If every element $a \in A$ is even, then why is it false that some $a \in A$ is even?

Question

Suppose $A$ is a set.

If every element $a \in A$ is even, then some $a \in A$ is even.

Why is this a false statement?

Consider $A=\varnothing$. – hmakholm left over Monica Mar 16 '16 at 14:17 — hmakholm left over Monica, Mar 16 '16 at 14:17

score 3 · Answer 1 · answered Mar 16 '16 at 14:18

3

Because it's not true for the empty set. Also, that is the only set it is not true for.

You may try to prove the latter (and the former too actually), then you will perfectly understand it.

answered Mar 16 '16 at 14:18

peter.petrov

12,568

score 3 · Answer 2 · answered Mar 16 '16 at 14:18

The statement is false, as $A = \emptyset$ is a counterexample: Every element of $\emptyset$ is even (as there aren't any elements), but there is no element of $\emptyset$ which is even.

The following statement is true:

Suppose, $A$ is a non-empty set. If every $a \in A$ is even, then some $a \in A$ is even.

If every element $a \in A$ is even, then why is it false that some $a \in A$ is even?

2 Answers2