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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.

Community
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Victor
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  • Well, clearly we have $|B-A| \leq |B|$, by inclusion. So what you need to do is find an injection $B \to (B-A)$. – Arthur Nov 29 '14 at 16:58
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    That depends a lot of what you already know about cardinals. You will need the Axiom of Choice somewhere either in the proof or in some prerequisite that you already know. – hmakholm left over Monica Nov 29 '14 at 16:59
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    Please provide more details, or answers may end up not being helpful since they may assume results or language you are unfamiliar with. What background results have you established at this point? Do you understand the definition of $|A|<|B|$ (please state it)? – Andrés E. Caicedo Nov 29 '14 at 17:24
  • Thanks for the edit. In this question it is shown that $A$ and $A\times{1,2}$ have the same size for any infinite set $A$. If you are familiar with the Bernstein-Cantor theorem that if $|C|\le |D|$ and $|D|\le|C|$ then $|C|=|D|$, this easily gives you what you need. – Andrés E. Caicedo Nov 29 '14 at 23:14
  • Sorry, @AndresCaicedo, I'm familiar with Cantor-Bernstein, but I don't understand how |A| = |A × {1, 2}| will help me... =/ – Victor Nov 29 '14 at 23:33
  • OK, let's try this: Are you familiar with the fact that any two infinite cardinals are comparable? (That is, if $C,D$ are infinite, then either $|C|\le|D|$ or $|D|\le|C|$.) – Andrés E. Caicedo Nov 29 '14 at 23:38

2 Answers2

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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

$$ |B| = |A| + |B\setminus A| = |B\setminus A| $$

Rudy the Reindeer
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  • Hi, Rudy, thanks for the answer. However, as we haven't seen the result κ+λ=max{κ,λ} in the course I'm taking, I will have to prove it in order to be able to use it in another proof... I have thought some time about the proof for it, but it doesn't seem trivial as well. – Victor Nov 30 '14 at 19:38
  • @Victor Ok, can you post the solution in an answer of your own then? By now you probably know the answer and I'd be interested to see it too. – Rudy the Reindeer Dec 06 '14 at 23:16
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@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):

  1. λ ≤ κ + λ, for any cardinals κ and λ;
  2. If κ1 ≤ λ1 and κ2 ≤ λ2, then κ1 + κ2 ≤ λ1 + λ2;
  3. 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.

Victor
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