# Isolated and discrete points

In the space of integers with usual topology, is it true that...

1. every point is isolated point for any set
2. no point is limit point for any set.

In real space and space of rationals...

1. no point is isolated from set containing it
2. every point in a set is a limit point

These are my first thoughts when I read the definition of isolated points and derived sets. Just wanted to verify if they are correct

• This may seem unfamiliar, but it is not difficult. What is the definition of isolated point? Now apply that to the integers. How might you choose $\epsilon$? Commented Apr 18, 2016 at 8:44

## 1 Answer

Your first line on integers is correct.

The set of integers $\mathbb Z$ is usually equipped with the discrete topology, which means that every subset of $\mathbb Z$ is an open set. Singletons are open sets and the statement that $\{x\}$ is an open set is exactly the same as the statement that $x$ is an isolated point of the whole space $\mathbb Z$.

Secondly if $x$ is an isolated point of the whole space, then automatically it is an isolated point of every set $S$ with $x\in S$.

Your second line on reals and rationals is not correct.

If $x\in X$ where $X=\mathbb Q$ or $X=\mathbb R$ then (if we are working with the usual topology on $X$) then singletons are evidently not open sets, or equivalently: no $x\in X$ is an isolated point of the whole space $X$.

Secondly (and this is what I mark as your incorrectness) if $S\subset X$ and $x\in S$ then quite well it can happen that $x$ is an isolated point of subset $S$. This is the case if and only if there is an open set $U$ such that $U\cap S=\{x\}$. Taking e.g. $S=\{0\}\cup\{z\in X\mid z>1\}$ we come to the conclusion that $0$ is an isolated point of $S$, since $U\cap S=\{0\}$ for open set $U=\{z\in X\mid z<1\}$.

• Thank you I understand my mistake is it correct that no point in R is isolated from an open set containing it Commented Apr 18, 2016 at 10:26
• If $U\subset\mathbb R$ is open then no $u\in U$ exists such that $u$ is an isolated point of $U$. If that's what you mean, then you are correct. Commented Apr 18, 2016 at 10:50
• You are very welcome. Commented Apr 18, 2016 at 11:21