Prove that 0 is the only accumulation point for set $\{1/n: n \in\mathbb{N}\}$ in the usual topology

Question

I don't understand Why 0 is the only accumulation point for set $\{1/n: n \in\mathbb{N}\}$ in the usual topology I think the derived set is the set including 0 according to the deffinition of accumulation point

Answer 1

Your definition was

$a$ is an accumulation point of $A$ $:\Longleftrightarrow$ $\forall ~ U \subseteq \mathbb{R}$ open with $a \in U$ : $~(A - \{a\}) \cap U \neq \emptyset$

and hence by negation

$a$ is no accumulation point of $A$ $:\Longleftrightarrow$ $\exists ~ U \subseteq \mathbb{R}$ open with $a \in U$ and $~(A - \{a\}) \cap U = \emptyset$

Now consider that $a \neq 0$ and $A = \{ \frac{1}{n} \mid n \in \mathbb{N}_{>0} \}$.

• If $a \in(-\infty , 0)$ we can choose $\varepsilon := \vert a\vert / 2$ such that $U :=B_\varepsilon$(a) only contains negative numbers and hence $(A - \{a\}) \cap U = \emptyset$ .

• If $a \in (1 , \infty)$ we can choose $\varepsilon := \vert a - 1 \vert / 2$ such that $U :=B_\varepsilon(a)$ only contains numbers bigger than $1$ and hence $(A - \{a\}) \cap U = \emptyset$ .

• If $a \in (0,1]$ we find an $n \in \mathbb{N}$ with $ \frac{1}{n+1} < a \leq \frac{1}{n} $ . If we then choose $\varepsilon := (n+1)^{-10000}$ (just to make sure), $U :=B_\varepsilon(a)$ will lay inbetwee the fractions and we have again $(A - \{a\}) \cap U = \emptyset$ .

This shows, that $a \neq 0$ can not be an accumulation point of $\{ \frac{1}{n} \mid n \in \mathbb{N}_{>0} \}$ .

Answer 2

I think that it is clear, that $0$ is an accumulation point of $A:=\{1/n: n \in \mathbb N\} .$

Now let $x \ne 0$. Then $|x|>0$, hence there is $N \in \mathbb N$ such that $1/n <|x|$ for all $n>N$. This gives: there is a neighborhood $U$ of $x$ such that $U$ contains at most finitely many elements of $A$. Therefore $x$ is not an accumulation point of $A$.

Answer 3

Note that an accumulation point of a set is the limit of some sequence of distinct elements of that set.

The only point which could be the limit of a sequence of distinct elements of $A:=\{1/n: n \in \mathbb N\}$ is $0$.

Thus $0$ is the only accumulation point of $A$