consider the space $\{0,1\}^{\mathbb{N}}$ of all infinite binary sequences, called the Cantor-Space. This space is metrizable with metric $$ d(u,v) = 2^{-(r-1)} \qquad \textrm{ where } r = \operatorname{min}\{n : u_n \ne v_n\}. $$ Therefore we have a topology. Let $X = \{0,1\}$. The open sets are sets of the form $W \cdot X^{\mathbb{N}}$ with $W \subseteq X^{*}$. The clopen sets are sets of the form $W \cdot X^{\mathbb{N}}$ with $W \subset X^{*}$ and $W$ finite. Now I conjecture:

If $E = W\cdot X^{\mathbb{N}}$ with $W$ infinite, then there exists a $w$ such that $w \in \overline{W}$ (i.e. there exists a sequence $w_i$ with $w_i \to w$) and $w \notin E$ (because then $w$ is open but not closed). For some examples I could construct such a sequence, but I am unable to find such a sequence for an arbitrary set of this form? Is it possible to construct such a sequence?

  • $\begingroup$ I don’t understand. The sets of the of form $W\times X^\mathbb N$ should be open only when $W$ is finite. Moreover, a cartesian product of closed sets should be closed. $\endgroup$ – Alex Ravsky Apr 16 '13 at 17:03
  • $\begingroup$ No, pick $w \in W \cdot X^{\mathbb{N}}$, then all word which have with $w$ a prefix in common are in $W \cdot X^{\mathbb{N}}$, and this set could be described by an open ball. It doesn't matter if $W$ is finite or infinite in this argument. $\endgroup$ – StefanH Apr 16 '13 at 17:07
  • $\begingroup$ The definitions and some basic properties are summarized in the first sections of this work: researchspace.auckland.ac.nz/handle/2292/10538 $\endgroup$ – StefanH Apr 16 '13 at 17:08

I came back to your question and it seems I got it this time (otherwise I still may try to take a look at it near 2021. :-)).

I’ll assume that $X^*$ is a set of finite words over $X$ and for each word $w\in X^*$ the set $w\cdot X^{\Bbb N}$ consists of all elements of $X^{\Bbb N}$ which starts from $w$.

Let $E = W\cdot X^{\mathbb{N}}$ with $W$ infinite. Consider a reduced set $$W^r=\{w\in W: w\mbox{ has no proper prefix in }W\}.$$ Then $W\cdot X^{\mathbb{N}}=W^r\cdot X^{\mathbb{N}},$ and we wish to exclude a case when the set $W^r$ is finite.

The by induction we can define a sequence $\{v_n:n\ge 0\}$ of words over $X$ such that $v_0=\varnothing$ and for each $n\ge 1$

(i) A word $v_n$ has length $n$,

(ii) a word $v_{n-1}$ is a prefix of $v_n$,

(iii) a set $W_n=\{w\in W_r: v_n\mbox{ is a prefix of }w \}$ is infinite.

Conditions i and ii implies that there exists an infinite word $w=\lim _{n\to\infty} v_n\in X^{\Bbb N}$. Conditions i and iii imply that $w\in\overline{W^r\cdot X^{\Bbb N}}$. Asssume that $w\in W^r\cdot X^{\Bbb N}$. Then $w$ has a prefix $w_n\in W^r$ of length $n$ for some $n$. Then $w_n=v_n$ and it is a prefix for each word from the set $W_{n+1}\subset W^r$, which contradicts the irreducibility of the set $W^r$.


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