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.