Pseudocomplete spaces It is well-known that both locally compact Hausdorff spaces and completely metrizable spaces are Baire (i.e. satisfiy the Baire category theorem). A common generalization of both classes while still being Baire is the class of Čech-complete spaces. A finite product of Baire spaces does not have to be Baire; any product of Čech-complete spaces is Baire but not necessarily Čech-complete.
In 1960 Oxtoby defined the class of pseudocomplete spaces. This is a nice property that captures the idea of the proof of the Baire category theorem. It is easy to see that every Čech-complete space is pseudocomplete, that every pseudocomplete space is Baire, and that every product of pseudocomplete spaces is pseudocomplete.
In the proof of the last fact there is a small issue for me, but first the definitions:


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*A topological space is quasiregular if every nonempty open set contains the closure of a nonempty open set.

*A family of nonempty open sets $\mathcal{B}$ is called a π-base (or pseudobase, but this name collides with the notion of pseudobase related to pseudocharacter) if every nonempty open set contains a member of $\mathcal{B}$.

*A topological space is pseudocomplete if it is quasiregular and there exist a sequence $\{\mathcal{B}_n: n ∈ ω\}$ of π-bases such that every sequence $B_0 ⊇ \overline{B_1} ⊇ B_1 ⊇ \overline{B_2} ⊇ B_2 ⊇ \cdots$ such that each $B_n ∈ \mathcal{B}_n$ has nonempty intersection.


In the proof of the productivity of pseudocomplete spaces it is needed that even sequences $B_{n_0} ⊇ \overline{B_{n_0 + 1}} ⊇ B_{n_0 + 1} ⊇ \overline{B_{n_0 + 2}} ⊇ B_{n_0 + 2} ⊇ \cdots$ where $B_n ∈ \mathcal{B}_n$ for $n ≥ n_0$ have nonempty intersection (i.e. we may start later in the sequence of $π$-bases). Oxtoby says that we may assume that $X$ is in every $\mathcal{B}_n$, but I do not see why it is the case. It seems that simply adding $X$ to the π-bases may destroy the property of the sequence. On the other hand, what seems to work is to remove some members of the π-bases. Namely, we put $\mathcal{B}'_0 := \mathcal{B}_0$ and $\mathcal{B}'_{n + 1} := \{B ∈ \mathcal{B}_{n + 1}: ∃C ∈ \mathcal{B}'_n: \overline{B} ⊆ C\}$. These will still be π-bases and for every $B_{n_0} ∈ \mathcal{B}'_{n_0}$ there is $B_{n_0 -1} ∈ \mathcal{B}'_{n_0 - 1}$ such that $B_{n_0 - 1} ⊇ \overline{B_{n_0}} ⊇ B_{n_0}$ and so on, i.e. we may extend the sequence at the beginning. This modified sequence satisfies the property we need and we may even add $X$ to these π-bases (unless $X = ∅$). But the argument seems to be nontrivial for a comment like “we may assume that $X ∈ \mathcal{B}_n$”. So I wonder, am I missing something?
Updated: Let me give a concrete example of the situation. Let's write $U \prec V$ for $\overline{U} ⊆ V$. Let $\{\mathcal{B}_n: n ∈ ω\}$ be a pseudocompleteness-witnessing sequence for $(0, 1)$ such that every member of $\mathcal{B}_n$ has diameter $< 2^{-(n + 2)}$, let $A_n := (0, 2^{-n})$, and let $\mathcal{B}'_n := \mathcal{B}_n ∪ \{A_n\}$.
The sequence $\mathcal{B}_0, \mathcal{B}'_1, \mathcal{B}'_2, …$ still witnesses that $(0, 1)$ is pseudocomplete since every compatible $\prec$-decreasing sequence $\{B_n: n ∈ ω\}$ is in fact compactible with the original system $\{\mathcal{B}_n: n ∈ ω\}$: if $B_{n + 1} = A_{n + 1}$, then $B_n = A_n$ since $A_n$ is the only member of $B'_n$ containing $A_{n + 1}$, but $A_0 = X ∉ \mathcal{B}_0$. On the other hand, if we add $X$ to $\mathcal{B}_0$, i.e. if we consider the system $\{\mathcal{B}'_n: n ∈ ω\}$, then we have the compatible sequence $A_0 \succ A_1 \succ A_2 \succ \cdots$ with empty intersection.
 A: $\newcommand{\cl}{\operatorname{cl}}$Your example shows that we cannot simply add $X$ to each of the $\pi$-bases, but only because the sets $A_n$ ($n\ge 1$) are basically irrelevant: they cannot appear in any sequence
$$B_0\supseteq\cl B_1\supseteq B_1\supseteq\cl B_2\supseteq B_2\supseteq\ldots$$
until $X=A_0$ is added to $\mathscr{B}_0$.
However, a simple preliminary step makes it possible to ensure that $X$ is in each $\pi$-base. Suppose that $\{\mathscr{B}_n:n\in\omega\}$ is a family of $\pi$-bases witnessing the pseudocompleteness of $X$. Let $\mathscr{B}_0'=\mathscr{B}_0$, and for $n\in\omega$ let
$$\mathscr{B}_{n+1}'=\{B\in\mathscr{B}_{n+1}:\exists U\in\mathscr{B}_n'\,(\cl B\subseteq U)\}\,.$$
Each $\mathscr{B}_n'$ is still a $\pi$-base for $X$, so $\{\mathscr{B}_n':n\in\omega\}$ still witnesses the pseudocompleteness of $X$, and it’s easy to see that $\{\mathscr{B}_n\cup\{X\}:n\in\omega\}$ also witnesses the pseudocompleteness of $X$.
A: It's really quite trivial that $X$ can be added to every pseudobase sequence of $\pi$-bases $\{\mathcal{B}_n: n \in \omega \}$ that witnesses pseudocompleteness. "It seems that simply adding $X$ to the $\pi$-bases may destroy the property of the sequence" as you say. I don't see that at all:
Suppose we add $X$ to all of them (of course keeps them all a $\pi$-base) and we'd suddenly have a counterexample to pseudocompleteness: a sequence $B_n \in \mathcal{B}_n, n \in \omega$ with $\forall n \in \omega: B_n \supseteq \overline{B_{n+1}}$ and such that $\bigcap_n B_n = \emptyset$. This means that we must have had $B_n = X$ infinitely many times (otherwise we could have started the sequence after the last occurrence of $X$, the only new set we added, and we cannot have $B_n \neq X$ for all $n$, because the original subbases do not have such a sequence) but then the decreasingness forces all $B_n = X$ and then the intersection is definitely not empty, but $X$. So we cannot add a bad sequence by adding $X$ to all $\pi$-bases, but it is convenient, because we then can use standard product base elements formed from the subbases to get the subbase for the product.
