Wallace's theorem on rectangular neighborhoods of compact rectangles
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Theorem. Let $Asubset X,Bsubset Y$ be compact subspaces of topological spaces $X,Y$. Let $Atimes Bsubset Wsubset Xtimes Y$ with $Wsubset Xtimes Y$ open. Then there exists opens $Asubset Usubset X,Bsubset Vsubset Y$ such that $Utimes Vsubset W$.
All proofs I could find are the same: we use the openness of $W$ to get for each point $(a,b)in Atimes B$ an open rectangle $U_{(a,b)}times V_{(a,b)}subset W$, and then draw a potato which leads us to apply the compactness of $A,B$ to produce a rectangle $Utimes V$.
Question. Is it possible to prove this theorem without mentioning points?
That $Wsubset Xtimes Y$ is open tells us that $Atimes B$ admits a basic open cover (i.e by open rectangles) which is contained entirely in $W$. However, without using the points of $Atimes B$ to index these open rectangles, I don't see how to proceed.
general-topology compactness
asked Dec 3 '18 at 22:48
ArrowArrow
5,10831445
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add a comment |
$begingroup$
Theorem. Let $Asubset X,Bsubset Y$ be compact subspaces of topological spaces $X,Y$. Let $Atimes Bsubset Wsubset Xtimes Y$ with $Wsubset Xtimes Y$ open. Then there exists opens $Asubset Usubset X,Bsubset Vsubset Y$ such that $Utimes Vsubset W$.
All proofs I could find are the same: we use the openness of $W$ to get for each point $(a,b)in Atimes B$ an open rectangle $U_{(a,b)}times V_{(a,b)}subset W$, and then draw a potato which leads us to apply the compactness of $A,B$ to produce a rectangle $Utimes V$.
Question. Is it possible to prove this theorem without mentioning points?
That $Wsubset Xtimes Y$ is open tells us that $Atimes B$ admits a basic open cover (i.e by open rectangles) which is contained entirely in $W$. However, without using the points of $Atimes B$ to index these open rectangles, I don't see how to proceed.
general-topology compactness
asked Dec 3 '18 at 22:48
ArrowArrow
5,10831445
$endgroup$
$begingroup$
What's the problem with a proof that mentions points? In fact Wallace holds for products of any number of spaces, not just for two. In its full form it's equivalent to AC, IIRC.
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– Henno Brandsma
Dec 4 '18 at 9:55
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@HennoBrandsma there is no problem. I'm just curious whether it's doable without points.
$endgroup$
– Arrow
Dec 4 '18 at 23:27
add a comment |
$begingroup$
Theorem. Let $Asubset X,Bsubset Y$ be compact subspaces of topological spaces $X,Y$. Let $Atimes Bsubset Wsubset Xtimes Y$ with $Wsubset Xtimes Y$ open. Then there exists opens $Asubset Usubset X,Bsubset Vsubset Y$ such that $Utimes Vsubset W$.
All proofs I could find are the same: we use the openness of $W$ to get for each point $(a,b)in Atimes B$ an open rectangle $U_{(a,b)}times V_{(a,b)}subset W$, and then draw a potato which leads us to apply the compactness of $A,B$ to produce a rectangle $Utimes V$.
Question. Is it possible to prove this theorem without mentioning points?
That $Wsubset Xtimes Y$ is open tells us that $Atimes B$ admits a basic open cover (i.e by open rectangles) which is contained entirely in $W$. However, without using the points of $Atimes B$ to index these open rectangles, I don't see how to proceed.
general-topology compactness
asked Dec 3 '18 at 22:48
ArrowArrow
5,10831445
$endgroup$
Theorem. Let $Asubset X,Bsubset Y$ be compact subspaces of topological spaces $X,Y$. Let $Atimes Bsubset Wsubset Xtimes Y$ with $Wsubset Xtimes Y$ open. Then there exists opens $Asubset Usubset X,Bsubset Vsubset Y$ such that $Utimes Vsubset W$.
All proofs I could find are the same: we use the openness of $W$ to get for each point $(a,b)in Atimes B$ an open rectangle $U_{(a,b)}times V_{(a,b)}subset W$, and then draw a potato which leads us to apply the compactness of $A,B$ to produce a rectangle $Utimes V$.
Question. Is it possible to prove this theorem without mentioning points?
That $Wsubset Xtimes Y$ is open tells us that $Atimes B$ admits a basic open cover (i.e by open rectangles) which is contained entirely in $W$. However, without using the points of $Atimes B$ to index these open rectangles, I don't see how to proceed.
general-topology compactness
general-topology compactness
asked Dec 3 '18 at 22:48
ArrowArrow
5,10831445
asked Dec 3 '18 at 22:48
ArrowArrow
5,10831445
asked Dec 3 '18 at 22:48
ArrowArrow
5,10831445
asked Dec 3 '18 at 22:48
ArrowArrow
5,10831445
asked Dec 3 '18 at 22:48
ArrowArrow
5,10831445
5,10831445
$begingroup$
What's the problem with a proof that mentions points? In fact Wallace holds for products of any number of spaces, not just for two. In its full form it's equivalent to AC, IIRC.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:55
$begingroup$
@HennoBrandsma there is no problem. I'm just curious whether it's doable without points.
$endgroup$
– Arrow
Dec 4 '18 at 23:27
add a comment |
$begingroup$
What's the problem with a proof that mentions points? In fact Wallace holds for products of any number of spaces, not just for two. In its full form it's equivalent to AC, IIRC.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:55
$begingroup$
@HennoBrandsma there is no problem. I'm just curious whether it's doable without points.
$endgroup$
– Arrow
Dec 4 '18 at 23:27
$begingroup$
What's the problem with a proof that mentions points? In fact Wallace holds for products of any number of spaces, not just for two. In its full form it's equivalent to AC, IIRC.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:55
$begingroup$
What's the problem with a proof that mentions points? In fact Wallace holds for products of any number of spaces, not just for two. In its full form it's equivalent to AC, IIRC.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:55
$begingroup$
@HennoBrandsma there is no problem. I'm just curious whether it's doable without points.
$endgroup$
– Arrow
Dec 4 '18 at 23:27
$begingroup$
@HennoBrandsma there is no problem. I'm just curious whether it's doable without points.
$endgroup$
– Arrow
Dec 4 '18 at 23:27
add a comment |
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$begingroup$
What's the problem with a proof that mentions points? In fact Wallace holds for products of any number of spaces, not just for two. In its full form it's equivalent to AC, IIRC.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:55
$begingroup$
@HennoBrandsma there is no problem. I'm just curious whether it's doable without points.
$endgroup$
– Arrow
Dec 4 '18 at 23:27