How can I prove this version of tube lemma for Tychonoff theorem?

Question

I am trying to prove this version of tube lemma for the Tychonoff theorem.

Lemma. Let $\mathscr{A}$ be a collection of basis elements for the topology of the product space $X \times Y$ , such that no finite subcollection of $\mathscr{A}$ covers $X \times Y$ . If X is compact, there is a point $x$ in X such that no finite subcollection of A covers the slice ${x} \times Y$ .

For solutions on the Internet, they claim that we can choose finitely many open sets of the form $U \times V$. (https://www2.math.ethz.ch/education/bachelor/lectures/fs2013/math/topo/soln09.pdf, 06-(a))

But I'm not sure that how can we choose the basis of the form like that. Although we know that $\mathscr{A}$ is a collection of basis element, It doesn't have to be a form of open boxes.

If anyone know that how can we choose such open boxes for arbitrary forms of bases?

Answer 1

Suppose, on the contrary, that for each $x\in X$ there is a finite $\mathscr{A}_x\subseteq\mathscr{A}$ that covers the slice $\{x\}\times Y$. Then there are an $n_x\in\Bbb Z^+$, open nbhds $U_k(x)$ in $X$ for $k=1,\ldots,n_x$, and open sets $V_k(x)$ in $Y$ for $k=1,\ldots,n_x$ such that

$$\mathscr{A}_x=\{U_k(x)\times V_k(x):k=1,\ldots,n_x\}\;.$$

For each $x\in X$ let

$$W(x)=\bigcap_{k=1}^nU_k(x)\;.$$

• Verify that $W(x)$ is an open nbhd of $x$.

For $x\in X$ let

$$\mathscr{A}_x'=\{W(x)\times V_k(x):k=1,\ldots,n_x\}\;.$$

• Verify that $\mathscr{A}_x'$ is an open cover of $\{x\}\times Y$.

Let $\mathscr{W}=\{W(x):x\in X\}$.

• Use compactness of $X$ to get a finite $\mathscr{W}_0=\{W(x_1),\ldots,W(x_m)\}\subseteq\mathscr{W}$ covering $X$ for some $m\in\Bbb Z^+$.
• Then show that $\bigcup_{k=1}^m\mathscr{A}_{x_k}$ is a finite subfamily of $\mathscr{A}$ covering $X\times Y$, thereby getting the desired contradiction.