# Cardinality of an infinite set minus an infinite subset

$\DeclareMathOperator{\card}{card}$ Suppose $\card(C)= \card(D)$, $D \subseteq C$, and $\card(C\setminus D)$ is infinite.

I want to show that $\card(C\setminus D) = \card(C)$.

This is easy enough for me to see in some specific cases. For example, $\card(\Bbb N=\card(\Bbb Q)$, $\Bbb N\subseteq\Bbb Q$, and $\card(\Bbb Q\setminus\Bbb N) = \card(\Bbb Q)$.

But because subtraction of two equal, infinite cardinals is not well defined, I am unsure of how to prove the general case. What would be a good method to go about it?

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$C = \mathbb{R}$, $D = \mathbb{R} \setminus \mathbb{Z}$. – WimC Dec 11 '12 at 20:10

## 2 Answers

It is not true. Let $C=\mathbb R, D=\mathbb R \setminus \mathbb Q$

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Either the question says that $C$ is countably infinite, or I'm assuming that you haven't learnt about uncountable sets yet. There are at least 2 types of basic infinities, so you should make this question more specific. In the example given by Ross Millikan, you have that $C$ is uncountably infinite, and so is $D$, but $C-D$ is only countably infinite. Look up the wikipedia articles on types of infinities if you're unfamiliar with them.

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Ah, no wonder I couldn't prove it, since it's false. What if you had a bijection f from C to D, so that you could write D = f ⟦C⟧. Would the above ever be true? – Rebecca Dec 11 '12 at 20:28
@Rebecca: You're already assuming that. You assume that $C$ and $D$ have equal cardinalities, which is just another way of saying that there's a bijection from $C$ to $D$. – Chris Eagle Dec 11 '12 at 20:37