Is a particular countable subset of the Cantor set Polish?

Question

Consider the Cantor space $\mathcal{C} := \{ 0, 1 \}^{\mathbb{N}}$ and the subset $\mathcal{T} \subseteq \mathcal{C}$ of sequences that start with $1$ and eventually "terminate" with $0$, i.e. $\mathcal{T} = \{ (1,0,0, \dots,), (1,1,0, \dots), \dots \} = \{ x \in \mathcal{C} \mid \exists K: \forall k \geq K: x_k = 0 \wedge \forall k < K: x_k = 1 \}$.

Is $\mathcal{T}$ Polish with the induced subspace topology of $\mathcal{C}$?

Properties of $\mathcal{T}$:

• $\mathcal{T}$ is countable and thus and $F_\sigma$-subset of $\mathcal{C}$

• $\mathcal{T}$ is not closed in $\mathcal{C}$

• $\mathcal{T}$ can also be seen as a disjoint union of Polish spaces (the singletons) and thus Polish as well. But I don't see how to write it as a $G_\delta$-subset of $\mathcal{C}$? Or is the disjoint union topology on $\mathcal{T}$ a different one as the subspace topology in this case (as is similar to the case of the rationals as a countable subset of the reals)?

Answer 1

HINT: As an alternative to Asaf's suggestion in the comments, observe that every Polish space is a Baire space (why?), and consider $\bigcap_{x\in\mathcal{T}}(\mathcal{T}\setminus\{x\})$.

Added: Oops! That hint was based on the same misreading that Asaf had, i.e., that $\mathcal{T}$ is the set of sequences beginning with $1$ that are eventually $0$. In fact your $\mathcal{T}$ is a discrete subset of the Cantor set: it's not hard to find nbhds of the points of $\mathcal{T}$ that witness this fact. (Alternatively, show that if $x_n$ is the member of $\mathcal{T}$ beginning with exactly $n$ ones, then the sequence $\langle x_n:n\in\Bbb Z^+\rangle$ converges in the Cantor set and therefore is discrete.)

Answer 2

You could use automata theory (see Thm I.6.2, I.9.9 and III.3.11 from this book) to solve your question.

Let $X$ be a subset of $\{0,1\}^\omega$. Then the following conditions are equivalent:

1. $X$ is a countable intersection of open sets,
2. There exists a subset $L$ of $\{0,1\}^+$ such that $X=\overrightarrow{\!L}$, where $$ \overrightarrow{\!L} = \{u \in \{0,1\}^\omega \mid u \text{ has infinitely many prefixes in }L\}, $$
3. $X$ is accepted by a deterministic Büchi automaton.

Since $1^+0^\omega = \overrightarrow{1^+0^+}$, the set $1^+0^\omega$ is a countable intersection of open sets and hence is Polish.

Remark. If $X$ is $\omega$-regular, like your example $1^+0^\omega$, you may assume in (2) that $L$ is a regular language and in (3) that $X$ is accepted by a finite deterministic Büchi automaton.