why$(\forall x \in U, P(x)) \implies (\exists x \in U, P(x))$ is false?

Question

may I have a complete proof of that "$(\forall x \in U, P(x)) \implies (\exists x \in U, P(x))$ " is false? thx guys

Answer 1

Observe that $\forall x\in U \; P(x)$ is defined as $\forall x\;(x\in U\implies P(x))$ and $\exists x\in U \; P(x)$ is defined as $\exists x\;(x\in U \text{ and } P(x))$.

So, when $U=\emptyset$, there is no $x$ such that $x\in U$ and so for all $x$, $x\in U$ is false. Thus the statement $x\in U\implies P(x)$ is vacuously true for any $x$ and thus $\forall x\in U \; P(x)$ is true.

It isn't hard to see that $\exists x\in U\; P(x)\equiv \exists x\;(x\in U \text{ and } P(x))$ is false since there is no $x$ with $x\in U$.

Hence for $U=\emptyset$, your statement is an implication with a true antecedent but a false consequnt.

For $U\not= \emptyset$, your statement is true.

Answer 2

It is true that "every purple elephant-sized duck owns polka dotted socks," for example. But there are no purple elephant-sized ducks.