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Standard definition of graph says that it is an ordered pair G=(N,L) where N is set of nodes and L is set of lines which connect the nodes.

From what I've read, the set L can be empty, but can set N be empty too?

I'm mainly asking this because for my exam, I found numerous problems that go like this: Graph X is given. How many subgraphs with property Y exist? Often the number depends on the definition of graph.

AndrejaKo
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Depends on your definition of a graph. This is addressed in a paper of Harary, Is the null-graph a pointless concept? The abstract is as follows:

The graph with no points and no lines is discussed critically. Arguments for and against its official admittance as a graph are presented. This is accompanied by an extensive survey of the literature. Paradoxical properties of the null-graph are noted. No conclusion is reached.

Personally, I think there's no reason not to admit it as a graph. There is a philosophy due to Grothendieck that it's better to work in a nice category with nasty objects than in a nasty category with nice objects, and the null graph makes the category of graphs nicer (giving it an initial object). As for the paradoxical properties, I haven't read the paper but they are likely due to the phenomenon that the nLab calls too simple to be simple. For example, the empty graph is not connected; it has zero connected components instead of one.

Qiaochu Yuan
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    @Qiaochu I agree completely... If you can have concepts like (-1)-coloring of graphs, then you should be able to have the concept of a null-graph. Math is the study of the abstract, and null-graphs definitely fall into that category. – Nicolas Villanueva Jun 08 '11 at 13:26
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    Isn't the connectivity of a graph defined in terms of there being a path between any pair of vertices? By this definition, the null graph is connected. Perhaps the paradoxical property is that the null graph is the only connected graph which does not have one connected component. –  Jun 08 '11 at 13:51
  • Never mind, I see that nLab considers the empty space to not be connected. So I guess I was using an old-fashioned definition of connectedness. –  Jun 08 '11 at 13:57
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    @Rahul: no. The empty graph is not connected for the same reason that $1$ is not prime: because you want a graph to have a unique decomposition into connected components, and uniqueness is impossible if you call the empty graph connected. (From the perspective of the nLab article the correct definition is "there exists a path between any pair of vertices, and there exists at least one such pair.") – Qiaochu Yuan Jun 08 '11 at 14:17
  • For whatever it's worth: in my current "global" conventions (scare quotes because global conventions should probably be time-independent as well, and I am far from confident of that), the empty space is connected but not irreducible. This has some pleasant consequences with regards to the spectrum of a commutative ring. What's a specific argument that the empty set should not be regarded as connected? – Pete L. Clark Jun 08 '11 at 14:18
  • @Pete: what I just said above. There is also another nice definition of connectedness from the nLab: in a sufficiently nice category, a connected object is an object $C$ such that $\text{Hom}(C, -)$ preserves coproducts, which is nicely intuitive, and this isn't true of the initial object. – Qiaochu Yuan Jun 08 '11 at 14:22
  • Hmm, I guess Qiaochu's latest comment also applies to topological spaces, but for some reason I am not yet convinced by it. There may just not be a convention regarding the connectedness of the empty space which is best in all situations... – Pete L. Clark Jun 08 '11 at 14:23
  • @Pete: can you explain what "pleasant consequences" you want from defining the empty space to be connected? That is, why do you want the spectrum of the zero ring to be connected? – Qiaochu Yuan Jun 08 '11 at 14:25
  • @Qiaochu: well yes, but as nLab points out, Top is not a sufficiently nice category in that sense: there exist totally disconnected nondiscrete spaces! – Pete L. Clark Jun 08 '11 at 14:26
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    @Qiaochu: well, I'm thinking about it. For one thing, I want $\operatorname{Spec} R$ to be connected iff $R$ has no nontrivial idempotents, but I know what you're going to say: we should distinguish between rings which have $2$ idempotents and rings which have only $1$ idempotent! Also: the definition of connectedness via continuous functions to ${0,1}$ makes the empty space connected (although again you could redefine a "constant function" here). I also want arbitrary products of connected spaces to be connected... – Pete L. Clark Jun 08 '11 at 14:31
  • Finally, let me mention that I just confirmed that Bourbaki's definition of connected does include the empty set. Not that Bourbaki is the mathematical bible, but it represents a group of talented mathematicians who put in an awful lot of time thinking about issues like this. It is true that Bourbaki's sophistication and erudition is not categorical in nature, to its detriment, but some of this is in the nature of general topology: if you took Grothendieck's maxim really seriously, you would not work in Top, since its categorical properties are not so great... – Pete L. Clark Jun 08 '11 at 14:37
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    @Pete: arbitrary products of connected spaces are connected either way! I'm not sure what you mean by the definition of connectedness via continuous functions to ${ 0, 1 }$, but this is the same issue as the idempotents: you want there to be no non-constant functions, as well as two trivial constant functions (another example of the nLab's "existence and uniqueness" observation). – Qiaochu Yuan Jun 08 '11 at 14:44
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    On a slightly different tack, connected plannar graphs have an Euler characteristic of 2, so it would be helpful if an empty graph was not connected, not plannar or not a graph. – Henry Jun 08 '11 at 14:47
  • @Henry: the empty graph is not planar. I know this sounds silly, but "planar graph" really means "triangulation of the sphere" (in a rough sense; perhaps I really mean "cell decomposition of the sphere") and a single $2$-cell is not a triangulation of the sphere in any reasonable sense. – Qiaochu Yuan Jun 08 '11 at 15:02
  • @Qiaochu: you're right about the products business: my mistake. For the other: a space is connected iff every continuous function to ${0,1}$ (discrete topology) is constant. Anyway, at the moment I would have to agree that I'm not being especially convincing...I'll let you know if I come up with / remember anything better. – Pete L. Clark Jun 08 '11 at 15:14
  • @Qiaochu: I'm convinced by your argument regarding connectedness. Speaking of planar graphs, if the triangle graph is planar and the empty graph is not, what is the status of the connected graphs with one and two vertices respectively? Also, does one add a caveat to Kuratowski's theorem so that only graphs with at least a certain number of vertices are considered planar? (Just so we're clear, these are genuine, not rhetorical, questions.) –  Jun 08 '11 at 15:45
  • @Rahul: hmm. I may have to retract my previous statement. I guess what I meant is "connected planar graph" turns out to be equivalent to "cell decomposition of the sphere" if the empty graph is not connected, and I guess "planar graph" is probably better off with its usual meaning. – Qiaochu Yuan Jun 08 '11 at 15:59
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    Added many years later: I no longer regard the empty space as connected: see e.g. p. 120 of http://math.uga.edu/~pete/pointset.pdf. – Pete L. Clark Nov 20 '20 at 19:31
  • The link to the paper by Harary & Read (not only Harary) at springerlink.com is dead. For future readers, the full citation is as follows: Harary, Frank; Read, Ronald C., Is the null-graph a pointless concept?, Graphs and Combin., Proc. Capital Conf., Washington, D.C. 1973, Lect. Notes Math. 406, 37–44 (1974). Zbl 0293.05101. – The Amplitwist May 19 '22 at 23:48