What is the cardinality of $P(\mathbb{R})\setminus P(\mathbb{Q})$?

Question

I'm a beginner in elementary set theory and I'm looking for a simple way (I can use facts from cardinal arithmetic) to show that:

$|P(\mathbb{R})\setminus P(\mathbb{Q})|=|P(\mathbb{R})|$

I would like to see several approaches for this. thank you.

Answer 1

Perhaps the simplest approach is to observe that $\wp(\Bbb R)\supseteq\big(\wp(\Bbb R)\setminus\wp(\Bbb Q)\big)\cup\{\varnothing\}\supseteq\wp(\Bbb R\setminus\Bbb Q)$, so it suffices to show that $|\wp(\Bbb R\setminus\Bbb Q)|=|\wp(\Bbb R)|$. This in turn is a straightforward consequence of the fact that $|\Bbb R\setminus\Bbb Q|=|\Bbb R|$: a bijection between sets $A$ and $B$ can easily be used to produce a bijection between $\wp(A)$ and $\wp(B)$.

Answer 2

Brian gave an excellent, axiom of choice free, method for proving this. But here is one which relies a little bit on the axiom of choice¹:

$$|\mathcal P(\Bbb R)|=|\mathcal P(\Bbb R)\setminus\mathcal P(\Bbb Q)|+|\mathcal P(\Bbb Q)|=\max\{|\mathcal P(\Bbb R)\setminus\mathcal P(\Bbb Q)|,|\mathcal P(\Bbb Q)|\}$$

But since $|\mathcal P(\Bbb Q)|=|\Bbb R|<|\mathcal P(\Bbb R)|$ we obtain the wanted equality. The axiom of choice is used in the fact that for infinite cardinals, $|A|+|B|=\max\{|A|,|B|\}$ which in the absence of choice need not be correct, or even well-defined.

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1. If one works hard enough, you can show that under some assumptions which are satisfied by $\Bbb R$, we can show that if $|A|+|\Bbb R|=|\mathcal P(\Bbb R)|$, then $|A|=|\mathcal P(\Bbb R)|$. For details, see this answer.