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I wish to prove that $|\mathbb{R}-\mathbb{S}|=2^{\aleph_0}$ when $\mathbb{S}\subset \mathbb{R}$ is countable.

I want to say that $|\mathbb{R}-\mathbb{S}|= |\mathbb{R}|-|\mathbb{S}|$ but we haven't studied yet what subtraction of cardinals means (I can guess, though).

How could I prove this using only basic cardinal properties?

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marked as duplicate by Asaf Karagila cardinals Jun 3 '15 at 12:09

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    $\begingroup$ I think it's true that if $A$ and $B$ are infinite, $|B|<|A|$, and $B\subset A$, then $|A\setminus B|=|A|$. $\endgroup$ – MPW Jun 3 '15 at 2:23
  • $\begingroup$ Sounds true. I'll try to prove it $\endgroup$ – Whyka Jun 3 '15 at 2:25
  • $\begingroup$ In fact, I think this may characterize infinite sets. $\endgroup$ – MPW Jun 3 '15 at 2:26
  • $\begingroup$ @MPW That claim is true in ZFC, but it uses the axiom of choice. It is clear for well-ordered sets, since if $\kappa$ is an infinite well-ordered cardinal and $\lambda<\kappa$, then the interval $[\lambda,\kappa)$ must have size $\kappa$. But without AC, it is no longer true, since if $A$ is an infinite Dedekind finite set, then even $A-\{a\}$ is strictly smaller, when $a\in A$. $\endgroup$ – JDH Jun 3 '15 at 2:30
  • $\begingroup$ @JDH: Thanks for the clarification. I'm admittedly weak in this area, I'm ashamed to say. $\endgroup$ – MPW Jun 3 '15 at 2:44
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$\newcommand\R{\mathbb{R}}$Suppose that $S\subset\R$ is a countable set of reals, and consider the complementary set $\R-S$. Since the unit interval is uncountable and more specifically contains uncountably many disjoint countably infinite sets (e.g. small translations of the rationals in some tiny interval), there is a countable set $T\subset[0,1]$ of the same size as $S$ that is disjoint from $S$. Thus, $\R-T$ is bijective with $\R-S$ by simply swapping elements of $S$ for $T$ and fixing all other reals. But $\R-T$ contains the interval $[2,3]$, and so $\R-S$ is at least as large as $[2,3]$, which has the same size as $\R$. And so $\R-S$ has size continuum.

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  • $\begingroup$ I'm not quite sure if this really follows. Consider the following argument. Since the unit interval $[0,1]\cap \mathbb{Q}$ contains a countable set, then $\mathbb{Q}\setminus \mathbb{S}$ is at least as large as $\mathbb{Q}\setminus([0,1]\cap \mathbb{Q})$ which has size at least the interval $[2,3]\cap \mathbb{Q}$ which has the same size as $\mathbb{Q}$ so $\mathbb{Q}\setminus\mathbb{S}$ has size same as $\mathbb{Q}$. $\endgroup$ – DRF Jun 3 '15 at 4:46
  • $\begingroup$ I edited to explain. It is important that the unit interval is uncountable, and the argument doesn't work with $\mathbb{Q}$, since one cannot necessarily find a permutation of $\mathbb{Q}$ that brings $S$ into the unit interval. But in $\mathbb{R}$, there is such a permutation, as I explain. $\endgroup$ – JDH Jun 3 '15 at 11:37
  • $\begingroup$ This version makes more sense, since you are now explicitly using the uncountability of $\mathbb{R}$. Originally that was IMO not obvious. $\endgroup$ – DRF Jun 3 '15 at 12:05

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