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## Main Question or Discussion Point

Which intervals [a,b] in the set of reals have the property that (the set of rationals and [a,b]) is both open and closed in the set of rationals? And the property closed but not open?

- Thread starter Ed Quanta
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Which intervals [a,b] in the set of reals have the property that (the set of rationals and [a,b]) is both open and closed in the set of rationals? And the property closed but not open?

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matt grime

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When are sets open and closed in the subspace topology? Ie look at the definition ofopen and closed for the subspace topology.

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Yes, you are right. I meant to say Q is a subspace of the Reals.

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We have the set S={x is an element of Q;2<=x^2<=3}

This set should be open in Q since for any a in Set S, we can find b>a such that b is also in S

This set is not open in the Reals since when a= 3^1/2, we cannot find b>a such that b is in S

This s et is closed in Q since taking the complement of set S, Q/S={(negative infinity, -(3^1/2)) V (-(2^1/2),2^1/2) V (2^1/2,infinity)], we can see Q/S is open

This set is not closed in the Reals

Did I mess this up at all?

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AKG

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This doesn't prove that it's open, but it's on the right track. You have to show something similar for being able to find a b < a, which you can easily do. But I would also try to be more precise. The above would be enough to show that it is open, but that's because it implies something else, which in turn, by definition, implies that the set is open. To be rigorous, you should show these implications (and it should be very easy).This set should be open in Q since for any a in Set S, we can find b>a such that b is also in S

But 3^1/2 is not in S. Note that S only contains rationals, and the root of 3 is irrational. This set is not open in the reals, but for a different reason. If S is open inThis set is not open in the Reals since when a= 3^1/2, we cannot find b>a such that b is in S

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S doesn't satisfy the requirements mentioned in the original question; you were looking for intervals in the reals, but S is only an interval in the rationals.Ed Quanta said:

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matt grime

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(and as in every topological space, Q itself is open and closed in the subspace topology, but neither in R).

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How do you prove that a set is open or closed without refering to the underlying topology?inquire4more said:

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Hurkyl

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How does "open" and "closed" even have any meaning without topology?

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