How can we show that any finite cover of A also covers the interval (0,1)?

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SUMMARY

The discussion centers on proving that any finite collection of intervals \( I_n \) that covers the set \( A \) of rational numbers between 0 and 1 must satisfy the inequality \( \sum I_n \geq 1 \). The outer measure of \( A \) is defined as the infimum of \( \sum I_n \) over all possible open intervals covering \( A \). Since \( A \) is countable, its outer measure is 0, but any finite cover must also encompass the entire interval (0,1), leading to the conclusion that the sum of the lengths of these intervals must be at least 1.

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Homework Statement


Let A be the set of all rational numbers between 0 and 1. Show that for any "finite" collection of intervals [tex]I_n[/tex] that cover A the following inequality holds: [tex]\sum I_n \geq 1[/tex].


Homework Equations



We are using the definition of the outer measure here. Where the outer measure of A is define as the infimum of [tex]\sum I_n[/tex] where the infimum is taken over all possible open intervals that cover A.


The Attempt at a Solution


I know that the outer measure of A is 0 because A is a countable set. If I consider finite covers of A, then the sum of their lengths obviously add up to 1 or greater. But I still have no sense of direction on where to continue with this problem.
 
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Ok, I have an idea on how to prove this now. Any finite cover of A must also cover the interval (0,1). Thus the sum of the lengths of the intervals that cover A must be greater than 1 since they also cover the interval (0,1). However, I still am having difficulty writing down an argument to show that any finite cover of A also covers the interval (0,1).
 

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