# How can knowing that $(0,1)$ is uncountable imply $R$ is also uncountable?

I've been considering a proof by contraposition, by stating that if $R$ is countable and $(0,1) \subset R$, then $(0,1)$ is countable. Seeing as we already know that $(0,1)$ is uncountable, that is enough to be a contradiction, is it not? I don't see any other way for this proof to work out....

• It's possible to construct a bijection (in fact a homeomorphism) $(0,1)\to\mathbb{R}$. Try working with tangent and arctangent. Commented Nov 29, 2016 at 5:41
• Yes, the fact that $(0,1)$ is uncountable and is a subset of $\Bbb R$ implies that $\Bbb R$ is uncountable. However, as others have pointed out, it is actually possible to construct a bijection between the two sets, thereby showing that they are even the same cardinality. (Otherwise the possibility would remain open that the cardinality of $(0,1)$ is less than that of $\Bbb R$.) Commented Nov 29, 2016 at 5:44
• $f(x) = {1 \over 2} ({2 \over \pi} \arctan x + 1)$ is a homeomorphism between $\mathbb{R}$ and $(0,1)$. Commented Nov 29, 2016 at 5:49
• Yes, that's good enough to show its uncountable. But that's not the only way. I) (0,1) ~ (n,n+1) so R = U {n,n+1}U Z is a countable union of uncountables. II) x -> x/(1+|x|) is the bijection you seek. Or one of them. There are others that will do. Commented Nov 29, 2016 at 6:53

All that's needed to show that two sets $A$ and $B$ have the same cardinality is a bijection $f:A\to B$. A bijection $f:\mathbb{R}\to(0,1)$ is given by $f(x) = \frac{e^x}{1+e^x}$.
• +1. Note for the OP that this bijection is not unique: we could also take, e.g., $g(x)={1+{2\over \pi}\arctan(x)\over 2}$, or many other options. Commented Nov 29, 2016 at 5:53