Mixing
Why the product of two manifolds is paracompact?
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Some authors define a manifold as a paracompact Hausdorff space that is locally Euclidean. Also it is said that a product of two manifolds is a manifold.
However, we know that product of a two paracompact spaces is not necessarily paracompact.
So how can we be sure that a product of two manifolds is also paracompact and thus is also a manifold? Is it somehow related to the second-countability property that is usually defined along paracompactness?
general-topology differential-geometry manifolds paracompactness
edited Jan 25 at 5:26
Eric Wofsey
190k14216347
asked Apr 8 '17 at 18:29
Sergey DyldaSergey Dylda
1466
$endgroup$
add a comment |
$begingroup$
Some authors define a manifold as a paracompact Hausdorff space that is locally Euclidean. Also it is said that a product of two manifolds is a manifold.
However, we know that product of a two paracompact spaces is not necessarily paracompact.
So how can we be sure that a product of two manifolds is also paracompact and thus is also a manifold? Is it somehow related to the second-countability property that is usually defined along paracompactness?
general-topology differential-geometry manifolds paracompactness
edited Jan 25 at 5:26
Eric Wofsey
190k14216347
asked Apr 8 '17 at 18:29
Sergey DyldaSergey Dylda
1466
$endgroup$
add a comment |
$begingroup$
Some authors define a manifold as a paracompact Hausdorff space that is locally Euclidean. Also it is said that a product of two manifolds is a manifold.
However, we know that product of a two paracompact spaces is not necessarily paracompact.
So how can we be sure that a product of two manifolds is also paracompact and thus is also a manifold? Is it somehow related to the second-countability property that is usually defined along paracompactness?
general-topology differential-geometry manifolds paracompactness
edited Jan 25 at 5:26
Eric Wofsey
190k14216347
asked Apr 8 '17 at 18:29
Sergey DyldaSergey Dylda
1466
$endgroup$
Some authors define a manifold as a paracompact Hausdorff space that is locally Euclidean. Also it is said that a product of two manifolds is a manifold.
However, we know that product of a two paracompact spaces is not necessarily paracompact.
So how can we be sure that a product of two manifolds is also paracompact and thus is also a manifold? Is it somehow related to the second-countability property that is usually defined along paracompactness?
general-topology differential-geometry manifolds paracompactness
general-topology differential-geometry manifolds paracompactness
edited Jan 25 at 5:26
Eric Wofsey
190k14216347
asked Apr 8 '17 at 18:29
Sergey DyldaSergey Dylda
1466
edited Jan 25 at 5:26
Eric Wofsey
190k14216347
asked Apr 8 '17 at 18:29
Sergey DyldaSergey Dylda
1466
edited Jan 25 at 5:26
Eric Wofsey
190k14216347
edited Jan 25 at 5:26
Eric Wofsey
190k14216347
edited Jan 25 at 5:26
Eric Wofsey
190k14216347
190k14216347
asked Apr 8 '17 at 18:29
Sergey DyldaSergey Dylda
1466
asked Apr 8 '17 at 18:29
Sergey DyldaSergey Dylda
1466
asked Apr 8 '17 at 18:29
Sergey DyldaSergey Dylda
1466
1466
add a comment |
add a comment |
2 Answers
2
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By the Smirnov metrization theorem, a paracompact Hausdorff space that is locally metrizable is metrizable. Therefore every manifold is metrizable, and hence so is the product of two manifolds. In particular the product is paracompact.
answered Apr 8 '17 at 18:35
Matt SamuelMatt Samuel
38.8k63769
$endgroup$
1
$begingroup$
IIRC this is called the Smirnov metrication theorem
$endgroup$
– Henno Brandsma
Apr 8 '17 at 18:41
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Thank you so much! If I understood it right, it is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Hausdorff & paracompact) -> Product of two metrizable spaces is metrizable -> Every metrizable space is paracompact -> Product is paracompact.
$endgroup$
– Sergey Dylda
Apr 8 '17 at 18:44
$begingroup$
@Sergey Yes that's right.
