Technique of decomposing a real interval into intervals

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The discussion focuses on the technique of decomposing real intervals into intervals with rational endpoints, specifically addressing half-open and closed cases. The user presents the decomposition of the interval [0,t) as the union of intervals [0,q) for all rational numbers q less than t. For the closed interval [0,t], the user suggests using an intersection of intervals [0,q) for all rational q greater than t, confirming that t is included in this intersection while no number greater than t is. This highlights the relationship between rational numbers and real intervals in mathematical analysis.

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wayneckm
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Hello all,

I always come across the technique of decomposing a real interval into intervals with rational end point, however, I am a bit confused with the half-open/half-closed cases. For example,

[tex][0,t) = \cup_{q < t, q \in \mathbb{Q}} [0,q)[/tex]. And for the case of [tex][0,t][/tex], we can only construct from using "outer sense", meaning that using all rational [tex]q > t[/tex]?

Also, what is the set of [tex]\cup_{q < t, q \in \mathbb{Q}} [0,q][/tex]?

Thanks.
 
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Actually I don't think it matters if you take [0, q) or [0, q] in the union, because there will always be a rational number arbitrarily close but smaller than t. In other words, for any [itex]q < t, q \in \mathbb{Q}[/itex] you can always find [itex]q' \in \mathbb{Q}[/itex] such that q < q' < t.

For the case of [0, t] I think you should be taking an intersection, like
[tex][0, t] = \cap_{q > t, q \in \mathbb{Q}} [0, q)[/tex]
(or, again, [0, q] will do).
You can check that t will be contained in the intersection (it is in every interval of the form [0, q) with q > t) but no number t' > t is (because you can always find a rational q such that t < q < t').
 

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