True, false, or meaningless?

Question

Are the following two assertions always true, always false or meaningless?

$\exists i \in \emptyset$

$\forall i \in \emptyset$

Because it seems that one encounters expressions of this kind fairly similar in mathematics, especially if we are dealing with degenerate cases of definitions. Let me give an example (in graph theory) of such a case. There, one can formalize the idea the two vertices are connected in the following way: Let $G=(V,E)$ be a graph and let $v,w\in V $. We define $v$ and $w$ to be "connected" if: $\exists n \in \mathbb{N}, \ \exists \alpha: \left\{ 1,\ldots,n \right\} \rightarrow V, \ \alpha_0=v \ \& \ \alpha_n=w \ \ \forall i\in \mathbb{N}, 0\leqslant i \ \& \ i<n:\ \left\{ \alpha_i, \alpha_{i+1} \right\} \in E$

Now, if we would ask, if every node is connected to itself (a fact which intuitively we would want to be true), we would have to exhibit an $n$ and a sequence $\alpha$, such that [bla bla...]. The obvious choice for $n$ is 0. But for this choice of $n$, no mattter which sequence $\alpha$ we would consider, the set of the $i$'s would be empty, because the set of the $i$'s is actually $\left\{ i\in \mathbb{N} | 0\leqslant j \ \& \ j<n \right\}$. But for $n=0$ this set is the empty set. Thus, because we quantify $\forall i \in \left\{ i\in \mathbb{N} | 0\leqslant j \ \& \ j<0 \right\}=\emptyset$, should the statement be true/false by definition?

EDIT: Sorry, for my unclear formulation and to everybody who on my fault interpreted it wrongly. The way Carl Mummert or Listing interpreted it, was what I meant.

Answer 1

When we translate mathematical statements into formal logic, the "bounded quantifiers" $(\exists x \in I) P(x)$ and $(\forall x \in I) P(x)$ are usually viewed as abbreviations, as follows:

• $(\exists x \in I) P(x)$ is an abbreviation of $(\exists x)(x \in I \land P(x))$.

• $(\forall x \in I) P(x)$ is an abbreviation of $(\forall x)(x \in I \to P(x))$.

With these conventions, the bounded quantifiers continue to make sense even when $I$ is empty. In that case, $(\exists x \in \emptyset) P(x)$ will always be false, and $(\forall x \in \emptyset) P(x)$ will always be true, regardless of the formula $P(x)$. So, for example, $(\exists x \in \emptyset)(i = i)$ is false and $(\forall x \in \emptyset)(i \not = i)$ is true.

A nice property of this definition of the bounded quantifiers is that it makes them dual in the sense that for any set $I$ (possibly empty) and any formula $P(x)$, we have

• $(\exists x \in I) P(x) $ if and only if $\lnot (\forall x \in I)\lnot P(x)$
• $(\forall x \in I)P(x) $ if and only if $ \lnot (\exists x \in I)\lnot P(x)$

These can be verified by direct calculation: $$ \begin{split} (\exists x \in I) P(x) & \Leftrightarrow (\exists x) (x \in I \land P(x)) \\ & \Leftrightarrow \lnot (\forall x) \lnot (x \in I \land P(x)) \\ & \Leftrightarrow \lnot (\forall x)(x \not \in I \lor \lnot P(x)) \\ & \Leftrightarrow \lnot (\forall x)(x \in I \to \lnot P(x))\\ & \Leftrightarrow \lnot (\forall x \in I)\lnot P(x) \end{split} $$ and $$ \begin{split} (\forall x \in I) P(x) & \Leftrightarrow (\forall x) (x \in I \to P(x)) \\ & \Leftrightarrow \lnot (\exists x) \lnot (x \not \in I \lor P(x))\\ & \Leftrightarrow \lnot (\exists x) (x \in I \land \lnot P(x))\\ & \Leftrightarrow \lnot (\exists x \in I) \lnot P(x) \end{split} $$

Answer 2

In this form, these are just beginnings of propositions. But you may have meant: $$ \exists i: i\in \{\} $$ $$ \forall i: i\in \{\} $$ The second one says that every $i$ is an element of the empty set. This proposition is false. It's enough to find one counterexample. Barack Obama is not an element of the empty set. So it's false because the condition ($i$ is an element of the empty set) doesn't hold for every $i$.

The first one is weaker but it is still false. To prove that it is true, we would have to find an element $i$ that is an element of the empty set. But because the empty set has no elements, you can't find any. :-) The situation when the condition (in this case, $i$ is an element of the empty set) has exactly zero solutions is the way - and the only way - how the existential quantifier may fail.

Your comments about the graphs are far more complicated than the two simple propositions above but it is true that one must safely understand the logic of the two propositions above to be sure that he can evaluate more complex existential and universal propositions about the graphs (and everything else in maths), too.