I was working on the following problem:

Find an example of a nonempty countable
set A such that every cluster point of A is in A and every point in A is
a cluster point of A.

And came up with this:

A set $U:=[0,1]\cap\mathbb{Q}$ contains all of its cluster points and every member of $U$ is a cluster point of $U$.

Proof:

* $[0,1]\cap\mathbb{Q}$ is closed in $\mathbb{Q}$.
* A set is closed iff it contains all of its cluster points.
* Since $[0,1]\cap\mathbb{Q}$ is closed in $\mathbb{Q}$, it must contain all of its cluster points.
* If $(x\_n)$ is any sequence of the form $\frac{p}{q}+\frac{1}{n}$, $\forall n\in\mathbb{N}$, $\frac{p}{q}\in[0,1]\cap\mathbb{Q}\Rightarrow\lim{(x\_n)}=\frac{p}{q}$.
* Therefore, every element in $[0,1]\cap\mathbb{Q}$ is a cluster point of $[0,1]\cap\mathbb{Q}$.

Is this sufficient? It seems to me to be a bit too easy. Am I doing something wrong here?