# An Open and Closed Interval in Q

• Bachelier
In summary, the interval ##(-\sqrt{2}, \sqrt{2})## is both open and closed in ##\mathbb{Q}##. This is because it has no limit points in ##\mathbb{Q}\setminus(-\sqrt{2}, \sqrt{2})##, which makes it closed, and its complement is also open in ##\mathbb{Q}##, which makes it open. Additionally, the fact that the closure of ##(-\sqrt{2}, \sqrt{2})## in ##\mathbb{Q}## is equal to the interval itself shows that it is a closed set. However, this is not true in ##\mathbb{R}##
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.

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

Bachelier said:
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.
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.

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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.

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jbunniii said:
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.

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.

Bachelier said:
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.
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}##.

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...

?

Bachelier said:
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...

?
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.

Bachelier said:
what you saying is that the rationals are the limit points of the reals
In fact, the exact opposite is true: the reals are the limit points of the rationals.

jbunniii said:
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}##.

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

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.

Bachelier said:
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 $\{-\sqrt{2}, \sqrt{2}\}$.

jbunniii said:
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.

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?

Bachelier said:
Cool. For instance, the boundary of the singleton S = {0} in ℝ under the discrete metric is the ∅ set.
Under the discrete metric, yes. Under the standard euclidean metric, the boundary would be {0}.

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

jbunniii said:
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}.

Nice

jbunniii said:
Under the standard euclidean metric, the boundary would be {0}.

So in this case int{0} = ∅

Bachelier said:
So in this case int{0} = ∅
Right.

## 1. What is an open interval in Q?

An open interval in Q is a set of rational numbers that lie between two specific values, where the two values are not included in the set. For example, the open interval (1, 5) in Q would include all rational numbers greater than 1 and less than 5, but would not include 1 or 5 in the set.

## 2. What is a closed interval in Q?

A closed interval in Q is a set of rational numbers that includes both of the specified values. For example, the closed interval [2, 6] in Q would include all rational numbers greater than or equal to 2 and less than or equal to 6.

## 3. How do open and closed intervals differ?

The main difference between open and closed intervals in Q is whether or not the specified values are included in the set. Open intervals do not include the specified values, while closed intervals do include them.

## 4. What is the notation used for open and closed intervals in Q?

Open intervals are typically represented using parentheses, such as (a, b), while closed intervals are typically represented using brackets, such as [a, b]. The notation can vary slightly depending on the context and the author.

## 5. Can an interval in Q be both open and closed?

No, an interval in Q cannot be both open and closed. An interval can only be one or the other, as they have different definitions and properties. For example, an open interval does not include its endpoints, while a closed interval does.

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