Why is the exterior set of $\mathbb R\setminus \mathbb Q$ a null set?

Question

Given the set $\mathbb R\setminus \mathbb Q.$

The interior set is the collection of all the interior points, where the interior point of a set $S$ from $\mathbb R,$ is a point $x \in S,$ such that there exists an $ \varepsilon >0 $ to make an open set U that looks like $(x - \varepsilon, x + \varepsilon)$ such that $x \in U$ and $U \subset S$.

My explanation for Interior set being a null set (Please review)

So, for any arbitrary irrational point in the given set, If I form an open interval around that point of size $|x|<\varepsilon$, but that interval has no other point than the irrational point itself, so therefore a neighbourhood does not exist for the irrational point.

Why would the Exterior set be a null set?

(Exterior Set - Collection of all the exterior points of set S)

(Exterior Point - A number $a \in\mathbb R$ is said to be an exterior point of a set $S$ from $\mathbb R$ if there exist a neighbourhood of a which is contained in $S^c$)

Answer 1

Your solution for the interior set is iffy - you start in the right way by examining each $x\in\mathbb R\setminus\mathbb Q$, but then you do something strange by writing $|x|<\varepsilon$ - since you are not allowed to control what $x$ is in an argument like this and then say that some interval consists only of $x$, which is false.

Rather, what you should ask is the following:

I've been given some $x\in\mathbb R\setminus\mathbb Q$. Is there any $\varepsilon>0$ such that the interval $(x-\varepsilon,x+\varepsilon)$ is a subset of $\mathbb R\setminus \mathbb Q$.

In simpler language, you are asking the following

Is there any $\varepsilon>0$ such that every element of $(x-\varepsilon,x+\varepsilon)$ is irrational?

The answer to this is "No" because every open interval contains a rational number. So, to write a proof that the interior is empty, you would start as follows:

We will show that the interior of $\mathbb R\setminus \mathbb Q$ is empty. To see this, fix $x\in\mathbb R\setminus \mathbb Q$. We claim that for any $\varepsilon>0$, the interval $(x-\varepsilon,x+\varepsilon)$ contains a rational number. ...

And then you would argue why this is true.

Note that the exterior of a set is just the interior of the complement - and to show that the interior of $\mathbb Q$ is empty, it suffices to show that every open interval contains an irrational number, which will follow very similar reasoning to the interior being empty.

Answer 2

$\mathbb R \setminus \mathbb Q$ has an empty interior because, for every $x\in\mathbb R \setminus \mathbb Q$, and every $\varepsilon>0$, $(x-\varepsilon,x+\varepsilon)\cap\mathbb Q \neq \emptyset$. This is because $\mathbb Q$ is dense in $\mathbb R$.

Your reasoning is incorrect because it is not true that you can find an interval around a point $x$ containing only the point $x$. In particular, the interval of size $|x|$ that you describe is given by $(x-|x|/2,x+|x|/2)$. This interval is non-empty as long as $x\neq 0$, in which case it must contain points in $\mathbb Q$ by the reasoning given above.

Similarly, if $x \in \mathbb R$, for any $\varepsilon$, $(x-\varepsilon,x+\varepsilon)\cap(\mathbb R \setminus \mathbb Q)\neq \emptyset$, so that $x$ cannot be an exterior point of $\mathbb R \setminus \mathbb Q$. Thus, the exterior of $\mathbb R \setminus \mathbb Q$ is empty.

Answer 3

It follows from the density of the rationals that the exterior of R/Q must be empty.

Answer 4

Your argument for why the interior for $\mathbb R \setminus \mathbb Q$ isn't quite right, and I don't think it is salvageable.

$\epsilon > 0$ is chosen to be arbitrarily small and unless $x=0$ we can not say $|x| < \epsilon$ as, being arbitrarily small, $\epsilon$ can be made to be less than $|x|$. And as $x$ is an arbitrary element of $S$ we are not allowed to make any assumptions about it. (And if $S =\mathbb R\setminus \mathbb Q$ we can't have $|x| = 0$.)

And even if we could set $|x| < \epsilon$ then $(x-\epsilon, x+\epsilon)$ still has infinite rational and irrational points. The only way to make $(x-\epsilon, x+\epsilon)$ to have only one point is ... well, it's impossible. I was going to say if $\epsilon =0$ but then $(x-\epsilon, x+\epsilon)=(x,x) = \{w| x < w < x\}= \emptyset$.

You are doomed.

......

Instead the argument is: For any $x\in\mathbb R$ and any $\epsilon > 0$ then the open interval $(x-\epsilon, x+\epsilon)$ will always contain rational points because $\mathbb Q$ is dense in $\mathbb R$ (if you haven't proven this you must). So $(x-\epsilon, x+\epsilon)\not \subset \mathbb R\setminus \mathbb Q$ and $x$ is not an interior point.

As $x$ and $\epsilon$ were arbitrary-- no neighborhood of any point is a subset of $\mathbb R\setminus \mathbb Q$ so no point is an interior point.

....

And proving no point is an exterior point is exactly the same.

(Note: a point is an exterior point if and only if it is an interior point of the complement.)

$(\mathbb R\setminus \mathbb Q)^c = \mathbb Q$.

And for any $x\in \mathbb R$ and any $\epsilon > 0$ then $(x-\epsilon, x+\epsilon)$ will contain infinitely many irrational points and $(x-\epsilon, x+\epsilon) \not \subset \mathbb Q = (\mathbb R\setminus \mathbb Q)^c$.

So no point is an exterior point of $(\mathbb R\setminus \mathbb Q)^c$