Prob. 14, Sec. 30, in Munkres' TOPOLOGY, 2nd ed: The product of a Lindelof space and a compact space is a Lindelof space

Question

Here is Prob. $14$, Sec. $30$, in the book Topology by James R. Munkres, 2nd edition:

Show that if $X$ is Lindelof and $Y$ is compact, then $X \times Y$ is Lindelof.

Here is my Math Stack Exchange post on this very problem.

In the present post, I seek to give an alternative proof.

My Attempt:

Let $\pi_1 \colon X \times Y \longrightarrow X$ and $\pi_2 \colon X \times Y \longrightarrow X$ be the projection mappings defined by $$ \pi_1 ( x \times y) := x \qquad \mbox{ and } \qquad \pi_2 ( x \times y) := y. $$ These two mappings are continuous open maps.

Let $\mathscr{A}$ be any open covering of $X \times Y$ by the basis elements for the product topology of the form $U \times V$, where $U$ is open in $X$, and $V$ is open in $Y$.

This open covering $\mathscr{A}$ yields the open coverings $\mathscr{A}_X$ and $\mathscr{A}_Y$ of $X$ and $Y$ respectively, given by $$ \mathscr{A}_X := \left\{ \pi_1 (A) \colon A \in \mathscr{A} \right\} \qquad \mbox{ and } \qquad \mathscr{A}_Y := \left\{ \pi_2 (A) \colon A \in \mathscr{A} \right\}. $$ Since $X$ is Lindelof and $Y$ is compact, some countable subcollection $$ \left\{ \pi_1 \left( A_1 \right), \pi_1 \left( A_2 \right), \pi_1 \left( A_3 \right), \ldots \right\} $$ of $\mathscr{A}_X$ covers $X$, and some finite subcollection $$ \left\{ \pi_2 \left( A_1^\prime \right), \ldots, \pi_2 \left( A_N^\prime \right) \right\} $$ of $\mathscr{A}_Y$ covers $Y$.

Therefore, the collection $$ \left\{ A_1, A_2, A_3, \ldots \right\} \cup \left\{ A_1^\prime, \ldots, A_N^\prime \right\} $$ is a countable subcollection of $\mathscr{A}$ that also covers $X \times Y$. Thus, it follows that $X \times Y$ is Lindelof.

Is this proof correct? If so, is my write-up of the proof clear enough too? Or, is there something wrong?

PS:

I think I can figure out where I have gone wrong.

As $$ X = \bigcup_{n=1}^\infty \pi_1 \left( A_n \right), $$ so we have \begin{align} X \times Y &= \pi_1^{-1} (X) \\ &= \pi_1^{-1} \left( \bigcup_{n=1}^\infty \pi_1 \left( A_n \right) \right) \\ &= \bigcup_{n=1}^\infty \pi_1^{-1} \left( \pi_1 \left( A_n \right) \right) \\ &\supset \bigcup_{n=1}^\infty A_n, \qquad \tag{1} \end{align} and since the mapping $\pi_1 \colon X \times Y \longrightarrow X$ need not be injective, therefore the inclusion in (1) above need not be an equality, thus showing that the sets $A_1, A_2, A_3, \ldots$ need not cover $X \times Y$.

A similar argument shows that the sets $A_1^\prime, \ldots, A_N^\prime$ need not cover $X \times Y$.

Am I right?

Answer 1

Your proof does not work.

Therefore, the collection $$ \left\{ A_1, A_2, A_3, \ldots \right\} \cup \left\{ A_1^\prime, \ldots, A_N^\prime \right\} $$ is a countable subcollection of $\mathscr{A}$ that also covers $X \times Y$.

The collection is indeed countable but there is no guarantee that it covers $X\times Y$.

For each point $(x,y)\in X\times Y$ we can find sets $A_i=U_i\times V_i\in\mathcal A$ and $A_j^{\prime}=U_j^\prime\times V_j^\prime\in\mathcal A$ such that $x\in\pi_1(A_i)=U_i$ and $y\in\pi_2(A_j^{\prime})=V_j^\prime$.

So $(x,y)\in U_i\times V_j^\prime$, but there is no guarantee that $U_i\times V_j^\prime\in\mathcal A$.