# Proving $|\mathbb{R}-\mathbb{S}|=2^{\aleph_0}$ when $\mathbb{S}\subset R$ is countable [duplicate]

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?

## marked as duplicate by Asaf Karagila♦ cardinals StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 3 '15 at 12:09

• I think it's true that if $A$ and $B$ are infinite, $|B|<|A|$, and $B\subset A$, then $|A\setminus B|=|A|$. – MPW Jun 3 '15 at 2:23
• Sounds true. I'll try to prove it – Whyka Jun 3 '15 at 2:25
• In fact, I think this may characterize infinite sets. – MPW Jun 3 '15 at 2:26
• @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 $a\in A$. – JDH Jun 3 '15 at 2:30
• @JDH: Thanks for the clarification. I'm admittedly weak in this area, I'm ashamed to say. – MPW Jun 3 '15 at 2:44

$\newcommand\R{\mathbb{R}}$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.
• 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 $\mathbb{Q}$. – DRF Jun 3 '15 at 4:46
• 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. – JDH Jun 3 '15 at 11:37
• This version makes more sense, since you are now explicitly using the uncountability of $\mathbb{R}$. Originally that was IMO not obvious. – DRF Jun 3 '15 at 12:05