I did not vote to close, because I'm not sure that there is nothing useful to be said. For example, you cannot say that the cardinality of a finite set is a natural number, unless you include 0 in $\Bbb N$. –  MJD May 14 '13 at 21:36
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4 Answers 4

up vote 10 down vote accepted

Let me ask a similar question in rebuttal:

Are finite sets countable?

The answer is simple. It depends on the context. Sometimes it's easier to have finite sets included in the definition of "countable" and sometimes it's easier to have just the finite set, and have "at most countable" for the term which includes finite sets as well.

I will give an argument why $\Bbb N$ should include $0$, though.

One can consider $0$ and $1$ as the basic atoms of the numbers we know. $\Bbb N$ is the set generated by $0,1$ using addition; and then $\Bbb Z$ is generated by adding additive inverses, and $\Bbb Q$ and then $\Bbb R$ and $\Bbb C$.

Of course that if you take your atomic set of numbers to be $\Bbb C$ or something else, then it might as well be redundant, but it's still a reasonable argument. With only some naive set theory, and axioms for addition and multiplication, we can create all the numbers we need! That's an incredible thing. And all just form the assumption that $0$ and $1$ exist.

On the other hand, in analysis it's often more convenient to have $0\notin\Bbb N$. For example when we say that $x^n$ is well defined for every $x\in\Bbb R$ and $n\in\Bbb N$. Or if we often talk about the sequence $\frac1n$, then we find it easier to write $\frac1n$ for $n\in\Bbb N$, rather than adding "...and $n>0$".

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the existence of $0$ is equivalent to the existence of a set, by separation axiom, and the existence of 1 isn't an assumption at all: $1:=\{ 0 \} $. Furthermore no binary operation such as addition is needed to define natural numbers, they're defined in a purely set-theoretical way, the only one possible in this case. –  Lano May 14 '13 at 21:33
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@Lana: No, $0$ is encoded as the empty set, and $1$ is encoded as $\{0\}$ (or any other way). But those are not the philosophical arguments and reasons why we should accept $0\in\Bbb N$. Those are just points which half-support the idea by using set theory as a foundation. –  Asaf Karagila May 14 '13 at 21:34
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(I tend to begin just about every class I teach with something like: "For me, the convention is that $0\in\mathbb N$".) –  Andres Caicedo May 14 '13 at 21:38
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@Andres, in the past three fall semesters when I TA'd freshmen I explained that in set theory $0$ is a natural number, and that they will be told otherwise in analysis, but that's a matter of convention, and analysts are wrong about that. :-) –  Asaf Karagila May 14 '13 at 21:41
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@Lana: I find your last two comments very rude. I think that I have studied and contemplated foundations of mathematics for the majority of the past three years, if not more during my undergrad. Don't try and lecture me on set theory as a foundation for mathematics, nor about how mathematics was developed. –  Asaf Karagila May 14 '13 at 21:43
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They should arguably include zero so that they are a monoid that gives rise to the integers, and all the other numbers, etc, like in Asaf's great response.

In my day to day life though only the cardinality of them matters, as they are really used purely as an index set. In this context it really is purely a matter of aesthetics. I would go back and forth including zero if I want to divide by them or not.

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My 2¢: Besides the in my opinion other good arguments why $\mathbb{N}$ should contain $0$, I argue for a purely notational one: since $\mathbb{Z}_+$ is a perfectly good name for the set $\{ 1, 2, 3, \ldots \}$ there is no need for a second name for this set. Hence $0 \in \mathbb{N}$.

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This is a strange argument. E.g., $\mathbb Z_{\geq 0}$ is a perfectly good name for $\{0,1,2,\ldots\}$ - should we forget about $\mathbb N$ at all? –  Artem May 14 '13 at 23:32
    
@Artem, I don't think it's a strange argument at all. We want symbols that minimize redundancy. I agree that $\mathbb{Z}_{\geq 0}$ also does the job, but I think that looks a lot uglier than $\mathbb{N}$. –  Christopher A. Wong May 14 '13 at 23:38
    
@ChristopherA.Wong What about $\mathbb Z_+$? The reason I am asking is that I know that some people (quite a few, actually) use the notation $\mathbb R_+$ to denote nonnegative reals. –  Artem May 14 '13 at 23:39
    
Using the notation $\Bbb Z_{>0}$ (or other subscripts) puts the spotlight on the integers, rather than the natural numbers. And one has to ask what is more fundamental, the integers or the natural numbers? –  Asaf Karagila May 14 '13 at 23:49
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Besides problems coming from notation (these are the only ones which distinguish $\mathbb{N}$ from $\omega$, the former used when treated as a monoid and the latter when thought as a ordinal) the reality is this: $0 \in \omega$ and $0 \subset \omega$ are both true ( remember that $0:= \emptyset$), so it's obviously contained, and it belongs to $\omega$ for lots of reasons (transitivity plus the fact that both are cardinals or by definition of $\omega $ as the set of natural numbers). The last is a good definition not given in an intuitive and inexistent way.

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I find this post rather incoherent (grammatically), and it has serious punctuation issues (I count 5 opening parentheses and only two closing ones!). –  tomasz May 15 '13 at 1:05
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