Closed set with rationals question

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SUMMARY

The discussion centers on the mathematical proof that if a closed set A contains every rational number in the closed interval [0,1], then [0,1] must be a subset of A. Participants clarify that the set of all rationals in [0,1] is not closed, as every point in this set is a boundary point, but this does not imply that [0,1] is contained within A. The key takeaway is that the closure of the rationals in [0,1] includes all irrational numbers in that interval, confirming that [0,1] is indeed a subset of A.

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Homework Statement


If A is a closed set that contains every rational number in the closed interval [0,1], show that [0,1] is a subset of A.

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The Attempt at a Solution


I'm confused because for the set A = all rationals in [0,1], every point is a boundary point so the set is closed. but clearly [0,1] is not a subset of A. This question is from Spivaks Calculus on Manifolds, question 1-19
 
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vancouver_water said:
every point is a boundary point so the set is closed
Though the boundary of a set is closed, that doesn't mean that a set A consisting only of boundary points is necessarily closed (the boundary of A could contain points outside of A too).
The set of all rationals in [0,1] is not closed.

You could try to prove that any (irrational) number in [0,1] is in the closure of the rationals in [0,1].
 
Last edited:
oh right I didn't realize the rationals was not closed, I think I got it now, Thanks!
 

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