Exercise: What is the closure in $\mathbb{R}$of each of the following set:
i) $\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}$
My attempt: I consider the topological space $\mathbb{R},\tau$, where $\tau$ is the following topology $\tau=\mathbb{R}\cup\emptyset\cup\{(a,b):\forall a,b\in\mathbb{R}\}$
The open sets can be defined as $\{A\in\tau:\text{such that} \forall a \in A\:\exists x,y\in\mathbb{R}, a\in(x,y)\subseteq A\}$
$\forall i\in \mathbb{N},\frac{1}{i}\in\mathbb{R}$
For elements $\frac{1}{i_1}$ and $\frac{1}{i_1}$ there exists $x,y\in\mathbb{R}$ such that $\frac{1}{i_1}\in(x,y)$ but $\frac{1}{i_1}\notin(x,y)$
Then $\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}$ has not limit points then $\overline{\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}}=\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}$
Question:
I am not really sure about the step "there exists $x,y\in\mathbb{R}$ such that $\frac{1}{i_1}\in(x,y)$ but $\frac{1}{i_1}\notin(x,y)$"
Is my proof right? If not is there any way to prove this?
Thanks in advance!
As I understood your definition of $\tau$ is standart topology on $\mathbb{R}$. And you may say the base of it must be $\{(a,b):\forall a,b\in\mathbb{R}\}$. And now say any $a\in \mathbb{R^+}$. By using precible of Archimedes, you can find $m>0$ integer such that $\frac{1}{m}<a$. So $\frac{1}{m}\in{\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}} $ and $\frac{1}{m}\in (-a,a)$. Since $(-a,a)$ is an arbitrary neighbourhood of $0$ then $0\in\overline{\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}}\neq\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}$