How to determine the closure of a subset of a topological space?

Question

What steps should one take in order to find out the closure of a subset in a topological space? For example given $(\mathbb{R}, \tau_{0})$ as the topological space and the subset $$A = \left \{ \frac{1 + (-1)^n}{2} + (-1)^n \frac{n - 2}{3n - 2}, n \epsilon \mathbb{N} \right \}.$$ How can one find the closure of $A$?

Answer 1

In general it depends entirely on the particular space and particular subset. Here the first step is to see just what $A$ looks like, where

$$A=\left\{\frac{1 + (-1)^n}{2} + (-1)^n \frac{n - 2}{3n - 2}:n\in\Bbb N\right\}\;.\tag{1}$$

It’s convenient to split $A$ into two subsets, one for even $n$ and one for odd $n$:

$$A_e=\left\{1+\frac{n-2}{3n-2}:n\in\Bbb N\text{ and }n\text{ is even}\right\}\;,$$ and

$$A_o=\left\{-\frac{n-2}{3n-2}:n\in\Bbb N\text{ and }n\text{ is odd}\right\}\;.$$

The sets $A_e$ and $A_o$ are actually the ranges of simple convergent sequences. What are $$\lim_{n\to\infty}\left(1+\frac{n-2}{3n-2}\right)\quad\text{ and }\quad\lim_{n\to\infty}\left(-\frac{n-2}{3n-2}\right)\;?$$ Can you see that these limits are limit points of $A_e$ and $A_o$, respectively?

Now make a sketch and decide whether $A$ has any other limit points. If so, identify them and explain why they are limit points; if not, try to show why no other point of $\Bbb R$ is a limit point of $A$.