# Construction of bijection

1. Sep 18, 2008

### soumyashant

Can you prove that $$\mathbf{R}$$ and $$\mathbf{R}-\mathbf{Q}$$ have same cardinality?

One way would be to say that $$\mathbf{R}-\mathbf{Q}$$ is not countable and must have cardinality <= $$\mathbf{R}$$ and invoke the Continuum Hypothesis to conclude that its cardinality is aleph-1 same as that of $$\mathbf{R}$$..

Somehow this does not look appealing...

Can you explicitly construct a bijection and help me to visualise the situation better??

Thanks.

2. Sep 18, 2008

### NoMoreExams

Well the union of Q and R\Q is R right? |Q| = \aleph_0 and |R| = c so if |R\Q| was aleph_0 then you have a contradiction i.e. isn't the union of countable sets, countable?

3. Sep 18, 2008

### HallsofIvy

Staff Emeritus
Yes, that proves that R\Q is not countable. But it does not prove that card(R\Q)= card(R). As soumyashant said, You would need the contiuum hypothesis, that there is no set of cardinality strictly between that of R and that of Q, to finish that proof.

4. Sep 19, 2008

### morphism

Try to prove the following fact: If A is an infinite set and B is a countable set, then $|A \cup B| = |A|$. You shouldn't need to invoke the CH.