A generalization of the generalized tube lemma

Question

I am trying to prove the following generalization of the generalized tube lemma:

Let $\{X_t\}_{t \in T}$ be a family of Hausdorff spaces and $\prod_{t \in T}A_t$ be a compact subset of $X=\prod_{t \in T}X_t$. If $W$ is an open subset of $X$ containing $\prod_{t \in T}A_t$ then for all $t \in T$ there exist open $W_t \subseteq X_t$ such that $\prod_{t \in T}A_t \subseteq \prod_{t \in T}W_t \subseteq W$.

It seems clear to me that the proof for the usual generalized tube lemma (where $T$ has two elements) just extended to an arbitrary product won't work, so it should rely on the properties of Hausdorff spaces in particular and the Tychonoff theorem. But beyond the fact that each $A_t$ is compact and closed and $\prod_{t \in T}A_t$ is closed, I don't see how to proceed. When I was searching to see if someone else had solved this problem, I noticed on the talk page for the Wikipedia entry on the tube lemma that someone mentioned that this generalization doesn't hold in general, but didn't provide a counter example.

Answer 1

Firstly, we don't need Hausdorffness of the $X_t$. Secondly, we can even strengthen it a bit: for all except finitely many $t \in T$ we can choose $W_t = X_t$, so $\prod_{t \in T} W_t$ is a canonical basic open subset of the product. This illustrates the intuitive idea that "compact sets behave like points" (e.g. in Hausdorff spaces we can separate two compact subsets by disjoint open sets, just like two points). This result is known as the Wallace Theorem.

Sketch of proof, following Engelking's "General Topology", Theorem 3.2.10:

1. Case of two spaces (classical), proof like the Tube lemma.
2. Case of finitely many spaces, by induction on the number of spaces and using 1.
3. For an arbitrary product, for each $a \in A = \prod_{t \in T} A_t$, we pick a canonical basic open subset $O_a$, depending on a finite subset $T_a$ of $T$, that contains $a$ and is contained in $W$. As $\prod_t A_t = A$ is compact by Tychonoff's theorem, finitely many of those cover $A$, so we have a finite subset $T'$ of $T$ such that, essentially, $\prod_{t \in T'} A_t \times \prod_{t \in T \setminus T'} A_t \subset W_1 \times W_2 \subset W$, where $W_1$ is a union of finite product of open sets (in the factors from $T'$), and $W_2$ is only a product of complete sets $X_t$ for the other $t$.
4. Finally we apply the case 2. again for $W_1$ and $\prod_{t \in T'} A_t$ to get the required canonical basic set.