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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 $U$.


  • $[0,1]\cap\mathbb{Q}$ 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?

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Only its cluster point in what space? In $[0,1]$ it is certainly false, but in $\mathbb Q$ it is certainly true. – Asaf Karagila Oct 20 '12 at 20:59
@AsafKaragila: The problem didn't specify, but since it's in a real analysis context, I'm going to assume that it might as well be in $\mathbb{R}$. – Joel Cornett Oct 20 '12 at 21:02
There has been some ambiguity in the use of the word "countable" recently - in your context does it mean that you can find a bijection with the set of positive integers (the "infinite only" definition) or with a subset of the positive integers (the "finite or countably infinite" definition)? – Mark Bennet Oct 20 '12 at 21:12
There is no such set in $\Bbb R$. Your example works if considered as a subset of $\Bbb Q$, but not as a subset of $\Bbb R$: in $\Bbb R$ every point of $[0,1]$ is a cluster point of the set, so it doesn’t contain all of its cluster points. – Brian M. Scott Oct 20 '12 at 21:12
@JoelCornett: The issue is that if we are working in the reals, $1/\sqrt{2}$ is a cluster point of your $U$ that is not in $U$. – André Nicolas Oct 20 '12 at 21:25
up vote 2 down vote accepted

$\{q\in\Bbb Q:0\le q\le 1\}$ 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.

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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.

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