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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.2012-01-15
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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).2012-01-15
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    As in David Mitra's answer, you need to specify "non-empty" open sets.2012-01-15
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    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.2012-01-15

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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, to $(0,1)$:

enter image description here

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https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument

Consider the above article. Historically the uncountability of intervals was actually used to prove it for the real numbers. Once you have it for (0,1) the order-preserving map shows it for each interval. It is trivial to extend this to unions of intervals.