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