Show $0$ is the only cluster point of the set $A = ${$\frac{1}{n} : n\in\mathbb N$}

Question

Need a check of working, I'm new to cluster points.

Define $a_n = \frac{1}{n}$ for all $n\in\mathbb N$. Clearly $A =$ {$\frac{1}{n}:n\in\mathbb N$} $=$ {$(a_n)$}. Now $a_n$ converges to $0$, thus every subsequence of $a_n$ also converges to $0$. Hence $0$ is the only cluster point of the set A.

Does that suffice, or is there something to add?

Answer 1

Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $A\setminus \{0\}$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $A\setminus \{0\}$ are subsequences of $(a_n)$.

To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 \in \mathbb R$, there exists $\varepsilon > 0$ such that $$[(x_0 - \varepsilon, x_0 + \varepsilon)\cap A]\setminus \{x_0\} = \varnothing.$$ You can do this easily by splitting into cases:

1. $x_0 < 0$
2. $x_0 > 1$
3. $x_0 \in (0,1]\cap A$
4. $x_0 \in (0,1]\setminus A$

I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.

Suppose $x_0 > 1$. Let $\varepsilon = \frac{x_0-1}{2}$. Then $[(x_0 - \varepsilon, x_0 + \varepsilon)\cap A]\setminus \{x_0\} = \varnothing$, so $A$ has no cluster points greater than $1$.

Answer 2

There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.

Answer 3

If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,

Suppose that $p \in \mathbb{R}$ is a cluster point of the set $A$. Then $\forall \epsilon >0$, the set $\{a_n \in A, |p-a_n|<\epsilon \}$ contains an infinite number of points. By the least upper bound property of $\mathbb{N}$, there is necessarily an index $i \in \mathbb{N}$ such that $a_i=\min \{a_n \in A, |p-a_n|<\epsilon \}$. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $\{a_n \in A, |p-a_n|<\epsilon \}$, ordered by their indices, is a subsequence of $\{a_n\}$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.