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This is not a homework question, although it may appear so from the title.
So, in Munkres, Theorem 26.7. says that a product of finitely many compact spaces is compact.
It is first proved for two spaces, the rest follows by induction.
Now, there's a point in the proof I don't quite understand.
The proof consists of two parts, in one part the tube lemma is proved, i.e. given spaces X and Y, with Y compact, and given x0 in X, if N is an open set in X x Y containing x0 x Y, then there is a neighborhood W of x0 in X such that N contains the set W x Y.
The tube lemma is used in the second part, and I don't quite see why, so I'd like to get this clarified.
The second part goes: Let X and Y be compact spaces, let A be an open covering of X x Y. Take x0 from X, the slice x0 x Y is compact (since it's homeomorphic to Y which is compact by hypothesis) and can therefore be covered by finitely many elements A1, ... ,Am of A. Their union N = A1 U ... U Am is an open set containing x0 x Y. Now the tube lemma is applied to N, and the proof is easily finished, i.e. there exists a neighborhood W of x0 such that W x Y is contained in N, and hence covered by finitely many elements of A1, ..., Am. Now for any x in X, take the neighborhood Wx, so we have an open cover for X, which has a finite subcover, since X is compact, and hence W1 x Y, ..., Wk x Y is a finite open cover X x .
The tube lemma allows us to find a neighborhood of x0, i.e. later on, for any x from X. So that the union of these neighborhoods covers X, and since X is compact, we can find a finite subcover consisting of these neighborhoods.
The proof is perfectly clear. But my question is, couldn't we have just said, before applying the tube lemma: "Take from N = A1 U ... U Am the set containing x0, and do this for every X. The collection of these sets is an open cover for X, and hence has a finite subcover."
Or it's simply the point that we don't know which set contains x0, so we can't specify which set to take?
So, in Munkres, Theorem 26.7. says that a product of finitely many compact spaces is compact.
It is first proved for two spaces, the rest follows by induction.
Now, there's a point in the proof I don't quite understand.
The proof consists of two parts, in one part the tube lemma is proved, i.e. given spaces X and Y, with Y compact, and given x0 in X, if N is an open set in X x Y containing x0 x Y, then there is a neighborhood W of x0 in X such that N contains the set W x Y.
The tube lemma is used in the second part, and I don't quite see why, so I'd like to get this clarified.
The second part goes: Let X and Y be compact spaces, let A be an open covering of X x Y. Take x0 from X, the slice x0 x Y is compact (since it's homeomorphic to Y which is compact by hypothesis) and can therefore be covered by finitely many elements A1, ... ,Am of A. Their union N = A1 U ... U Am is an open set containing x0 x Y. Now the tube lemma is applied to N, and the proof is easily finished, i.e. there exists a neighborhood W of x0 such that W x Y is contained in N, and hence covered by finitely many elements of A1, ..., Am. Now for any x in X, take the neighborhood Wx, so we have an open cover for X, which has a finite subcover, since X is compact, and hence W1 x Y, ..., Wk x Y is a finite open cover X x .
The tube lemma allows us to find a neighborhood of x0, i.e. later on, for any x from X. So that the union of these neighborhoods covers X, and since X is compact, we can find a finite subcover consisting of these neighborhoods.
The proof is perfectly clear. But my question is, couldn't we have just said, before applying the tube lemma: "Take from N = A1 U ... U Am the set containing x0, and do this for every X. The collection of these sets is an open cover for X, and hence has a finite subcover."
Or it's simply the point that we don't know which set contains x0, so we can't specify which set to take?