Can Open Subsets of Real Numbers Form Countable Unions of Intervals?

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Open subsets of real numbers can be expressed as disjoint unions of open intervals, and these can indeed form countable unions. A bijection can be established between the natural numbers and the open intervals within the set, demonstrating that the cardinality of the open subset matches that of the natural numbers. This implies that any open subset E is countable. The discussion also briefly touches on the density of rational numbers within the real numbers, reinforcing the concept of countability. Overall, the conclusion is that open subsets of real numbers are countable unions of disjoint open intervals.
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Question:
Given that any open subset E of the set of real numbers is a disjoint union of open intervals.
Is E a countable union of disj. opn intervls.

Answer:

Yes it is. to show this we need to find a Bijection from the set of natural numbers to E.

E = disjoint U_(i in N) of (a_j , b_j) with j in N and a_j , b_j in R

consider then g: N to E with f(n) = i

this is surely a bijection. Hence |E| = |N| hence E is countable.

?

thanks
 
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Hint: The rationals are dense in R.

Bachelier said:
Question:
Given that any open subset E of the set of real numbers is a disjoint union of open intervals.
Is E a countable union of disj. opn intervls.

Answer:

Yes it is. to show this we need to find a Bijection from the set of natural numbers to E.

E = disjoint U_(i in N) of (a_j , b_j) with j in N and a_j , b_j in R

consider then g: N to E with f(n) = i

this is surely a bijection. Hence |E| = |N| hence E is countable.

?



thanks

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