Cardinality of set difference How to prove this: given an infinite set $B$ and $A\subset B$ such that $|A| < |B|$, then $|B-A| = |B|$?
Progress
So, I do understand the definition of $|A| < |B|$ (it means $|A| \le |B|$ and $|A| \ne |B|$, i.e., there's a bijective function from $A$ to a subset of $B$, but not to $B$).
I've already thought of constructing a bijective function from $B\setminus A$ to $B$, but I haven't come up with any useful idea as to how to construct such function...
I also know the axiom of choice, and the basics about cardinals and ordinals.
 A: Maybe you could use that (assuming the axiom of choice) we have 
$$ \kappa + \lambda = \max \{\kappa , \lambda \}$$
if at least one of the two cardinals is infinite (see here).
You don't mention whether you assume the axiom of choice but you probably do, so I will assume that you do. 
Then if $|B\setminus A| \le |A|$ it follows from $ \kappa + \lambda = \max \{\kappa , \lambda \}$ and cardinal addition $|X|+|Y|=|X\cup Y|$ that  $|B|=|A|+ |B\setminus A| = |A|$. But by assumption $|A| <|B|$ so this is a contradiction hence we must have $|B\setminus A| > |A|$.
If $|B\setminus A| > |A|$ then again from $ \kappa + \lambda = \max \{\kappa , \lambda \}$ and cardinal addition it follows that
$$ |B| = |A| + |B\setminus A|  = |B\setminus A| $$
A: @Rudy,
Thanks, I solved my problem using your answer and the proof for "κ + λ = max{κ, λ} if at least one of κ and λ is infinite". I proved this using the following properties (which have been proved during our classes, so they could be used as premises):


*

*λ ≤ κ + λ, for any cardinals κ and λ;

*If κ1 ≤ λ1 and κ2 ≤ λ2, then κ1 + κ2 ≤ λ1 + λ2;

*κ + κ = κ, if κ is infinite.


So, here is the proof:
If κ ≤ λ, as λ ≤ λ, we get κ + λ ≤ λ + λ, and then κ + λ ≤ λ (because λ has to be infinite). From this and property 1 above, κ + λ = λ = max{κ, λ}.
The proof is analogous if λ ≤ κ.
