What is the adherent value of $(\frac1n,1)$ in this topology?

Question

The question is to find the adherent value of $(\frac1n,1)$ in the toplogy $\{\mathbb{R}^2,\emptyset, (B_r)_{r>0}\}$

where $B_r=\{(x,y)\in\mathbb{R}^2, (x-3)^2+y^2<r^2\}$

That is for any $r>0$ we have $$card\{n\in\mathbb{N}, (\frac1n,1)\in B_r\}=+\infty$$

$(\frac1n,1)\in B_r\Rightarrow (3-\frac1n)^2<r^2-1$ for $r>1$ we have $|3-\frac1n|<\sqrt{r^2-1}$

how to continue please

Answer 1

[Update] OP wanted the set of $\omega$-accumulation points, neither accumulation points nor adherent points.

An $\omega$-accumulation point of a set $A$ in a topological space $X$ is a point $x$ that every open set containing x contains infinitely many points of $A$. [Wikipedia: Limit point]

Let $A=\{(\frac{1}{n},1)\mid n\in\mathbb{N}\}\subset\mathbb{R}^2$. Then the set of $\omega$-accumulation points of $A$ is $$ \mathbb{R}^2\setminus B_\sqrt{10} $$

1. Let $d_n=\sqrt{(3-\tfrac{1}{n})^2+1}$ be a sequence of real numbers. Then $d_n$ converges to $\sqrt{10}$.

2. For any $x\in B_\sqrt{10}$, we have $d_x=\|(3,0)-x\|<\sqrt{10}$. Since $d_n$ converges to $\sqrt{10}$, there exists a natural number $N$ such that $d_x<d_n<\sqrt{10}$ for all $n\geq N$. Consider the open set $B_{d_N}$. Since $d_x<d_N$, we have $x\in B_{d_N}$, but $A\cap B_{d_N}$ is finite. Thus $x$ is not an $\omega$-accumulation point of $A$.

3. However, for $x\notin B_\sqrt{10}$ we have $d_x=\|(3,0)-x\|\geq\sqrt{10}$ so that if any open ball $B_r$ contains $x$ then $r$ should be greater than $\sqrt{10}$. When $r>\sqrt{10}$, note that $B_r$ contains all points of $A$. Therefore $x$ is an $\omega$-accumulation point of $A$.

A limit point (or accumulation point) of a set $A$ in a topological space $X$ is a point $x$ that every neighbourhood of $x$ with respect to the topology on $X$ also contains a point of $A$ other than $x$ itself. [Wikipedia: Limit point]

Let $A=\{(\frac{1}{n},1)\mid n\in\mathbb{N}\}\subset\mathbb{R}^2$. Then the set of limit (or accumulation) points of $A$ is $$ \mathbb{R}^2\setminus \bigl( B_\sqrt{5} \cup \{(1,1)\} \bigr) $$

1. The closest point in $A$ from the center $(3,0)$ of $B_r$ is $(1,1)$ and the distance between the two points is $\sqrt{5}$. Thus $A\cap B_\sqrt{5}=\varnothing$.

2. For any $x\in B_\sqrt{5}$, we have an open set $B_\sqrt{5}$ containing $x$ but $A\cap B_\sqrt{5}=\varnothing$. Thus there is no limit (or accumulation) point of $A$ in $B_\sqrt{5}$.

3. Note that $A\cap B_\sqrt{6}=\{(1,1)\}$ since $\|(3,0)-(\tfrac{1}{n},1)\|=\sqrt{(3-\tfrac{1}{n})^2+1}>\sqrt{6}$ for all $n\geq2$. Thus $(1,1)$ is not a limit (or accumulation) point of $A$.

4. If $x\notin B_\sqrt{5}$ and $x\neq(1,1)$, then the distance between $x$ and the center $(3,0)$ of $B_r$ is greater than or equal to $\sqrt{5}$. What is an open set $B_r$ containing $x$? That is, $B_r$ for $r>\sqrt{5}$, and moreover, $(1,1)\in A\cap B_r$. Therefore, $x$ is a limit (or accumulation) point of $A$.

An adherent point of a subset $A$ of a topological space $X$, is a point $x$ in $X$ such that every open set containing $x$ contains at least one point of $A$. [Wikipedia: Adherent point]

Let $A=\{(\frac{1}{n},1)\mid n\in\mathbb{N}\}\subset\mathbb{R}^2$. Then the set of adherent points of $A$ is $$ \mathbb{R}^2\setminus B_\sqrt{5} $$

1. For any $x\in B_\sqrt{5}$, we have an open set $B_\sqrt{5}$ containing $x$ but $A\cap B_\sqrt{5}=\varnothing$. Thus there is no adherent point of $A$ in $B_\sqrt{5}$.

2. If $x\notin B_\sqrt{5}$, the distance between $x$ and the center $(3,0)$ of $B_r$ is greater than or equal to $\sqrt{5}$. What is an open set $B_r$ containing $x$? That is, $B_r$ for $r>5$, and moreover, $(1,1)\in A\cap B_r$. Therefore, $x$ is an adherent point of $A$.