Sufficient proof that $[0,1]\cap\mathbb{Q}$ contains only its cluster points. 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?
 A: $\{q\in\Bbb Q:0\le q\le 1\}$ is indeed a countable subset of the metric space $\Bbb Q$ that is equal to the set of its cluster points in that space. However, no such set exists in $\Bbb R$. Suppose that $A\subseteq\Bbb R$ is equal to the set of its own cluster points. Then $A$ is closed, since it contains all of its cluster points, and it has no isolated points. A closed set with no isolated points is sometimes called a perfect set. This answer gives a proof that a perfect set in $\Bbb R$ (or in fact any complete metric space) has cardinality $2^\omega=\mathfrak c$; in particular, it’s uncountable.
A: Your proof is correct if you state your conclusion as

Let $A = [0,1] \cap \mathbb{Q}$ ($=\{q \in \mathbb{Q}\,:\, 0 \leq q \leq 1\}$) be a subset of the normed space $(\mathbb{Q},|.|)$. Then every $x \in A$ is a cluster point of $A$ and every cluster point $y$ of $A$ lies in $A$.

The important point being that $A$ is considered to be a part of the topological space $\mathbb{Q}$, not of $\mathbb{R}$.
You make use of that restriction when you assert that $A$ is closed. $A$ is not closed as a subset of $\mathbb{R}$ - in $\mathbb{R}$, $\bar{A} = [0,1]$.
Wheter or not your proof is sufficient thus depends on whether you were supposed to find a suitable $A$ in $(\mathbb{R}, |.|)$, or in any topological space of your liking. Note that in the first case, you'll actually have to show that no such set exists, since you won't be able to find one.
