1
$\begingroup$

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 $A$?

  • 0
    It would have been clearer to just say "given $\Bbb R$ as the topological space".2012-11-09

1 Answers 1

4

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 $A$.

  • 0
    @Thomas: I forgot to add the qualifiers after the colon. (I didn’t want to make the substitutions $n=2m$ and $n=2m+1$.)2012-11-09