0

How can I show that sets $| \mathbb R-\mathbb Q |$ and $ | \mathbb R |$ have same cardinality.
My Solution: As long as for element in one set I can match it with unique element
in another set they have same cardinality:
After removing $\mathbb Q $ from $\mathbb R$ I can shift elements of Irrational numbers to correspond to elements of two sets match like this: $ \begin{array}{c|c|c|c|c|c} \mathbb R&0&1&\cdots&n&\cdots\\\hline \mathbb R\setminus\mathbb Q & \sqrt{2} &\sqrt{5} &\cdots & \sqrt{n+1} &\cdots \end{array} $
Note that I use n as Real number for simplicity.

Even if this solution is correct, I feel like Professor won't accept it on exam.
How would formal solution look like?

Baimyrza Shamyr
  • 211
  • 2
  • 12
  • 2
    What is supposed to be the image of $2$ with your shift? – md5 Jan 06 '17 at 13:21
  • 2
    Your problem is a special case of the one dealt with in [this question](http://math.stackexchange.com/questions/355049/let-a-be-any-uncountable-set-and-let-b-be-a-countable-subset-of-a-prove). – Brian M. Scott Jan 06 '17 at 13:24
  • I'm afraid you are right -- your attempt is very far from being a proof. For instance, what is the image of $\sqrt 2$ ? (Perhaps this is what @md5 meant to ask.) It can't be $\sqrt 2$, because that is already the image of $0$. – TonyK Jan 06 '17 at 13:25
  • @md5 shift is just an example. I agree it is bad example, but the best came to my mind. – Baimyrza Shamyr Jan 06 '17 at 13:27
  • 2
    @AlexMacedo it's not a duplicate of that one. –  Jan 06 '17 at 13:28
  • The number of algebraic irrationals is countable, the big group is the number of the irrationals that are transcendental. – Masacroso Jan 06 '17 at 13:47

2 Answers2

3

Define $$f(x)=\begin{cases}x+\sqrt 2\text{ if } x-n\sqrt 2\in\Bbb Q\text{ for some integer }n\ge 0 \\ x\text{ otherwise}\end{cases}$$

ajotatxe
  • 62,632
  • 2
  • 52
  • 101
-2

Clearly, $\mathbb{R} \setminus \mathbb{Q}$ has at most the same cardinality as $\mathbb{R}$ and is an infinite set. The set can therefore be either countable or of the same cardinality as $\mathbb{R}$.

Assuming the continuum hypothesis, if $|\mathbb{R} \setminus \mathbb{Q}| \neq |\mathbb{R}|$, we must then have that $\mathbb{R} \setminus \mathbb{Q}$ is countable.

But the union of two countable sets is countable, and $(\mathbb{R} \setminus \mathbb{Q}) \cup \mathbb{Q} = \mathbb{R}$. Therefore $\mathbb{R} \setminus \mathbb{Q}$ must have the same cardinality as $\mathbb{R}$.

Hans Hüttel
  • 4,098
  • 2
  • 12
  • 22