Can someone explain to me why the interior of rationals is empty? That is $\text{int}(\mathbb{Q}) = \emptyset$?

The definition of an interior point is "A point $q$ is an interior point of $E$ if there exists a ball at $q$ such that the ball is contained in $E$" and the interior set is the collection of all interior points.

So if I were to take $q = \frac{1}{2}$, then clearly $q$ is an interior point of $\mathbb{Q}$, since I can draw a ball of radius $1$ and it would still be contained in $\mathbb{Q}$.

And why can't I just take all the rationals to be the interior?

So why can't I have $\text{int}\mathbb{(Q)} = \mathbb{Q}$?

  • $\begingroup$ Hint: what does contained in mean? $\endgroup$ – Alex Oct 30 '12 at 4:42
  • $\begingroup$ notice that "the ball is contained in $E$" means all of the points in the ball is contained in $E$, but any "ball" centered at a rational point contain some irrational point, which is not in $\mathbb{Q}$ $\endgroup$ – TTY Oct 30 '12 at 4:44
  • $\begingroup$ @Tao, but why do we even need to check those irrationals?Our ball will have holes, but we can't have that? $\endgroup$ – Hawk Oct 30 '12 at 4:47
  • 1
    $\begingroup$ I'm assuming you are using the standard definition, a "ball" on the real line, centered at a point $x$ with radius $r$ is defined (as long as I know)to be the set $\{y: |y-x|<r\}$, certainly in this definition we cannot have "holes" $\endgroup$ – TTY Oct 30 '12 at 4:53

If the whole set is the $\mathbb{Q}$, then $int\mathbb{Q}=\mathbb{Q}$,

If the whole set is the $\mathbb{R}$ or $\mathbb{R}^n$, then $int\mathbb{Q}=\emptyset$,

because, $\forall q\in \mathbb{Q}, and \,\forall \epsilon>0, B_\epsilon(q)=\{x\in\mathbb{R}:|x-q|<\epsilon\}$ contains irrational numbers, which are not in the $\mathbb{Q}$, so $q$ is not a interior point of $\mathbb{Q}$.

the statement is proved.

the problem depends on the whole set you are talking about.

  • $\begingroup$ Oh okay, that makes more sense! $\endgroup$ – Hawk Oct 30 '12 at 5:36

I'm assuming since you're using the Euclidean Metric that you're viewing $\mathbb{Q}$ as a subset of $\mathbb{R}$. The emptiness of the interior follows from the density of the rationals in the reals. So in fact, you can't actually take an open set around a rational number and stay within the rationals because real numbers will always get in your way.

  • $\begingroup$ Empty interior doesn't follow from being dense. $\endgroup$ – Nick Matteo Sep 12 '14 at 0:29
  • $\begingroup$ The empty interior follows from $\mathbb{R}-\mathbb{Q}$ being dense in $\mathbb{R}$ and as Kundor remarked not from $\mathbb{Q}$'s density in $\mathbb{R}$. A thorough answer to this question and afew others can be found here: math.stackexchange.com/questions/1306770/… $\endgroup$ – AIM_BLB May 31 '15 at 17:51

Think of it this way: we say $A$ is a subset of $B$ if for all $a \in A$, $a$ is in $B$.

Now, we say that the interior of some set $S$ is the set of all of its interior points. A point is an interior point if there exists a neighborhood $N$ of $x$, for some $x \in S$, such that $N$ is a subset of $S$.

Now, keeping in mind the definition of a subset, there exists no $\epsilon$ such that an irrational number is not contained within the neighborhood $(x - \epsilon, x + \epsilon)$ - hence, since there exists an $x \in N$ such that $x$ is not in $S$, given $I = \mathbb R \setminus \mathbb Q$, then there can be no such $N$ such that $N$ is a subset of $S$ - it fails to meet the definition of a subset.

When you consider what is required for a point to be considered an interior point, you can see that no such interior points exist, and thus $\operatorname{int}(\mathbb Q) = \emptyset$.


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