Talk:Tube lemma

Confusion in the proof

I think there is some confusion in the proof: in the definition of $\pi _{Y}C$, the $\exists x$ should be deleted: it is the x in the lemma. Second, $\pi _{Y}$ is used with two different meanings.

I would write something like:

Let $\pi _{Y}:X\times Y\longrightarrow Y$ be the projection on the second factor. If $(U_{\alpha })_{\alpha \in A}$ is an open cover (in $X\times Y$) of $\{x\}\times Y$, then $(\pi _{Y}(U_{\alpha }))_{\alpha \in A}$ is an open cover of the compact space $Y$, from which we can extract a finite open subcover, which corresponds to a finite family $(U_{\alpha })_{\alpha \in B}$, which covers $\bigcap _{\alpha \in B}\pi _{X}(U_{\alpha })\times Y$. The intersection being a finite intersection of neighborhoods of $\{x\}$, the lemma is proved.

82.242.170.86 18:23, 9 February 2007 (UTC)

Deletion

Dear All,

I believe that this article should be either nominated for deletion or restarted (deleted and rewritten) for the following reasons:

1. The article does not provide a purpose or reason for proving the tube lemma.

2. The proof is correct but not appropriately structured.

3. A much simpler proof exists.

4. The article does not give the general form of the tube lemma which is a much stronger result than the tube lemma. This is another reason why the article should be restarted.

5. The article should properly define terminology such as what a slice is and what a tube is in dot points.

6. This is not the exact form of the tube lemma; the following is:

Let N be an open set containing {x} X Y for some x belonging to X where X and Y are topological spaces. If Y is compact, then there exists U open in X such that the tube U X Y contains {x} X Y and is contained in N.

Note the difference: N need not cover X X Y.

7. A proper proof should not require the use of projection maps. The result: If X is compact and Y is homeomorphic to X, then Y is compact, should be used in order to provide an easier proof. Basically, you don't need to go into projections.

8. The generalized tube lemma will give some purpose to this article; without it the article is not complete.

9. The article doesn't state that the purpose of the tube lemma is to prove that the product of finitely many compact spaces is compact and that the tube lemma cannot be generalized to arbitrary products.

10. The article doesn't supply any links to the Tychonoff theorem. It doesn't even hint at any relation between the tube lemma and compactness.

I hope you agree with all the reasons I have put forth. I am going to put up and AFD tag for the above reasons. Basically, my purpose is to remove this article and add a new article on the generalized tube lemma with applications to compactness.

Topology Expert (talk) 10:55, 23 June 2008 (UTC)

I disagree with your point 8, unless you find some application for this generalization. It seems to appear only as an exercise in some textbooks. I regret that the emphasis on this generalization makes the proof of the mere Tube lemma less readable. Anne Bauval (talk) 19:10, 21 June 2015 (UTC)

Some issues

There are some problems with this article. First, in the proof, $C$ is used but not defined. Second, it is not stated if this proves the Tube lemma or the generalized tube lemma. Example 1 makes no sense to me. Also, can you explain the reference to the maximum principle? It seems completely bogus. Oded (talk) 18:12, 26 June 2008 (UTC)

Axiom of choice

It appears that this (rather vague) proof implicitly uses the axiom of choice (choice of V_a,b), which is not necessary or appropriate—with the axiom of choice, Tychonoff's theorem is available. Planetmath gives a more reasonable proof. Dfeuer (talk) 20:47, 15 November 2013 (UTC)