# An Open and Closed Interval in Q

1. Jan 27, 2013

### Bachelier

Why is the interval $(-√2,√2)$ closed in $\mathbb{Q}$

I know why it is open, but do we consider it closed because it has no limit points in $\mathbb{Q}$, thus vacuously it is closed.

2. Jan 27, 2013

### Bachelier

In other words since the boundary of $(-\sqrt 2,\sqrt 2)$ does not exist in $\mathbb{Q}$, then it is closed.

3. Jan 27, 2013

### jbunniii

It certainly has limit points in $\mathbb{Q}$. For example, 0 is a limit point because the sequence (1/n) has 0 as a limit. Similarly, any rational number contained in $(-\sqrt{2}, \sqrt{2})$ is a limit point of that interval. The important thing is that there are no limit points of $(-\sqrt{2}, \sqrt{2})$ in $\mathbb{Q}\setminus (-\sqrt{2}, \sqrt{2})$. (Proof?)

Another way to see that $(-\sqrt{2}, \sqrt{2})$ is closed in $\mathbb{Q}$ is that its complement $(-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$ is open in $\mathbb{Q}$, for essentially the same reason that $(-\sqrt{2},\sqrt{2})$ is open.

Last edited: Jan 27, 2013
4. Jan 27, 2013

### jbunniii

A third way to see it is that the closure of $(-\sqrt{2}, \sqrt{2})$, which is nominally $[-\sqrt{2}, \sqrt{2}]$, is actually the same as $(-\sqrt{2}, \sqrt{2})$ because, as you said, $-\sqrt{2}$ and $\sqrt{2}$ are not elements of $\mathbb{Q}$. Since the interval equals its closure, it must be a closed set. You can make this argument more rigorous if you know a bit about subspace topologies.

Last edited: Jan 27, 2013
5. Jan 27, 2013

### Bachelier

Since $\mathbb{Q}$ is dense in $\mathbb{R}$ with infinitely many empty interior points then the boundary of $(-\sqrt 2, \sqrt2) =[-\sqrt{2}, \sqrt{2}]$ which happens to be its closure as well.

6. Jan 27, 2013

### jbunniii

I'm not sure what you mean by "empty interior points." As a subset of $\mathbb{R}$, in fact $\mathbb{Q}$ has no interior points.

Here is what I was getting at when I referred to subspace topologies. If we view $\mathbb{Q}$ as a subspace of $\mathbb{R}$, then a subset $A \subset \mathbb{Q}$ is open in $\mathbb{Q}$ if and only if it is possible to write it as $A = U \cap \mathbb{Q}$, where $U$ is an open subset of $\mathbb{R}$. Also, a subset $A \subset \mathbb{Q}$ is closed in $\mathbb{Q}$ if and only if it is possible to write it as $A = F \cap \mathbb{Q}$, where $F$ is a closed subset of $\mathbb{R}$.

If I may introduce the notation $(-\sqrt{2}, \sqrt{2})_\mathbb{R}$ to mean $\{x \in \mathbb{R} : -\sqrt{2} < x < \sqrt{2}\}$ and $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ to mean $\{x \in \mathbb{Q} : -\sqrt{2} < x < \sqrt{2}\}$, then we have
$$(-\sqrt{2}, \sqrt{2})_\mathbb{Q} = (-\sqrt{2}, \sqrt{2})_\mathbb{R} \cap \mathbb{Q}$$
which shows that $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ is an open subset of $\mathbb{Q}$, and
$$(-\sqrt{2}, \sqrt{2})_\mathbb{Q} = [-\sqrt{2}, \sqrt{2}]_\mathbb{R} \cap \mathbb{Q}$$
which shows that $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ is a closed subset of $\mathbb{Q}$.

7. Jan 27, 2013

### Bachelier

let me understand something:
You said: "The important thing is that there are no limit points of $(-\sqrt{2}, \sqrt{2})$ in $\mathbb{Q}\setminus (-\sqrt{2}, \sqrt{2})$."

what you saying is that the rationals are the limit points of the reals. I always thought that the reals tend to solve the problem of convergence of the rationals:

think $\pi$ and the sequence. 3, 31/10, 314/100, 3141/1000....

?

8. Jan 27, 2013

### jbunniii

Yes, but if we are talking about $(-\sqrt{2}, \sqrt{2})$ as a subset of $\mathbb{Q}$, then there are no irrational numbers available, either in $(-\sqrt{2}, \sqrt{2})$ or in its complement, $\mathbb{Q}\setminus(-\sqrt{2}, \sqrt{2})$. The sequence 3, 31/10, 314/100, 3141/1000, ... does not have a limit in this space even though the elements are getting closer and closer together. To use the appropriate technical jargon, this is a Cauchy sequence but not a convergent sequence: the limit to which it would have to converge, namely $\pi$, is missing from the space because it is irrational. In a complete space such as $\mathbb{R}$, all Cauchy sequences converge, but $\mathbb{Q}$ is not complete, meaning it has "holes" (all the irrationals are missing), so not every Cauchy sequence converges.

