# Thread: The closure of Q is Q?

1. ## The closure of Q is Q?

I was asked to verify whether or not the closure of Q is R.

I said, it sure is, and gave what I thought was sound reasoning.

Then I was presented with a counterexample:

Since Q is countable, denote each element of Q as $x_n$. Then, consider Q $\cup [x_n - 2^{-n},x_n +2^{-n}].$ This set is clearly closed, and since $\sum_{n=0}^{\infty} \frac{2}{2^n} = 2$, the set must be a proper subset of R. I cannot refute this, but it doesn't seem right to me.

2. Originally Posted by davismj
I was asked to verify whether or not the closure of Q is R.

I said, it sure is, and gave what I thought was sound reasoning.

Then I was presented with a counterexample:

Since Q is countable, denote each element of Q as $x_n$. Then, consider Q $\cup [x_n - 2^{-n},x_n +2^{-n}].$ This set is clearly closed, and since $\sum_{n=0}^{\infty} \frac{2}{2^n} = 2$, the set must be a proper subset of R.

What set exactly are you talking about here? You had $\mathbb{Q}\cup [x_n-2^{-n},x_n+2^{-n}]$....so? Apparently you added the lengths of all the

intervals $[x_n-2^{-n},x_n+2^{-n}]^\infty_{n=0}$ and you got, wrongly by the way, 2 (it is 4, but it never minds)...so, again? What is your point?

Tonio

I cannot refute this, but it doesn't seem right to me.
.

3. Are you sure that set is closed?

4. I don't see how this contradicts the fact that the closure of Q is R. How can it constitute a counterexample? You've merely found a proper subset of R that contains Q. By the way, the measure of Q $\cup [x_n - 2^{-n},x_n +2^{-n}]$ is <= 4, since the small intervals are not disjoint, but is certainly positive, since $\ [x_1 - 1/2,x_1 + 1/2]$ has measure 1.

Proof that the clousre of Q is R:

We already know that the closure of Q is contained in R; the converse follows from the fact that Q is dense in R: fix an x in R, then for each positive integer n, there is a rational number $\ a_n$ in $\ (x - 1/n,x + 1/n)$, so this sequence $\ {{a_n}}$ of rational numbers converges to x, and hence every x in R is a limit point of Q => R is contained in the closure of Q.

5. By the way, a countable union of closed sets need not be closed. Is your set closed? It is not clear (to me) what it looks like. For example can it contain open intervals? Certainly it is a proper subset of R, but how does one go about finding a point in its complement?

6. Originally Posted by Alchemist
By the way, a countable union of closed sets need not be closed. Is your set closed?
Surely, it is not closed.

It is not clear (to me) what it looks like.
Never mind that .
Since the closure of $\mathbb{Q}$ is $\mathbb{R}$ the set in question cannot be closed: the closure of a set is the smallest superset that is closed, so any set $X$ with $\mathbb{Q}\subseteq X\subsetneq \mathbb{R}$ cannot be closed.

7. Originally Posted by Failure
Surely, it is not closed.

Never mind that .
Since the closure of $\mathbb{Q}$ is $\mathbb{R}$ the set in question cannot be closed: the closure of a set is the smallest superset that is closed, so any set $X$ with $\mathbb{Q}\subseteq X\subsetneq \mathbb{R}$ cannot be closed.
Thank you everyone. I don't agree with the counterexample, but I'm not sure how to disprove it.

I think we have that since the proposed counterexample is a proper subset of R, we know that there is at least one irrational number not in the set. However, this irrational number is certainly (by the density of Q in R) the limit of sequence of rational numbers, and therefore a limit point of Q. Therefore, the proposed set is not closed. Is that about right?

8. Not every proper subset of R misses an irrational number--the irrationals themselves form a proper subset of R. What are you trying to do--prove that Q-closure is R, or just prove that the counterexample given is invalid? To do the latter, all you need to say is that the set is not "clearly closed", because only finite unions of closed sets are necessarily closed. If it's the former, there are lots of ways, but if it were me I'd just note that one definition of the reals is equivalence classes of Cauchy sequences in Q.

9. Originally Posted by Failure
Surely, it is not closed.

Since the closure of $\mathbb{Q}$ is $\mathbb{R}$ the set in question cannot be closed: the closure of a set is the smallest superset that is closed, so any set $X$ with $\mathbb{Q}\subseteq X\subsetneq \mathbb{R}$ cannot be closed.
Ah, that's right! This is the most direct argument.

Originally Posted by davismj
I think we have that since the proposed counterexample is a proper subset of R, we know that there is at least one irrational number not in the set. However, this irrational number is certainly (by the density of Q in R) the limit of sequence of rational numbers, and therefore a limit point of Q. Therefore, the proposed set is not closed. Is that about right?
Yeah I guess this is another way to show that the set is not closed. Since it is not closed, it cannot be a counterexample.

10. Originally Posted by Alchemist
Ah, that's right! This is the most direct argument.

Yeah I guess this is another way to show that the set is not closed. Since it is not closed, it cannot be a counterexample.
Uh, well I don't think "since the closure of Q is R" is a good assumption to make in the question "is R the closure of Q?"

11. Originally Posted by davismj
Uh, well I don't think "since the closure of Q is R" is a good assumption to make in the question "is R the closure of Q?"
I don't understand the problem. Suppose that $\mathbb{Q}=\overline{\mathbb{Q}}$ then we see that $\sqrt{2}\in\left(\mathbb{R}-\mathbb{Q}\right)^{\circ}$. But, let $\varepsilon>0$ be given. Note that the sequence $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)\ to\sqrt{2}$ and so there exists some $q\in\left\{x_n:n\in\mathbb{N}\right\}$ such that $|\sqrt{2}-q|<\varepsilon$. But, $q\in\mathbb{Q}$ and since it was arbitrary $\sqrt{2}\notin\left(\mathbb{R}-\mathbb{Q}\right)^{\circ}$, contradiction.

12. Can't you just say that by density property a sequence in Q (rationals) can converge to some X in the irrationals, and since now every sequence in Q converges in Q, Q is not closed?

13. Originally Posted by cp05
Can't you just say that by density property a sequence in Q (rationals) can converge to some X in the irrationals, and since now every sequence in Q converges in Q, Q is not closed?
If you're not sure if $\overline{\mathbb{Q}}=\mathbb{Q}$ you probably can take for granted that you don't know that $\overline{\mathbb{Q}}=\mathbb{R}$