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?