Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x \in A \implies \forall \epsilon: U_\epsilon(x) \cap A \neq \emptyset$

share|improve this question
2  
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
1  
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
add comment

1 Answer

up vote 10 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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.