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

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 $$S^c$$)

• Some annoying pedantry best left as a comment: $$\mathbb R\setminus \{\mathbb Q\} = \mathbb R$$ Sep 12 '19 at 16:41
• Even if you were allowed to control the value of $x$ (you are not) so that $|x| < \epsilon$ even still the neighborhood of $(x-\epsilon, x+\epsilon)$ will still contain infinitely many rational and irrational points. Sep 12 '19 at 16:42
• To explain @AlvinLepik comment: $\{\mathbb Q\} \ne \mathbb Q$. $\{\mathbb Q\}$, a set with a single element, exists completely outside and is disjoint of the real numbers. So $\mathbb R\setminus \{\mathbb Q\} = \mathbb R$ in the same way that $\mathbb R\setminus \{babar the elephant\} = \mathbb R$. What one means is $\mathbb R\setminus \mathbb Q =$ the irrational numbers. (No set brackets.) Sep 12 '19 at 16:47

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.

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

$$\mathbb R \setminus \mathbb Q$$ 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.

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