Showing that $\mathbb{Z}\cap(-10, 10)$ is open in $\mathbb{R}$ Let $\mathbb{Z}\cap(-10, 10)$ be a subset of $\mathbb{R}$. Pick any $x\in\{x_i\}_{i=-9}^9$, define $\varepsilon :=10-|x|$. Then pick $y\in B_\varepsilon(x)$ so that $|y-x|<\varepsilon$. We need to show that $|y|<10$. Now, $|y|=|y-x+x|\le |y-x|+|x|<\varepsilon+|x|=10-|x|+|x|=10$.
Do you think this proof is correct?
 A: No, your definition of an open set is not correct. Given any point in the set, you need to be able to find an $\epsilon$ so that the whole $\epsilon$ ball around the point is in your set. $0$ is in your set. What ball centered on $0$ is also in your set?
A: It looks like you've proven $(-10,10)$ is open by showing that for any $x\in (-10,10)$ the ball $B_{10-|x|}(x)$ is contained in the set. Or, rather, you've shown that around each point in $\mathbb Z \cap (-10,10)$, there is an open ball contained in $(-10,10)$. The problem is that this isn't what you need to prove to show openness - you'd need to show that, around each point in $\mathbb Z \cap (-10,10)$ there is an open ball contained in $\mathbb Z\cap (-10,10)$. In particular, you'd need to additionally show that the $y$ in your proof is an integer.
(This is problematic because in every open ball, there is a non-integer, meaning the set is actually not open, nor is any non-empty subset of $\mathbb Z$)
A: You don't just need to show that $|y|<10$ which implies $y\in(-10,10)$, you also need to show that $y\in\mathbb{Z}$ so that $y\in\mathbb{Z}\cap(-10,10)$ holds (which is not the case).
A: The complement of the set $\mathbb{Z}\cap (-10,10)=\{-9,\ldots,9\}$ is a union of open sets and so is itself open. A set is closed if its complement is open.
Edit: assuming "standard topology", i.e. not a space with the discrete metric.
