Cardinality of sets: prove equality

[Dustinsfl]

$$A=\mathbb{R}-0$$

$$(0,1)\subseteq \mathbb{R}-0$$

Assume A is countable.

Since A is countable, then $$A\sim\mathbb{N}$$.
Which follows that $$(0,1)\sim\mathbb{N}$$.

However, (0,1) is uncountable so by contradiction, Card(A)=c

Correct?

 

[Hurkyl]

The fact that A is not countable, together with the fact it is a subset of R only proves that

$$\aleph_0 < \mathop{Card}(A) \leq c$$​

 

[HallsofIvy]

So it looks like you will have to use the Continuum Hypothesis!

 

[Hurkyl]

Or find a better lower bound.

 

[Martin Rattigan]

Think about the function that maps $n\mapsto n+1$ for $n\in \mathbb{N}$ where $\mathbb{N}$ is regarded as a subset of $\mathbb{R}$.

 

[Dustinsfl]

Thanks for the help