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Here is the question I am trying to answer: Let $\mathbb{R}$ denote the reals and suppose A is a countable subset of $\mathbb{R}$. Show that $\mathbb{R} \sim \mathbb{R} \setminus A $.

My attempt:

Because $\mathbb{R}$ is uncountable and the subset $A$ of $\mathbb{R}$ is countable, $\mathbb{R} \setminus A $ is uncountable. Therefore $ \mathbb{R} \sim \mathbb{R} \setminus A $ .

My question: Is this specific enough or do I need to find a better way to show that $\mathbb{R} \sim \mathbb{R} \setminus A $? If I need to be more specific, do I have to come up with a function $f$ such that $f$ maps $\mathbb{R}$ onto $ \mathbb{R} \setminus A$ such that the function is 1-1 and onto? If I do need to create a function, any helpful hints to help me create this function?

Thanks a lot.

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    It is not enough just to show that $\mathbb R \setminus A$ is uncountable. We do not know whether $\aleph_0 < |X|\le |\mathbb R| \implies|X|=|\mathbb R|$. This is the celebrated [Continuum Hypothesis](https://en.wikipedia.org/wiki/Continuum_hypothesis). – 2017-02-25
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    Try to prove that if $|a|\leq |b|$, then $|a\cup b|=|b|$. I think it will be useful in your future. – 2017-02-25

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The first step is to find a countably infinite set $B\subseteq\mathbb R\setminus A.$ If $A$ is finite this is easy, just let $B=\mathbb N\setminus A.$ If $A$ is countably infinite, you can adapt Cantor's diagonal argument to construct an infinite sequence of distinct elements of $\mathbb R\setminus A.$

Let $S=\mathbb R\setminus(A\cup B).$ Then $\mathbb R$ is the disjoint union of $S$ and $A\cup B,$ while $\mathbb R\setminus A$ is the disjoint union of $S$ and $B.$ Now all you have to do is find a bijection between $A\cup B$ and $B.$

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    I think you must mean $S=\mathbb R\setminus(A\cup B)$. – 2017-02-25
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    Thanks a lot! That was really helpful – 2017-03-01