Limit point $p$ of $A \subset 2^\mathbb{R}$ such that no sequence in $A$ converges to $p$. Can $A$ be countable?

Question

The problem here is to find a subset $A \subset 2^\mathbb{R}$ and a limit point $p$ of $A$ such that no sequence in $A$ converges to $p$. Here, $2^\mathbb{R}$ is the product $\prod_{\lambda \in \mathbb{R}}\{0, 1\}$ equipped with the product topology.

I think I have an example for when $A$ is not countable. Suppose $A$ is the set of all points $(x_i)_{i \in \mathbb{R}}$ such $$ (x_i) = 1 \hspace{0.5cm} \text{ if } i \in \mathbb{Q} \text{ and } \hspace{0.5cm}(x_i) = 0 \text{ or } 1 \hspace{0.5cm} \text{ if } i \in \mathbb{R}\setminus\mathbb{Q}. $$ Consider the point $p \in 2^\mathbb{R}$ such that $$ p(i) = \begin{cases} 1 & \text{for all } i \in \{\pi + q \mid q \in \mathbb{Q} \}\\ 0 & \text{otherwise } \end{cases} $$ Here $p(i)$ indicates the $i$-th coordinate (where $i$ is a real number) of the point $p \in 2^\mathbb{R}$. $\pi$ was chosen randomly; I just needed some irrational.

Now any open set of $p$ must be of the form $$ \{(x_i) \in 2^\mathbb{R} \mid x_{\pi + q_1} = x_{\pi + q_2} = \cdots = x_{\pi + q_n} =1 \text{ where } q_1, q_2, \dots, q_n \in \mathbb{Q}\} $$ which will always intersect $A$. Hence, $p$ is a limit point of $A$.

By a theorem in topology, if there exists a sequence of points in $A$ which converge to $p$, then $p \in \overline{A}$. And, I already proved that every open set in $2^X$ is closed for any $X$, so $A = \overline{A}$; however $p \not\in A$. Therefore there is no sequence.

My questions:

1. Is my work correct?
2. Is it possible to find an example for when $A$ is countable? I've tried but can't seem to find one that works. I think it's supposed to be possible.

Answer 1

For notational simplicity identify points in $2^\mathbb R$ with subsets of $\mathbb R$ in the natural way, i.e., identify a set with its characteristic function.

Let $A$ be the set of all $x\in2^\mathbb R$ such that (1) $x$ is the union of finitely many open intervals with rational endpoints and (2) the measure of $x$ is less than $1$.

Then $A$ is countable, and $\mathbb R$ is a limit point of $A$, but no sequence in $A$ converges to $\mathbb R$. (Each element of $A$ is a set of measure at most $1$; it follows from basic properties of Lebesgue measure that the limit of a sequence in $A$, if it exists, has measure at most $1$.)

Answer 2

In this answer I give a sequence $(f_n)_n$ in $\{0,1\}^C$ (where $C= \{0,1\}^\mathbb{N}$ is the Cantor space, which has the same cardinality as $\Bbb R$ so that $\{0,1\}^\mathbb{R}$ is homeomorphic to $\{0,1\}^C$) that has no convergent subsequence (showing that this large product is not sequentially compact).

This defines a countable set that has (lots of) limit points by compactness but no sequence converging to that limit point.