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Statement: Suppose $X$ is Lindelöf and $Y$ is compact, then $X \times Y$ is Lindelöf.

I know that the real proof should be using the tube lemma, but what is the problem with the following proof?

Pick an open cover $\mathcal{U}$ of $X\times Y$. Since the projection map $\pi_1$ and $\pi_2$ are open maps, the two collections $\mathcal{A} = \{\pi_1(U)| U\in \mathcal{U}\}$ and $\mathcal{B} =\{\pi_2(U)| U\in \mathcal{U}\}$ form open coverings for $X$ and $Y$, respectively. Since $X$ is Lindelof, there exists a countable subcovering $\mathcal{U_X} = \{\pi_1(U_1), \pi_1(U_2), \cdots, \pi_1(U_n), \cdots\} \subset \mathcal{A}$ that covers $X$. Since $Y$ is compact, there exists a finite subcovering $\mathcal{U_Y} = \{\pi_2(U_1'), \pi_2(U_2'), \cdots, \pi_2(U_n')\}\subset \mathcal{B}$ that covers $Y$. Then the collection of sets $\mathcal{U'} = \mathcal{U_X} \cup \mathcal{U_Y}$ forms a countable covering $X\times Y$.

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Firstly, as a notation issue $\mathcal{U}_X$ is a cover of $X$ and likewise $\mathcal{U}_Y$ is one for $Y$. We are looking for a subcover of $\mathcal{U}$ which are open sets in $X \times Y$, while your $\mathcal{U}_X$ are open sets in $X$, not of $\mathcal{U}$.

You probably mean the sets in $\mathcal{U}$ they are projections of, so if we have $\mathcal{A}' = \{\pi_1[U_1], \ldots, \pi_1[U_n], \ldots, \}$ as a countable subcover of $\mathcal{A}$ you define $\mathcal{U}_X$ as $\{U \in \mathcal{U}: \pi_1[U] \in \mathcal{A}'\}$, so at least all the $U_n$ etc. Similarly for $\mathcal{U}_Y$ and a finite subcover $\mathcal{B}'$ of $\mathcal{B}$, of course.

How would a "proof" that this is in fact a subcover go? Suppose $(x,y) \in X$. Then $x$ is covered by some $\pi_1[U(x)] \in \mathcal{A}'$, and also by some $\pi_1[U(y)] \in \mathcal{B}'$. Nothing garantuees you that $(x,y)$ is in either $U(x)$ or $U(y)$ as the first only has a condition only on $x$ and the second only on $y$, not on them both combined.

You might think, define my subcover to be all products $\{U \times V: U \in \mathcal{A}', B \in \mathcal{B}'\}$. This is not necessarily a subcover either: e.g. suppose we were covering with open circles in the plane, then the products of projections give open squares (so a totally different open cover).

The final idea you could have (also doesn't work): start with a cover $\{U_i \times V_i: i \in I\}$ of basic open sets (these suffice for showing Lindelöfness), find countably many $J \subset I$ such that $\{U_i: i \in J\}$ cover $X$ and finitely many $J'$ such that $\{V_j: j \in J'\}$ covers $Y$. Then form all $\{U_i \times V_j: i \in I', j \in J'\}$. These now do form a countable cover of $X \times Y$, but not necesarilly a subcover as the index sets $I',J'$ could be totally disjoint (we get them independently of each other), so we are making new sets here that were not in the original cover, which is what the point was.

It's good to think about proofs like this; they don't work, and you see why you need another idea (here the tube lemma).

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I've found it useful, when thinking about a proof, to carry it to the extreme. For example, your "proof".does not need compactness, it works for X and Y Lindelof. In fact, it "shows" for any space X that any cover of XxX which includes the diagonal has a finite subcover. Considering those cases should lead you to the error.

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  • $\begingroup$ Why was this down-voted? Is it incorrect? (Or did someone just not think this was sufficient for an "answer")? $\endgroup$ – Max von Hippel Apr 29 at 22:11

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