So I am solving the following question in apostol Determine all accumulation points of the following sets in $R^1$ and decide whether the sets are open or closed (or neither).
All integers:
The set of integers doesn't contain an accumulation point of Z I will do it by contradiction suppose x $\in \mathbb{R}$ is an accumulation point so we must have all balls of radius r > 0 to have points in common with integers in particular consider B(x,x/2) we have $(B(x,x/2) – {x}) \cap \mathbb{Z} = \emptyset$, so set $\mathbb{Z}$ doesn't contain an accumulation point.
Now I want to prove that our set Z is open I proved it is closed since if we consider $\mathbb{R} – \mathbb{Z} = \bigcup_{n \in \mathbb{N}}(n, n + 1)$ is union of open sets which is open. I already see the logic of how to prove that our set Z isn't open since for every open ball we will always have points in irrational numbers but I can't seem to be able to prove it rigorously.
Best Answer
To prove $\Bbb{Z}$ is not open, you could show that 1) $\Bbb{R} \setminus \Bbb{Z}$ is not closed or that 2) $\Bbb{Z}$ consists of isolated points.
1) 0 is an accumulation point of $\Bbb{R} \setminus \Bbb{Z}$ because it is the limit of the sequence {1/n}, but since $0 \in \Bbb{Z}$, it is not in $\Bbb{R} \setminus \Bbb{Z}$.
2) Let $a \in \Bbb{Z}$ and consider the open ball $B(a,1/2)=(a-1/2,a+1/2)$. Since $\Bbb{Z}\cap B(a,1/2)=\{a\}, a$ is an isolated point of $\Bbb{Z}$, so it is not an interior point of $\Bbb{Z}$.
The argument with the irrational numbers that you mention is another way of showing no point is interior; I think you could say that any open ball around $a \in \Bbb{Z}$ contains irrational numbers because the set of irrational numbers is dense in $\Bbb{R}$, so the ball is not fully contained in $\Bbb{Z}$.