Closed sets in the product topology

Question

Let $X=\Pi_{n\in \mathbb{N}}X_n$, where $X_n=\{0,1,...,k\}$ and $k\in \mathbb{N}$. Let $K\subset \bigcup_{n\in \mathbb{N}}\{0,1,...,k\}^n$ be arbitrary. $X$ is like the set of infinite words constructed from the alphabet $\{0,1,...,k\}$ and $K$ is a set of finite words. Define $$X_K=\{x\in X:\text{no x contains any finite word of K at any position}\}$$ If we put the product topology on $X$ I want to know if we can show that $X_K$ is closed in $X$? Any help is appreciated. Thank you!

Answer 1

In the product topology (where each $X_n$ is discrete, I assume, as this is common in such a setting) one commonly used base for the open subsets of $X$ consists of all sets of the form $O(w,m) = \{x \in X: x_i = w_i, i = m,\ldots, m + l(w)-1\}$, where $w$ is a finite word of length $l(w)$ from the alphabet $\{1,\ldots,k\}$, all infinite words that agree with a given finite word $w$ at some index $m$. It's clear that these sets are open as standard basic products $X_1 \times \ldots \times X_{m-1} \times \{w_1\} \times \{w_2\} \times \ldots \{w_{l(w)}\} \times X_{m+l(w)} \times \ldots$. (This uses that $w$ has finite length and all $X_m$ are discrete).

The complement of $X_K$ can be written as $\bigcup \{O(w,m): m \in \mathbb{N}, w \in K\}$, all words that do agree with a word from $K$ somewhere. As a union of (basic) open sets, the complement is thus open and the set $X_K$ closed (for any sized $K$).