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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?

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    $\begingroup$ 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. $\endgroup$
    – lulu
    Sep 14 at 18:11
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    $\begingroup$ @lulu Why are you answering in a comment? $\endgroup$
    – Arthur
    Sep 14 at 18:14
  • $\begingroup$ You are confusing $\forall$ with $\exists$. The incorrect one is $\forall$, while the correct one is $\exists$. $\endgroup$ Sep 14 at 18:28
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$$A=\{n \text{ is an integer : for all integers } a, n=2a\}\\ =\{n \in\mathbb Z: \forall a\in\mathbb Z\;\,n=2a\}\\ =\text{the set of integers such that each one is double }\textit{every }\text{ integer}\\ =\emptyset.$$

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.
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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 $$A=\emptyset.$$

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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.

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