# Can someone explain why this set is wrong in describing even numbers?

This is not the set of even nummbers: $$\{n \text{ is an integer : for all integers } a, n=2a\}.$$

But this is: $$\{n \text{ is an integer : there exists an integer } a \text { such that }n=2a\}.$$

I was told that the number 6 is in the second set, but not the first. I've been thinking this example over, but I don't understand. Could someone explain?

• The first would require that $n=2a$ for all integers a. There is no integer $n$ that satisfies that, so the first set is empty. $6$, for example, may be $2\times 3$ but it is not $2\times 100$ so it fails the test.
– lulu
Sep 14 at 18:11
• @lulu Why are you answering in a comment? Sep 14 at 18:14
• You are confusing $\forall$ with $\exists$. The incorrect one is $\forall$, while the correct one is $\exists$. Sep 14 at 18:28

$$A=\{n \text{ is an integer : for all integers } a, n=2a\}\\ =\{n \in\mathbb Z: \forall a\in\mathbb Z\;\,n=2a\}\\ =\text{the set of integers such that each one is double }\textit{every }\text{ integer}\\ =\emptyset.$$

Since no integer is simultaneously twice of $$-5,$$ twice of $$0,$$ twice of $$71,$$ etc., the set $$A$$ has no member.

$$B=\{n \text{ is an integer : there exists an integer } a \text { such that }n=2a\}\\ \{n\in\mathbb Z: \exists a\in\mathbb Z\;\,n=2a\}\\ =\text{the set of integers such that each one is double }\textit{some }\text{ integer}\\ =\{\ldots,-6,-4,-2,0,2,4,6,8,\ldots\}\\ =2\mathbb Z.$$

Set $$B$$ is populated precisely with the even integers:

• take some (any) integer, then double it; the result is a member of set $$B$$;
• repeat infinitely.

Let $$A=\{\,n\in\Bbb Z\mid \forall a\in\Bbb Z\colon n=2a\,\}.$$ Assume $$n\in A$$. Then $$\forall a\in\Bbb Z\colon n=2a.$$ By specialization $$a\leftarrow 21$$, $$n=2\cdot 21=42$$ By an alternative specialization $$a\leftarrow 333$$, $$n=2\cdot 333=666.$$ From these, $$42=666,$$ contradiction. We conclude that $$n\notin A$$. In other words $$A=\emptyset.$$

Given how the sets are expressed, take the first one.

$$\{n\in\Bbb Z|\forall a\in \Bbb Z, n=2a\}=\{ \hspace{2mm}\}$$ meaning that the set is empty.