Vtyjkyuk

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

Define $a_n = frac1n$ for all $ninmathbb N$. Clearly $A =$ $frac1n:ninmathbb N$ $=$ $(a_n)$. 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?

Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ 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 $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.

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 $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. 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, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.

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.