Confirm the following: if $\forall x \, P(x)$ then $\exists x\ P(x)$

Question

I tried to find the answer to this question, but surprisingly, I was unable to. I just want to confirm that if something is true for all objects in a set, it is certainly true for some objects in a set.

Mostly, I want to make sure that I am correctly understanding the meaning of "there exists" in the logical context. I am interpreting "there exists" as meaning "there exists a a subset of elements that..."

So, to provide an illustration:

Where the color green means that whatever element is found at a particular location is "true" for $P(x)$.

"There exists x" is strictly talking about the truth/false evaluation of whatever is within the boundary of the subset, correct? No comment is being made about anything outside of that boundary, yes?

Any insight is greatly appreciated!

As a silly example:

If all dogs are mammals, then some dogs are mammals.

Answer 1

It's just one of those quirks of standard first-order logic. A non-empty domain of discussion is usually assumed though this assumption is never made explicit in proofs. In mathematics, quantifiers are usually restricted to one or more sets which may or may not be empty.

Example: For an arbitrary set $S$, it does not follow from $\forall x \in S: P(x)$ that $\exists x\in S:P(x)$ since $S$ may be empty.

Answer 2

Mostly, I want to make sure that I am correctly understanding the meaning of "there exists" in the logical context. I am interpreting "there exists" as meaning "there exists a a subset of elements that..."

Almost. It means exactly: "there is a non-empty subset of elements that..."

Why, then, should we have that every element in $S$ will satisfy the predictate $P$, therefore it shall follow that the subset of elements in $S$ that satisfies the predicate will not be empty unless $S$ itself is empty.

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In order to show $\exists x{\in}S~.P(x)$ is true, we must demonstrate that the subset $\{x\in S: P(x)\}$ is not empty (there is at least one thing in $S$ that satisfies $P$).   When every element in $S$ does satisfy the predicate, then that subset is in fact $S$ itself.

In order to show $\forall x{\in}S~.P(x)$ is true, we need demonstrate that the subset $\{x\in S: \lnot P(x)\}$ is empty (there is nothing in $S$ that does not satisfy $P$).

So let us consider the emptyset.   $\forall x{\in} \emptyset~.P(x)$ is true and $\exists x{\in}\emptyset~.P(x)$ is false because the only subset of the emptyset is the emptyset.

Answer 3

Your understanding for the $\forall$ and $\exists$ quantifier is correct and also the analogy which you have created through Venn diagrams.

The First Order Formula $\forall x P(x) \implies \exists x P(x)$ is a valid formula if the domain is non-empty.

Case 1: Domain D $\in \emptyset$:
If the domain is empty then:

- The $\forall$ is true for any Predicate/Parameterised Proposition $P(x)$ because there is no element in D for which $P(x)$ will be false. i.e $\forall x (x \in D \implies P(x))$
- The $\exists$ is false for any Predicate/Parameterised Proposition $P(x)$ because there is no element for which $P(x)$ is true. i.e You cannot find a subset for which $P(x)$ will be true.

Therefore, $\forall x P(x) \implies \exists x P(x)$ will be false if the domain is empty because hypothesis/premise $\forall x P(x)$ is true but the conclusion $\exists x P(x)$ is false, hence implication will be false.

Case 2: Domain is non-empty:
If the domain is non-empty then $\forall x P(x) \implies \exists x P(x)$ is a valid formula.

A formula involving predicate variables is valid if it is true for every domain no matter how the predicate variables are interpreted.

Suppose take the interpretation for $P(x) :\text{'x' is a Prime number}$; Domain: $N$.
$\forall x P(x)$ says that all natural numbers are prime numbers, which is false i.e $\forall x P(x)$ is false.
$\exists x P(x)$ says that there is a sub-set of natural numbers which are prime numbers. i.e ${2,3,5,7,11,...} \subset N$, $\exists x P(x)$ is true.
Therefore, $\forall x P(x) \implies \exists x P(x)$ is vacuously true, as the implication is true when the hypothesis is false.

Suppose take another interpretation for $P(x) : \text{'x = x'}$; Domain: $Z$.
$\forall x P(x)$ says that all intergers are equal to itself which is true i.e $\forall x P(x)$ is true.
$\exists x P(x)$ says that there is a subset of intergers in $Z$ which are equal to itself which is true i.e $\exists x P(x)$ is true.
Therefore, $\forall x P(x) \implies \exists x P(x)$ will be true, as the implication is true when hypothesis and conclusion are true.

Answer 4

@DanChristensen gave a good answer, and I like it. Only due to some participants' misgivings I've decided to expand on Dan's answer.

First, before a strict answer, let me mention that in the everyday mathematical practice, as opposed to mathematical logic, we never truly write $\ \forall_x\,P(x)\ $ even when we do. When we do then we truly mean something like $\forall_x\,((x\in A)\Rightarrow P(x)).$ This is abbreviated to $\ \forall_{x\in A}\,P(x).\ $

For instance, $\ \forall_n\,P(n) \ $ often means $\ \forall_n\,((n\in\mathbb N)\Rightarrow P(n))\ $ or similar.

Now, back to the strict answer. Mathematical logicians and model theorists consider general theories while there is only one trivial empty theory. Thus then, it makes sense to talk about meaningful theories only which have a non-empty domain. The exclusion of this one empty model simplifies logicians' life, they are happy to write $ \forall_x\,P(x)\Rightarrow\exists_x\,P(x)\ $ -- it simply means that logicians assume that the domain is non-empty. You never discuss or build a theory which has empty domain hence you should not complain.

(There seems to be also a historical reason too but let me skip it).