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 &gt;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 $|x|&lt;\varepsilon$, 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 $S^c$)
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 $|x|&lt;\varepsilon$ - 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 $\varepsilon&gt;0$ 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 $\varepsilon&gt;0$ 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 $\varepsilon&gt;0$, 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.
It follows from the density of the rationals that the exterior of R/Q must be empty.