For each $I\in\mathbb{N}$ let $C_i$ be a closed subset of a topological space $(X_i\tau_i)$. Prove that $\prod_\limits{i=1}^{\infty}C_i$ is a closed subset of $\prod_\limits{i=1}^{\infty}(X_i,\tau_i)$.
$\prod_\limits{i=1}^{\infty}(X_i,\tau_i)\setminus\prod_\limits{i=1}^{\infty}C_i=\bigcup_\limits{i=1}^{\infty}\prod_\limits{i=1}^{\infty} (X_i\setminus C_i)$ since the union of open sets is open.
Question:
Is this reasoning acceptable for infinite product topological spaces?
Thanks in advance!
Best Answer
Your formula makes no sense because of the two conflicting $i$ indices.
We can correct it as follows (for arbitrary index set):
$$\prod_{i \in I} X_i \setminus \prod_{i \in I} C_i = \bigcup_{j\in I} \prod_{i \in I} O(i,j)$$ where $O(i,j) = X_i$ for $i \neq j$ and $O(i,j) = X_j \setminus C_j$ for $i=j$. We could write that product of $O(i,j)$ for fixed $j$ as $\pi_j^{-1}[X_j \setminus C_j]$ as well.
This holds as $f \notin \prod_{i \in I} C_i$ iff there exists some $j$ such that $f(j) \notin C_j$.
At any rate, the complement of $\prod_i C_i$ is writeable as a union of products of (sub)basic open sets and hence open.