Discrete mathematics subsets Suppose I have two sets A and B:
$$ A = \lbrace 2k-1 : k \in \mathbb{Z}\rbrace$$ 
$$ B = \lbrace 2l+1 : l \in \mathbb{Z}\rbrace$$ 
I need to prove that A = B. 
I know that to prove equality between two sets I need to prove both:
$$ A \subseteq B $$
and
$$ A \supseteq B $$
I tried to start with something like :
Suppose x is an element of A, then 
$$ x = 2k - 1 $$
EDIT : Which we can rewrite as
$$ x = 2k + 1 - 2$$
$$ x  = 2(k-1) + 1$$
Because $$ (k-1) \in \mathbb{Z} $$
We know that x is also in B.
Is this the correct way of approaching this problem?
 A: An alternate way of doing the same thing is to use the set definitions :  
$A=\{2k-1:k\in \mathbb{Z}\}$  
$ $ $  $ $ $ $ $ $=\{2(l+1)-1:l\in \mathbb{Z}\}$  
$ $ $  $ $ $ $ $ $=\{2l+2-1:l\in \mathbb{Z}\}$  
$ $ $  $ $ $ $ $ $=\{2l+1:l\in \mathbb{Z}\}$ 
$ $ $  $ $ $ $ $ $=B$  
This actually takes care of both $A\subseteq B$ and $B\subseteq A.$
A: Hint: 


*

*$2k-1=2l+1$ where $l=(k-1)$;

*$2l+1=2k-1$ where $k=l+1$.

A: Your approach is correct, but just to help give you an example in writing out your mathematical argument to give you some confidence, I would write it like this:
Suppose $x \in A$. Then there exists a $k \in \mathbb{Z}$ such that $x = 2k - 1 = 2l + 1$ where $l = k - 1$ (following the hint above/what you wrote). Since $k \in \mathbb{Z}$, $l \in \mathbb{Z}$, so $x \in B$.
See if you can write out the $B \subseteq A$ direction. Hope that helps!
A: Let $x\in A.$ Then $x=2k-1$ for some $k\in \mathbb{Z}.$ Also since $k\in \mathbb{Z}, k=l+1$ for some $l\in \mathbb{Z}.$  
So $x=2(l+1)-1=2l+2-1=2l+1$ so that $A\subseteq B.$ Similarly,  
let $x\in B,$ then $x=2l+1$ for some $l\in \mathbb{Z}.$ Again by same arguement as above we can show that $B\subseteq A$ and hence $A=B.$
