Proof that the even natural numbers are a subset of $\mathbb{N}$ I'm doing some self-studying on discrete mathematics and so far its going well however I came upon a question that whilst I can make sense of and I know that set $A$ is a subset of set $B$, I cannot think of how to express it and the book does not feature an answer for this question.
The question is, prove that $A$ is a subset of $B$, where
$$A = \{ 2n \mid n\in\mathbb{Z}^+\},\quad B = \{ n \mid n\in\mathbb{Z}^+\}.$$
I am somewhat at a loss to the logic of this, both sets are identified as they both have elements that contain positive integers and $n$ is half of $2n$, but then isn't if we were to use numbers: $n = 1$, $2n = 2$ etc... therefore the two are not equal as they will never contain the same numbers, I believe I am thinking about this wrong but am somewhat at a loss with this simple question! Any help is appreciated, thanks!
 A: Recall that a set $S$ is a subset of a set $T$, which we express as $S\subseteq T$, when
$$x\in S\implies x\in T.$$
Thus, in your problem, to show that $A\subseteq B$, what you must show is that, for any $n\in\mathbb{Z}^+$, we have that $2n$ is among the elements of $B$. Note that, because $B$ has precisely the same elements as $\mathbb{Z}^+$, we simply have $B=\mathbb{Z}^+$.
But it's true that $2n\in\mathbb{Z}^+$; we know that $\mathbb{Z}^+$ is closed under multiplication by $2$! Thus we are done.
The user copper.hat explains it quite well in the comment above; in set-builder notation, the variables are dummy variables. That is, they do not continue to have a fixed meaning outside of the set-builder expression. Thus, for example,
$$\mathbb{Z}^+=\{n\mid n\in\mathbb{Z}^+\}=\{m\mid m\in\mathbb{Z}^+\}=\{\xi\mid \xi\in\mathbb{Z}^+\}=\{\star\mid \star\in\mathbb{Z}^+\}$$
and, when you want to prove that $A$ is a subset of $B$, the correct interpretation of what you must show is that
$$\text{for any }x\in A\text{, we have }x=y\text{ for some }y\in B$$
$$\text{for any }n\in\mathbb{Z}^+\text{, we have }2n=y\text{ for some }y\in B$$
$$\text{for any }n\in\mathbb{Z}^+\text{, we have }2n=m\text{ for some }m\in\mathbb{Z}^+$$
A: Ok if $A\subset B$ you have to show that $x\in A \implies x \in B$.
$$x\in A \implies \exists n \in \mathbb{Z}: x=2 n$$ 
Because of $2 \in \mathbb{Z}^+$ and $n \in \mathbb{Z}^+$ and $(\mathbb{Z}^+, \cdot)$ is a semi-group (you just need to say that when $a\in \mathbb{Z}^+$ and $b\in \mathbb{Z}^+$ than 
$$a\cdot b\in \mathbb{Z}^+.$$ 
Because of this there exists a $m$ so that $ 2 n = m \in \mathbb{Z}$ and thats why $A\subset B$