9. Jan 27, 2013

### jbunniii

In fact, the exact opposite is true: the reals are the limit points of the rationals.

10. Jan 27, 2013

### Bachelier

Thanks. I fully understand why the interval is both open and closed.

My previous statement about the density of $\mathbb{Q}$ came from a wikipedia page:

I guess I should specify the Metric Space that we are working with. respect to the statement:

∂(-√2,√2) = [-√2,√2] is true in $\mathbb{R}$ but not in $\mathbb{Q}$ as ∂(-√2,√2) = ∅. Am I correct?

btw Thank you for explaining the subspace topology. There is a good wiki article about it.

11. Jan 27, 2013

### jbunniii

That's a good sign!
OK, that would be this line: "These last two examples illustrate the fact that the boundary of a dense set with empty interior is its closure." However, the context here is that they are working with subsets of $\mathbb{R}$.

This is an important point: you can't talk about whether a set is open or closed, or what its boundary or interior are, without specifying which space you are working with. Here are some concrete examples.

Let us define, as I did in a previous post, $(-\sqrt{2}, \sqrt{2})_\mathbb{Q} = \{x \in \mathbb{Q} : -\sqrt{2} < x < \sqrt{2}\}$ and $(-\sqrt{2}, \sqrt{2})_\mathbb{R} = \{x \in \mathbb{R} : -\sqrt{2} < x < \sqrt{2}\}$. Also define $[-\sqrt{2}, \sqrt{2}]_\mathbb{Q} = \{x \in \mathbb{Q} : -\sqrt{2} \leq x \leq \sqrt{2}$ and $[-\sqrt{2}, \sqrt{2}]_\mathbb{R} = \{x \in \mathbb{R} : -\sqrt{2} \leq x \leq \sqrt{2}\}$.

Of course, $(-\sqrt{2}, \sqrt{2})_\mathbb{Q} = [-\sqrt{2}, \sqrt{2}]_\mathbb{Q}$, but $(-\sqrt{2}, \sqrt{2})_\mathbb{R} \neq [-\sqrt{2}, \sqrt{2}]_\mathbb{R}$.

Then, AS SUBSETS OF $\mathbb{Q}$:
* $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ is both open and closed
* Every point of $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ is an interior point
* Every point of $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ is a limit point, and there are no other limit points of this set
* The boundary of $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ is empty.

Whereas, AS SUBSETS of $\mathbb{R}$:
* $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ is neither open nor closed
* $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ has no interior points
* The set of limit points of $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ is $[-\sqrt{2}, \sqrt{2}]_\mathbb{R}$
* The boundary of $(-\sqrt{2}, \sqrt{2})_\mathbb{Q}$ is also $[-\sqrt{2}, \sqrt{2}]_\mathbb{R}$.

And, AS SUBSETS OF $\mathbb{R}$:
* $(-\sqrt{2}, \sqrt{2})_\mathbb{R}$ is open but not closed
* Every point of $(-\sqrt{2}, \sqrt{2})_\mathbb{R}$ is an interior point
* The set of limit points of $(-\sqrt{2}, \sqrt{2})_\mathbb{R}$ is $[-\sqrt{2}, \sqrt{2}]_\mathbb{R}$
* The boundary of $(-\sqrt{2}, \sqrt{2})_\mathbb{R}$ is $\{-\sqrt{2}, \sqrt{2}\}$.

Finally, AS SUBSETS OF $\mathbb{R}$:
* $[-\sqrt{2}, \sqrt{2}]_\mathbb{R}$ is closed but not open
* The set of interior points of $[-\sqrt{2}, \sqrt{2}]_\mathbb{R}$ is exactly $(-\sqrt{2}, \sqrt{2})_\mathbb{R}$
* The set of limit points of $[-\sqrt{2}, \sqrt{2}]_\mathbb{R}$ is $[-\sqrt{2}, \sqrt{2}]_\mathbb{R}$
* The boundary of $[-\sqrt{2}, \sqrt{2}]_\mathbb{R}$ is $\{-\sqrt{2}, \sqrt{2}\}$.

12. Jan 27, 2013

### Bachelier

Cool. For instance, the boundary of the singleton S = {0} in ℝ under the discrete metric is the ∅ set.

But if we specify the topology as being that {0} has no open subsets, then the ∂S is S itself.

Am I correct?

13. Jan 27, 2013

### jbunniii

Under the discrete metric, yes. Under the standard euclidean metric, the boundary would be {0}.

Well, {0} has only two subsets: itself and ∅. We don't get a choice about ∅: it is open in any topology. So I guess you mean we specify the topology such that {0} is not open. Then you are correct: any open set containing {0} must also contain other points of $\mathbb{R}$, so 0 is a boundary point. However, there might also be other boundary points. For example, if we use the trivial topology such that the only open sets are $\emptyset$ and $\mathbb{R}$, then every point of $\mathbb{R}$ is a boundary point of {0}.

14. Jan 27, 2013

### Bachelier

Nice

So in this case int{0} = ∅

15. Jan 27, 2013

Right.