Originally Posted by

**james121515** I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?

$\displaystyle \mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $\displaystyle B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $\displaystyle A=B$

$\displaystyle \mbox{\textbf{Proof.}}$ Let $\displaystyle x\in A$. then $\displaystyle \exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently,

$\displaystyle \Longrightarrow x=2k+1+1-1$

$\displaystyle \Longrightarrow x=2k+2-1$

$\displaystyle \Longrightarrow x=2(k+1)-1$

Since $\displaystyle k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $\displaystyle x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $\displaystyle A\subseteq B$