Empty domains and the vacuous truth

[Danijel]

So, here's my question. I read somewhere that all universal truths on empty domains are vacuously true, whereas all existential are false. However, if all statements of the form (∀x∈A)(P(x)) , where A is an empty set, are vacuously true, then the statement (∃x∈A)(P(x)) should also be true, because if something holds for all x, then there obviously exists an x for which the statement holds (in fact, it holds for every x). Am I wrong?

 

[PeroK]

So, here's my question. I read somewhere that all universal truths on empty domains are vacuously true, whereas all existential are false. However, if all statements of the form (∀x∈A)(P(x)) , where A is an empty set, are vacuously true, then the statement (∃x∈A)(P(x)) should also be true, because if something holds for all x, then there obviously exists an x for which the statement holds (in fact, it holds for every x). Am I wrong?

Yes, you are wrong! The point about the empty set is that it has no members, so no existential proposition can hold.

Your logic that "all" implies "at least one" is true in all cases except when dealing with the empty set.

 

[PeroK]

PS there is only one empty set. You might want to use the existential and universal logic to prove that. You shouldn't really talk about "an" empty set, therefore!

 

[fresh_42]

So, here's my question. I read somewhere that all universal truths on empty domains are vacuously true, whereas all existential are false. However, if all statements of the form (∀x∈A)(P(x)) , where A is an empty set, are vacuously true, then the statement (∃x∈A)(P(x)) should also be true, because if something holds for all x, then there obviously exists an x for which the statement holds (in fact, it holds for every x). Am I wrong?

Yes, you're wrong.
All Marsians have blue eyes is true, because you cannot find a counterexample.
There is a Marsian is already wrong, regardless of which property will follow.
So in order for $(\exists x\in A)(P(x))$ to be true, $A\neq \emptyset$ has first to be true, because it is equivalent to $A \cap \{x\, : \,P(x)\} \neq \emptyset$ so both sets have to be non-empty. $(\forall x \in A)(P(x))$ is equivalent to $A \subseteq \{x\, : \,P(x)\}$ and $\emptyset \subseteq S$ for all sets $S$, more or less per definition.

 

[Danijel]

So, then, should the statement (∀x∈A)(∃y∈A)((x,y)∈R) , where A is the empty set and R a relation on the empty set (hence, empty relation), also be false? Or do we ignore everything that comes after ∀x∈A and consider the part (∃y∈A)((x,y)∈R) as some P(x,y)?

 

[Danijel]

Also, is saying there exists A with property B, the same as, for some A holds the property B?

 

[fresh_42]

So, then, should the statement (∀x∈A)(∃y∈A)((x,y)∈R) , where A is the empty set and R a relation on the empty set (hence, empty relation), also be false?

No. $(\forall x \in \emptyset\, : \, \text{ false }) \text{ true }$

Or do we ignore everything that comes after ∀x∈A and consider the part (∃y∈A)((x,y)∈R) as some P(x,y)?

Yes.

Also, is saying there exists A with property B, the same as, for some A holds the property B?

Yes. $\exists A \in \mathcal{S}\, : \, B \Longleftrightarrow \{S \in \mathcal{S}\, : \,B\} \neq \emptyset$

 

[Danijel]

Thank you.

 

[TeethWhitener]

It might also help to consider that $\exists x {}Px$ is equivalent to $\neg \forall x {}\neg Px$. Thus $\exists$ and $\forall$ must have opposite truth values for an arbitrary predicate over an empty domain.

 

[Mark44]

Marsians

In English, we write Martians, but not for any reason I understand.

 

[Erland]

This could be wieved as a consequence of the way understand implication in formal logic: recall that the implication $P\rightarrow Q$ is defined to be true if $P$ is false.

Now, $(\forall x \in A)\, P(x)$ is considered an abbreviation of $\forall x (x\in A \rightarrow P(x))$. Thus, if $A=\varnothing$, then $x\in A$ is false for all $x$, and hence $x\in A \rightarrow P(x)$ is true for all $x$, which means that $\forall x (x\in A \rightarrow P(x))$ is true, no matter what $P(x)$ stands for.

On the other hand, $(\exists x \in A)\,P(x)$ is considered as an abbreviation of $\exists x (x \in A \land P(x))$, so if $A=\varnothing$, then $x \in A \land P(x)$ is false for all $x$, which means that $(\exists x \in A)\,P(x)$ is false.

So, $(\exists x \in A)\,P(x)$ is not a logical consequence of $(\forall x \in A)\, P(x)$, since if $A=\varnothing$, then the latter is true and the former is false.