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Why are open sets in $\mathbb{R}$ uncountable?

We've proven that $\mathbb{R}$ is uncountable and our definition for the open set A is $x \in A \implies \forall \epsilon: U_\epsilon(x) \cap A \neq \emptyset$

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This is a wrong definition, according to it, $A=\{0\}$ is an open set. Instead, use $\exists \epsilon > 0. U_{\epsilon}(x) \subseteq A$, and it follows that any nonempty open set contains an interval, so it is uncountable. – sdcvvc Jan 15 '12 at 11:30
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In fact, not only is the definition wrong, but it is also a bit silly: every $A \subseteq \mathbb R$ passes as an open set according to it (because, if $x \in A$, then $U_\varepsilon(x) \cap A$ contains at least $x$, and is hence nonempty). – Srivatsan Jan 15 '12 at 11:39
As in David Mitra's answer, you need to specify "non-empty" open sets. – Mark Bennet Jan 15 '12 at 11:43
Can you find a bijection from (a,b) to (-1,1)? A bijection from (-1,1) ro R is x/(x^"+1); off the top of my head so confirm this. – Adam Jan 15 '12 at 12:05

1 Answer

up vote 9 down vote accepted

Every non-empty open set in $\Bbb R$ contains an open interval. Given an open interval $O$, there is a bijection from $O$ to $(0,1)$ (or $\Bbb R$; use the inverse tangent function appropriately altered), which is uncountable.

Here, informally, is a bijection from $(0,1)$ (represented by the semicircle) to $\Bbb R$ (represented by the line):

enter image description here

And one from $(a,b)$, with $0<b-a<1$, to $(0,1)$:

enter image description here

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