2 corrected three errors

The whole discussion seems to devolve on whether the empty graph (or empty space) should be considered "connected". Angelo and I are of the school that it should not, but this should be explained since some of the traditional definitions of "connected" apparently allow the empty space to be connected.

A general abstract context is as follows. Let $C$ be a category with finite coproducts with the property that for any two objects $a$, $b$ (whose coproduct is denoted $a+b$), the canonical functor

$$C/a \times C/b \to C/(a+b): (x \to a, y \to b) \mapsto (x + y \to a + b)$$

is an equivalence. Such a category is said to be extensive. The category of topological spaces is extensive, the category of graphs is extensive, any topos is extensive, and there are many, many other examples.

Now, say an object $a$ in an extensive category to be is connected if the functor

$$\hom(a, -): C \to Set$$

preserves binary coproducts (whence it can be shown to preserve finite coproducts). This is a fundamental definition; see the nLab for an extended discussion. Under this definition, the empty space (the empty graph, etc.), i.e., the initial object, is not connected.

An equivalent definition is to say $c$ is connected if, whenever $c \cong a + b$, exactly one of $a, b$ is inhabited. If one insists that the empty space should be inhabitedconnected, then change the word "exactly" to "at most", and instead of saying the canonical map $\hom(c, x) + \hom(c, y) \to \hom(c, x + y)$ is an isomorphism, say it is merely surjective. However, most results come out more cleanly by working with the definition above, which disqualifies the empty set.

Compare the notion of prime ideal: working in the lattice of ideals of a commutative ring p.i.d. $R$ where $\leq$ is given by reverse inclusion, the coproduct or join of ideals $a, b$ is $ab$, the initial ideal is $R$, and we say an ideal $p$ is prime if $p \neq R$ and $p \leq ab$ implies $p \leq a$ or $p \leq b$. The condition $p \neq R$ is considered fundamental to the definition of prime. Without it, we no longer have e.g. unique decomposition of integers into prime factors (compare the fact that every graph is uniquely a coproduct of connected graphs under our definition, but this is not so if the empty graph is considered to be connected). See also the numerous examples in the nLab discussion "too simple to be simple"; for example, $1$ is too simple to be a prime, and the zero module is considered too simple to be a simple module.

Every acyclic graph (a forest) is uniquely a coproduct of acyclic connected graphs (i.e., trees) under our definition of connectedness. This includes the empty forest. So a forest can be empty, but a tree cannot.

1

The whole discussion seems to devolve on whether the empty graph (or empty space) should be considered "connected". Angelo and I are of the school that it should not, but this should be explained since some of the traditional definitions of "connected" apparently allow the empty space to be connected.

A general abstract context is as follows. Let $C$ be a category with finite coproducts with the property that for any two objects $a$, $b$ (whose coproduct is denoted $a+b$), the canonical functor

$$C/a \times C/b \to C/(a+b): (x \to a, y \to b) \mapsto (x + y \to a + b)$$

is an equivalence. Such a category is said to be extensive. The category of topological spaces is extensive, the category of graphs is extensive, any topos is extensive, and there are many, many other examples.

Now, say an object $a$ in an extensive category to be connected if the functor

$$\hom(a, -): C \to Set$$

preserves binary coproducts (whence it can be shown to preserve finite coproducts). This is a fundamental definition; see the nLab for an extended discussion. Under this definition, the empty space (the empty graph, etc.), i.e., the initial object, is not connected.

An equivalent definition is to say $c$ is connected if, whenever $c \cong a + b$, exactly one of $a, b$ is inhabited. If one insists that the empty space should be inhabited, then change the word "exactly" to "at most", and instead of saying the canonical map $\hom(c, x) + \hom(c, y) \to \hom(c, x + y)$ is an isomorphism, say it is merely surjective. However, most results come out more cleanly by working with the definition above, which disqualifies the empty set.

Compare the notion of prime ideal: working in the lattice of ideals of a commutative ring $R$ where $\leq$ is given by reverse inclusion, the coproduct or join of ideals $a, b$ is $ab$, the initial ideal is $R$, and we say an ideal $p$ is prime if $p \neq R$ and $p \leq ab$ implies $p \leq a$ or $p \leq b$. The condition $p \neq R$ is considered fundamental to the definition of prime. Without it, we no longer have e.g. unique decomposition of integers into prime factors (compare the fact that every graph is uniquely a coproduct of connected graphs under our definition, but this is not so if the empty graph is considered to be connected). See also the numerous examples in the nLab discussion "too simple to be simple"; for example, $1$ is too simple to be a prime, and the zero module is considered too simple to be a simple module.

Every acyclic graph (a forest) is uniquely a coproduct of acyclic connected graphs (i.e., trees) under our definition of connectedness. This includes the empty forest. So a forest can be empty, but a tree cannot.