Proving Two Sets are Equal - Infinite Sets - Example Let
$$A = \{x | x = 2n+1, n\in\mathbb{Z}\}$$ and
$$B = \{x | x = 2m-21, m\in\mathbb{Z}\}.$$ 
I am trying to prove $A =B.$ I understand that I need to prove  $A\subseteq B$ and $B\subseteq A$; But my difficulty is how to prove it.
Any help would be appreciated!
 A: Let me solve one of the directions. I would like to prove that $A\subseteq B$. How do I do it? I take some arbitrary element of $A$ and show that it must also be an element of $B$.
So take some $x\in A$. By definition of $A$, I can write $x=2n+1$ for some integer $n$. If I can somehow also write $x$ as $x=2m-21$ for an integer $m$, then I know that $x\in B$ by definition of $B$.
To do this, I rewrite $x=2n+1$ as follows, where I underway use the trick of adding $0$ (which does nothing), but I disguise $0$ as $21-21$:
$$x=2n+1=2n+1+(21-21)=2n+(1+21)-21=2n+22-21=2(n+11)-21.$$
But $m=n+11$ is an integer! So if I plug $m=n+11$ into this expression, I get
$$x=2m-21,$$
which exactly means that $x$ is an element of $B$.
A: $2n+1=2n+1+(21-21)=2n+22-21=2(...)-21$ so $A\subset B$. Other side is similar.
A: If $x\in A$, $x=2n+1=2(n+11)-21\in B$ $\Rightarrow \forall x\in A, x\in B \Rightarrow A\subseteq B$.
If $x\in B$, $x=2m-21=2(m-11)+1\in A$ $\Rightarrow \forall x\in B, x\in A \Rightarrow B\subseteq A$.
A: Do you know any number theory?
You can prove the sets are equal if you can prove that $x = 2n + 1$ and $x = 2m - 21$ have the same solutions.
Consider the diophantine equation:
$$x - 2n = 1$$
By Bezout's lemma, this has a solution as long as $\operatorname{gcd}(x,-2) | 1$. In other words, as long as $x$ is odd. 
Etc.
