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As part of a solution to a question I'm doing right now, I need to prove the theorem stated in the title. While it seems quite obvious, I don't know how to actually prove it.
Any help will be very appreciated! Thanks in advance.

Edit: I've tried finding an invertible function from $X \setminus Y$ to $X$, without great success.

Martin Sleziak
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tcb93
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    What have you tried?Please include your attempt in the question. More context would be helpful too. – Vivaan Daga Aug 20 '22 at 18:54
  • First try getting a bijection between $\mathbb{R}$ and $\mathbb{R}\setminus\mathbb{N}$. Hint: Use a bijection between $[0,1)$ and $(0,1)$ to construct the bijection between \mathbb{R}$ and $\mathbb{R}\setminus\mathbb{N}$. Then you can use this bijection to complete the proof. – C Squared Aug 20 '22 at 21:02
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    From how to ask a good question: Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. – jjagmath Aug 21 '22 at 10:40

1 Answers1

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You can use these two observations:

  • Under the given assumptions, you have $|X\setminus Y|\ge \aleph_0$.
  • If $|A|\ge\aleph_0$, then $|A|+\aleph_0=|A|$.

We have $|X\setminus Y|\ge \aleph_0$. If $|X\setminus Y|<\aleph_0$ we would get that $|X|=|Y|+|X\setminus Y|\le \aleph_0+\aleph_0=\aleph_0$, a contradiction. So we have the opposite inequality - $|X\setminus Y|\ge\aleph_0$.

Notice that here we're using that cardinalities of any two sets are comparable - which is basically axiom of choice. (Of course, if you are working with some specific $X$ and $Y$, it is possible that you can immediately see that $|X\setminus Y|\ge\aleph_0$.)

You can also have a look at this post: Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice? (Especially if you want an argument avoiding AC.) Several questions linked there are close to yours. (Some of them for some specific sets $X$ and $Y$ rather than in full generality.)

$|A|+\aleph_0=|A|$ whenever $|A|\ge\aleph_0$. This is basically the "Hilbert hotel" argument. The inequality $|A|\ge\aleph_0$ means that we have a "copy" of $\mathbb N$ inside $A$. For this copy we can find a bijection with "two copies" of the same set. (This is the same proof as $\aleph_0+\aleph_0=\aleph_0$.)

Martin Sleziak
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