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Is there a way of proving that if $|X|=2^{\aleph_0}$ and $Y\subseteq X$ such that $|Y|=\aleph_0$ then $|X-Y|=2^{\aleph_0}$ without using axiom of choice or continuum hypothesis. Most of the proofs I've seen for this rely on the fact that if $X-Y$ does not have the cardinality of $\mathbb{R}$, then it must be countable (because it is clearly infinite). I was wondering if there is any way around this.

xu_pi
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  • If $Y$ is countable and $X\setminus Y$ is countable, then $X$ is countable. You don't need the axiom of choice to prove that the union of two countable sets is countable. Then $|X|=|X\setminus Y| +\aleph_0=|X\setminus Y|$. Am I missing something? I'm no set theorist, but this seems simple. – saulspatz Jun 10 '20 at 01:54
  • But without the continuum hypothesis how can you know that if $X\setminus Y$ is not uncountable then it is countable. Couldn't it be one of the infinite cardinals inbetween? – xu_pi Jun 10 '20 at 01:59
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    @IanXulBelaustegui: Uncountable simply means not countable; it does not mean of cardinality $2^{\aleph_0}$. If a set is not uncountable, it is by definition countable. – Brian M. Scott Jun 10 '20 at 02:02
  • @BrianM.Scott Yes, sorry. What I meant is if it doesn't have the cardinality of $2^{\aleph_0}$. – xu_pi Jun 10 '20 at 02:07
  • @IanXulBelaustegui: At most you need to know that $X\setminus Y$ contains a countably infinite subset; then you can use the trick described in this question to show that $|X\setminus Y|=|X|$. – Brian M. Scott Jun 10 '20 at 02:12
  • @BrianM.Scott And can we know that such a subset exists? I'm just struggling with the limits of what can be done without axiom of choice. I know we can find a countably infinite subset of any infinite set which is well ordered. And the well ordering theorem is equivalent to the axiom of choice. So axiom of choice implies there is such a set. But is it necessary? – xu_pi Jun 10 '20 at 02:18

2 Answers2

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Yes, it is possible to prove this in just ZF. Let us identify our set $X$ with $\mathbb{R}^2$ (via some bijection, since we know $|X|=|\mathbb{R}^2|=2^{\aleph_0}$). Consider the first projection map $p:Y\to\mathbb{R}$ defined by $p(x,y)=x$. Since $Y$ is countable, $p$ cannot be surjective, so there exists some $x\in\mathbb{R}$ such that $(x,y)\not\in Y$ for all $y\in\mathbb{R}$. That is, $\{x\}\times\mathbb{R}\subseteq X-Y$. Thus $|X-Y|\geq |\{x\}\times\mathbb{R}|=2^{\aleph_0}$ and $|X-Y|\leq|X|=2^{\aleph_0}$, so $|X-Y|=2^{\aleph_0}$.

More generally, a similar argument shows that if $X$ is a set such that $|X\times X|=|X|$, then for any $Y\subseteq X$ such that there is no surjection from $Y$ to $X$, we must have $|X-Y|=|X|$.

Eric Wofsey
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If $X\setminus Y$ contains a countably infinite subset, you can use the trick described in this question to show that $|X\setminus Y|=|X|$. That’s automatic if every Dedekind infinite is infinite, something that follows from but is strictly weaker than the countable axiom of choice, which is a rather weak form of the axiom of choice. It can also be a consequence of the reason for which you know that $|X|=2^{\aleph_0}$. For instance, if you know this because you have a bijection between $X$ and $\Bbb R$ or $\wp(\omega)$, then you have access to many countably infinite subsets of $X$ via the bijection.

Brian M. Scott
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