# Closed and Open Sets Question

## 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?

## Answers and Replies

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matt grime
Homework Helper
Could you rephrase the question please. I'll guess that you want to consider [a,b]nQ, the interesction of the interval with the rationals, and that you mean the subspace topology on Q.

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

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

Ok, so maybe looking at this concrete example will help clear things up for me.

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?

AKG
Homework Helper
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 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 is not open in the Reals since when a= 3^1/2, we cannot find b>a such that b is 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 in R, then for each x in S, there is an open interval I such that x is in I, and I is a subset of S. Note that an open interval in Q is different from an open interval in R. An open interval of Q around each x in S exists such that Q is a subset of S. An open interval of R around some x in S would contain rationals and irrationals, but since S contains only rationals, this open interval cannot be a subset of S.

Good point. I forgot that 3^1/2 wouldnt be included in my set since it is irrational. How do I write down the complements for set S then? Can I still write what I previously wrote down?

Ed Quanta said:
Good point. I forgot that 3^1/2 wouldnt be included in my set since it is irrational. How do I write down the complements for set S then? Can I still write what I previously wrote down?
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.

matt grime
Homework Helper
I really think you should look caerfully at the definition of subspace topology. U is open in Q iff there is an open set V in R such that QnV=U. Similarly for closed subsets. If you just think about the definitions the answer is quite easy.
(and as in every topological space, Q itself is open and closed in the subspace topology, but neither in R).

I get the impression he's trying to approach it from a purely set theoretic treatment, such as for a class, and may not have the benefit of treating it topologically.

inquire4more said:
I get the impression he's trying to approach it from a purely set theoretic treatment, such as for a class, and may not have the benefit of treating it topologically.
How do you prove that a set is open or closed without refering to the underlying topology?

Hurkyl
Staff Emeritus