Topology $\tau(\frac{1}{n})$

Question

Defined as $\tau(\frac{1}{n}) = \{A\subset \mathbb{R} : (\exists N \in \mathbb{N}) \forall(n\in \mathbb{N},n\geq N) \frac{1}{n} \in A\} \cup \{\emptyset \}$ was previously proven to be a topology.

Consider $\mathbb{R}$ with the topology $\tau(\frac{1}{n})$. Take $A = [-1,\frac{1}{5}]$, find $A^{o},A^{a},\overline{A}$. Take $A = (1,2)$, find $A^{o},A^{a},\overline{A}$. (A-interior, A-accumulation, A-closure)

My definitions: Let G be an open set

$x\in A^{o} \iff \exists G\in \tau$ $: x\in G\subset A$

$x\in A^{a} \iff \forall(G\in \tau,x\in G)$ $\exists y \neq x : y\in G\subset A$

$x\in \overline{A} \iff \forall(G\in \tau,x\in G)$ $\exists y : y\in G\cap A$

I know that there is an n$\in\mathbb{N}$ (namely n$\geq 5$) that makes $A$ an open topology so I can see that I WTS that $A\subset A^{o}$. I also have a BIG Theorem (Sister) that I can use which states that $A$ is open $\iff A^{o} = A$ which may be the quick route here since I'm pretty sure that $A^{o} = [-1,\frac{1}{5}]$ but please correct me if I am wrong.

For $A^{a}$ I understand the definition and I know that I need to let $G$ be an open set of unions with a tail end of $\frac{1}{n}$ and then intersect that with $A$ to create an open set $G$ where $y\in G$ and $x\in G$ but $y\neq x$ but I am struggling to do this so any help is appreciated.

The closure seems that it would be done similary to the accumulation point although that it just states that there $\exists x,y\in G$ so I was thinking those could be equal and that ultimately $A=\overline{A}$.

Any help is appreciated. Thanks!

Answer 1

For obvious reasons, let $A=[-1,1/5]$ and $B=(1,2)$.

Notice that $A\in\tau(\frac{1}{n})$ and therefore $A^{\circ}=A=[-1,1/5]$.

Besides, let $x\in\mathbb R$ and let $G\in\tau(\frac{1}{n})$ be an neighborhood of $x$. Then $1/n\in G\cap A$ for some (and thus every) large enough $n$ and therefore $x\in A^a$. Thus $A^a = \mathbb R$ and $\overline{A}=\mathbb R$.

For $B$, notice that $\mathbb R\setminus B =(-\infty,1]\cup[2,\infty)$ belongs to $\tau(\frac{1}{n})$, so $B$ is closed and therefore $\overline{B}=B=(1,2)$.

Also, if $x\in B^{\circ}$, then there exists $G\in\tau(\frac1n)$ such that $x\in G\subseteq B$ and therefore $1/n\in B$ for some (and thus every) large enough $n$, which is a contradiction. So $B^{\circ} = \emptyset$.

Can you show that $B^a = \emptyset$ using analogous arguments?

For $x\in\mathbb R$ we can always let $G = \{x\} \cup \{1/n : n\geq 2\} \in \tau(\frac1n)$ and notice that $G\cap B\subseteq \{x\}$. Thus $x$ is not $B$-accumulation.