4

I don't understand why the interior of $\mathbb{Q}$ in $\mathbb{R}$ is empty, since, for every ball with the center being a rational number, given an $\epsilon>0$, I can find an infinite sequence of rational numbers that approach this point. For example, take $\frac{1}{2}$. The sequence $\frac{1}{2}+\frac{1}{n}$ can be made as close as I want to the number $\frac{1}{2}$, therefore I can always have open balls with center $\frac{1}{2}$ such that there are rationals inside it.

Eric Wofsey
  • 330,363
Guerlando OCs
  • 6,584

4 Answers4

8

It doesn’t matter that there are rationals inside the ball: what matters is that your open ball is not a subset of $\Bbb Q$. In order for $\Bbb Q$ to be open in $\Bbb R$, for each $q\in\Bbb Q$ there would have to be an open ball $B(q,\epsilon)$ about $q$ such that $B(q,\epsilon)\subseteq\Bbb Q$, and that is never the case: every open interval in $\Bbb R$ contains irrational numbers.

Brian M. Scott
  • 616,228
  • but why $0$ in ${0,1,1/2,...,1/n}$ is isolated? – Guerlando OCs Apr 03 '16 at 02:34
  • hmmm, I understood. $\mathbb{Q}$ has no isolated points in $\mathbb{R}$ because any open subset is not contained in $\mathbb{Q}$, because there are irrationals between the rationals. Now, for the set in my comment above, any subset is contained in it. – Guerlando OCs Apr 03 '16 at 02:36
  • @Guerlando: Do you mean the finite set with $n+1$ elements, or did you mean ${0}\cup\left{\frac1n:n\in\Bbb Z^+\right}$? – Brian M. Scott Apr 03 '16 at 02:36
  • If I recall correctly, $0$ is not an isolation point of $A={1/n: n\in \mathbb{Z}^+}$ nor is it an isolated point of $A\cup {0}$. For any neighborhood of $0$ has to contain some $1/n$. However, It would be an isolated point of the finite set ${0,1,1/2,\dotsc, 1/n}$ because you can find some neighborhood that misses all of $1,\dotsc,1/n$, and its a finite set! – Nap D. Lover Apr 03 '16 at 02:46
  • @LoveTooNap29: It’s not a point of your set $A$ at all. The rest of what you say is correct and is why I asked the OP whether he meant what he wrote, or whether he really meant your ${0}\cup A$. – Brian M. Scott Apr 03 '16 at 02:50
  • Taking advantage of the discussion, I have another question: what about the interior of $Q$ in $Q$? I Always have doubts about it. – Cleto Pereira Oct 24 '19 at 19:00
  • @CletoPereira (very belatedly): For any space $X$ the set $X$ is open, so it is its own interior. In particular, $\Bbb Q$ is an open set in the space $\Bbb Q$, so its interior in that space is $\Bbb Q$. – Brian M. Scott Nov 09 '21 at 17:34
5

For $x$ to be an interior point, it is not enough that there exist some rational in the ball around $x$, the definition of interior point require that there is some ball that every points in it are rational.

Hint: Can you see that your proposed construction of the ball always contain some irrational points?

BigbearZzz
  • 15,084
3

Let $x\in\mathbb{Q}$ Consider any open ball $\mathbb{B}(x,r)$ irrespective of how small you choose the radius $r$ the open ball $\mathbb{B}(x,r)\not\subset\mathbb{Q}$.Because between any two rations there are infinitely many irrations. That is why $int\mathbb{Q}=\phi$ in$\mathbb{R}$.

Rayees Ahmad
  • 1,325
0

It is an immediate consequence of Baire's category theorem. Every singleton $\{q\}$ with $q$ rational number is a closed set with empty interior, therefore also the union of all these singletons, i.e., $\mathbb{Q}$, has empty interior.

Kosh
  • 1,490