$endgroup$
– Matt Samuel
Apr 8 '17 at 18:45
add a comment |
$begingroup$
A connected manifold is paracompact iff it is second-countable, so a general paracompact manifold is just a (possibly uncountable) disjoint union of second-countable manifolds. So, if you take a product of two such spaces, you get a disjoint union of products of two second-countable manifolds. A product of two second-countable spaces is second-countable, so each connected component of the product is second-countable and thus paracompact. So, the product is a disjoint union of paracompact spaces and thus itself paracompact.
answered Jan 25 at 5:26
Eric WofseyEric Wofsey
190k14216347
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
By the Smirnov metrization theorem, a paracompact Hausdorff space that is locally metrizable is metrizable. Therefore every manifold is metrizable, and hence so is the product of two manifolds. In particular the product is paracompact.
answered Apr 8 '17 at 18:35
Matt SamuelMatt Samuel
38.8k63769
$endgroup$
1
$begingroup$
IIRC this is called the Smirnov metrication theorem
$endgroup$
– Henno Brandsma
Apr 8 '17 at 18:41
$begingroup$
Thank you so much! If I understood it right, it is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Hausdorff & paracompact) -> Product of two metrizable spaces is metrizable -> Every metrizable space is paracompact -> Product is paracompact.
$endgroup$
– Sergey Dylda
Apr 8 '17 at 18:44
$begingroup$
@Sergey Yes that's right.
$endgroup$
– Matt Samuel
Apr 8 '17 at 18:45
add a comment |
$begingroup$
By the Smirnov metrization theorem, a paracompact Hausdorff space that is locally metrizable is metrizable. Therefore every manifold is metrizable, and hence so is the product of two manifolds. In particular the product is paracompact.
answered Apr 8 '17 at 18:35
Matt SamuelMatt Samuel
38.8k63769
$endgroup$
1
$begingroup$
IIRC this is called the Smirnov metrication theorem
$endgroup$
– Henno Brandsma
Apr 8 '17 at 18:41
$begingroup$
Thank you so much! If I understood it right, it is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Hausdorff & paracompact) -> Product of two metrizable spaces is metrizable -> Every metrizable space is paracompact -> Product is paracompact.
$endgroup$
– Sergey Dylda
Apr 8 '17 at 18:44
$begingroup$
@Sergey Yes that's right.
$endgroup$
– Matt Samuel
Apr 8 '17 at 18:45
add a comment |
$begingroup$
By the Smirnov metrization theorem, a paracompact Hausdorff space that is locally metrizable is metrizable. Therefore every manifold is metrizable, and hence so is the product of two manifolds. In particular the product is paracompact.
answered Apr 8 '17 at 18:35
Matt SamuelMatt Samuel
38.8k63769
$endgroup$
By the Smirnov metrization theorem, a paracompact Hausdorff space that is locally metrizable is metrizable. Therefore every manifold is metrizable, and hence so is the product of two manifolds. In particular the product is paracompact.
answered Apr 8 '17 at 18:35
Matt SamuelMatt Samuel
38.8k63769
edited Apr 8 '17 at 18:43
answered Apr 8 '17 at 18:35
Matt SamuelMatt Samuel
38.8k63769
answered Apr 8 '17 at 18:35
Matt SamuelMatt Samuel
38.8k63769
answered Apr 8 '17 at 18:35
Matt SamuelMatt Samuel
38.8k63769
38.8k63769
1
$begingroup$
IIRC this is called the Smirnov metrication theorem
$endgroup$
– Henno Brandsma
Apr 8 '17 at 18:41
$begingroup$
Thank you so much! If I understood it right, it is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Hausdorff & paracompact) -> Product of two metrizable spaces is metrizable -> Every metrizable space is paracompact -> Product is paracompact.
$endgroup$
– Sergey Dylda
Apr 8 '17 at 18:44
$begingroup$
@Sergey Yes that's right.
