(R, d) as a metric space, where d(x, y) = |x − y| . What are the interior, closure and accumulation points of? I understand that interior points are ones which are fully included in the set e.g. [0,1) we say 0 is not an interior point. The closure is the point(s) added to close the set and that the accumulation point is the limit however I'm struggling with these three examples any help?



R \ Q

  • $\begingroup$ Hey there, consider formatting your question with MathJax to make it more readable! It is ok to struggle with maths problems (that is why they are problems :P ) but we would be able to better help you if you provided us with your insights on the specific examples. $\endgroup$
    – RGS
    Oct 18, 2017 at 8:57

1 Answer 1


$M_1:=\cup_{n\in\mathbb{N}}(n,n+1)$: Since every $(n,n+1)$ is open, the union is open as well. Hence the interior points of $M_1$ are $M_1$. The closure of $M_1$ is just $\{x\in\mathbb{R}:\ x\geq 0\}$ (if you define the natural numbers with zero), since you can approximate every natural number $n$ by the sequence $(n+\frac{1}{k})_{k\in\mathbb{N}}$, which lies in $M_1$. With the same argument you get the accumulation points as well.

$M_2:=\mathbb{Q}\cap([1,3]\cup[-\pi,0])\cup\{4,5,6\}$: $M_2$ has no interior points, since every $q\in M_2$ is rational. Therefore for every $\varepsilon>0$ you can find an irrational number in the ball $B_\varepsilon(q)$. This means $q$ is not an interior point. The closure can be dealt with as follows. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ you have $\overline{M_2}=[1,3]\cup[-\pi,0]\cup\{4,5,6\}$. The accumulation points are $\mathbb{Q}\cap([1,3]\cup[-\pi,0])$, since again $\mathbb{Q}$ is dense. Therefore for every $\varepsilon>0$ and every $q\in \mathbb{Q}\cap([1,3]\cup[-\pi,0])$ the set $(B_{\varepsilon}(q)\setminus q)\cap M_2$ is not empty. The other points $\{4,5,6\}$ can be dealt by choosing $\varepsilon=\frac{1}{2}$. For this $\varepsilon$ you would obtain $B_{\varepsilon}(4)\setminus\{4\}\cap M_2=\emptyset$.

I leave the last one as an exercise, since it can be dealt with similar arguments.

  • $\begingroup$ Thank you so much! I look forward to tackling the last one on my own given your guidance on the other two! $\endgroup$
    – Emma.Dolan
    Oct 18, 2017 at 12:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .