2

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.
trujello
  • 1,218
  • 1
    Not every open set in $2^X$ is closed, nor the other way around. There is a base of clopen sets though, but that's all. – Henno Brandsma Sep 01 '19 at 10:59
  • Oh I see. Thank you! Does relate to the fact that basic open sets are used to create open sets but closed sets cannot necessarily be used to create closed sets in the same fashion? (i.e., clopen sets can be unioned to create open sets but not closed ones?) – trujello Sep 03 '19 at 03:36
  • yes, unions of clopen sets (from the base) need no longer be closed, nor are their intersections still open in general. But a clopen base is a common and useful thing. – Henno Brandsma Sep 03 '19 at 04:17

2 Answers2

3

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 $1$.)

bof
  • 78,265
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.

Henno Brandsma
  • 242,131