$endgroup$
– Matt Samuel
Apr 8 '17 at 18:45
add a comment |
1
$begingroup$
IIRC this is called the Smirnov metrication theorem
$endgroup$
– Henno Brandsma
Apr 8 '17 at 18:41
$begingroup$
Thank you so much! If I understood it right, it is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Hausdorff & paracompact) -> Product of two metrizable spaces is metrizable -> Every metrizable space is paracompact -> Product is paracompact.
$endgroup$
– Sergey Dylda
Apr 8 '17 at 18:44
$begingroup$
@Sergey Yes that's right.
$endgroup$
– Matt Samuel
Apr 8 '17 at 18:45
1
1
$begingroup$
IIRC this is called the Smirnov metrication theorem
$endgroup$
– Henno Brandsma
Apr 8 '17 at 18:41
$begingroup$
IIRC this is called the Smirnov metrication theorem
$endgroup$
– Henno Brandsma
Apr 8 '17 at 18:41
$begingroup$
Thank you so much! If I understood it right, it is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Hausdorff & paracompact) -> Product of two metrizable spaces is metrizable -> Every metrizable space is paracompact -> Product is paracompact.
$endgroup$
– Sergey Dylda
Apr 8 '17 at 18:44
$begingroup$
Thank you so much! If I understood it right, it is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Hausdorff & paracompact) -> Product of two metrizable spaces is metrizable -> Every metrizable space is paracompact -> Product is paracompact.
$endgroup$
– Sergey Dylda
Apr 8 '17 at 18:44
$begingroup$
@Sergey Yes that's right.
$endgroup$
– Matt Samuel
Apr 8 '17 at 18:45
$begingroup$
@Sergey Yes that's right.
$endgroup$
– Matt Samuel
Apr 8 '17 at 18:45
add a comment |
$begingroup$
A connected manifold is paracompact iff it is second-countable, so a general paracompact manifold is just a (possibly uncountable) disjoint union of second-countable manifolds. So, if you take a product of two such spaces, you get a disjoint union of products of two second-countable manifolds. A product of two second-countable spaces is second-countable, so each connected component of the product is second-countable and thus paracompact. So, the product is a disjoint union of paracompact spaces and thus itself paracompact.
answered Jan 25 at 5:26
Eric WofseyEric Wofsey
190k14216347
$endgroup$
add a comment |
$begingroup$
A connected manifold is paracompact iff it is second-countable, so a general paracompact manifold is just a (possibly uncountable) disjoint union of second-countable manifolds. So, if you take a product of two such spaces, you get a disjoint union of products of two second-countable manifolds. A product of two second-countable spaces is second-countable, so each connected component of the product is second-countable and thus paracompact. So, the product is a disjoint union of paracompact spaces and thus itself paracompact.
answered Jan 25 at 5:26
Eric WofseyEric Wofsey
190k14216347
$endgroup$
add a comment |
$begingroup$
A connected manifold is paracompact iff it is second-countable, so a general paracompact manifold is just a (possibly uncountable) disjoint union of second-countable manifolds. So, if you take a product of two such spaces, you get a disjoint union of products of two second-countable manifolds. A product of two second-countable spaces is second-countable, so each connected component of the product is second-countable and thus paracompact. So, the product is a disjoint union of paracompact spaces and thus itself paracompact.
answered Jan 25 at 5:26
Eric WofseyEric Wofsey
190k14216347
$endgroup$
A connected manifold is paracompact iff it is second-countable, so a general paracompact manifold is just a (possibly uncountable) disjoint union of second-countable manifolds. So, if you take a product of two such spaces, you get a disjoint union of products of two second-countable manifolds. A product of two second-countable spaces is second-countable, so each connected component of the product is second-countable and thus paracompact. So, the product is a disjoint union of paracompact spaces and thus itself paracompact.
answered Jan 25 at 5:26
Eric WofseyEric Wofsey
190k14216347
answered Jan 25 at 5:26
Eric WofseyEric Wofsey
190k14216347
answered Jan 25 at 5:26
Eric WofseyEric Wofsey
190k14216347
answered Jan 25 at 5:26
Eric WofseyEric Wofsey
190k14216347
190k14216347
add a comment |
add a comment |
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