Under what hypotheses do all bounded sets have “area”?Lebesgue measuarable sets under a differentiable bijectionArea under a curveBounded sets have finite measureArea under the curve?Area bounded under curvesUnder what hypotheses is a solution to the Lagrangian multiplier equations automatically a global minimum?Bounded and unbounded setssets in $sigma$-algebra have preimage measurable if the sets generating the $sigma$-algebra is measurable under preimageSum of measurable functions is measurable: countable choice required?Area under Curve Limits
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Under what hypotheses do all bounded sets have “area”?
Lebesgue measuarable sets under a differentiable bijectionArea under a curveBounded sets have finite measureArea under the curve?Area bounded under curvesUnder what hypotheses is a solution to the Lagrangian multiplier equations automatically a global minimum?Bounded and unbounded setssets in $sigma$-algebra have preimage measurable if the sets generating the $sigma$-algebra is measurable under preimageSum of measurable functions is measurable: countable choice required?Area under Curve Limits
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In "Infinitesimal Calculus" by Henle and Kleinberg, page $52$, the statement is made that, "[U]nder some strange hypotheses, all bounded sets can be assigned a numerical 'area'."
I haven't been able to locate anything about this by Googling. Can some elucidate the statement, or give me a reference?
Full statement: https://books.google.com/books?id=mMPCAgAAQBAJ&pg=PA52
Incidentally, it has also been shown that under some strange hypotheses, all bounded sets can be assigned a numerical "area." The universe under these hypotheses is a very odd and interesting one, but we are not interested in it here, for in that universe the hyperreal numbers do not exist!
calculus measure-theory
edited 8 hours ago
Chris Culter
21.6k4 gold badges38 silver badges89 bronze badges
asked 8 hours ago
saulspatzsaulspatz
21.1k4 gold badges16 silver badges38 bronze badges
$endgroup$
add a comment |
$begingroup$
In "Infinitesimal Calculus" by Henle and Kleinberg, page $52$, the statement is made that, "[U]nder some strange hypotheses, all bounded sets can be assigned a numerical 'area'."
I haven't been able to locate anything about this by Googling. Can some elucidate the statement, or give me a reference?
Full statement: https://books.google.com/books?id=mMPCAgAAQBAJ&pg=PA52
Incidentally, it has also been shown that under some strange hypotheses, all bounded sets can be assigned a numerical "area." The universe under these hypotheses is a very odd and interesting one, but we are not interested in it here, for in that universe the hyperreal numbers do not exist!
calculus measure-theory
edited 8 hours ago
Chris Culter
21.6k4 gold badges38 silver badges89 bronze badges
asked 8 hours ago
saulspatzsaulspatz
21.1k4 gold badges16 silver badges38 bronze badges
$endgroup$
$begingroup$
What does 'bounded' mean in context (e.g. bounded as in a metric space, topological space, order, etc.)?
$endgroup$
– R. Burton
8 hours ago
$begingroup$
@R.Burton I don't know. The statement appears as an aside, in a historical discussion at the beginning of the integral calculus portion of the book, right after a mention of Lebesgue.
$endgroup$
– saulspatz
8 hours ago
add a comment |
$begingroup$
In "Infinitesimal Calculus" by Henle and Kleinberg, page $52$, the statement is made that, "[U]nder some strange hypotheses, all bounded sets can be assigned a numerical 'area'."
I haven't been able to locate anything about this by Googling. Can some elucidate the statement, or give me a reference?
Full statement: https://books.google.com/books?id=mMPCAgAAQBAJ&pg=PA52
Incidentally, it has also been shown that under some strange hypotheses, all bounded sets can be assigned a numerical "area." The universe under these hypotheses is a very odd and interesting one, but we are not interested in it here, for in that universe the hyperreal numbers do not exist!
calculus measure-theory
edited 8 hours ago
Chris Culter
21.6k4 gold badges38 silver badges89 bronze badges
asked 8 hours ago
saulspatzsaulspatz
21.1k4 gold badges16 silver badges38 bronze badges
$endgroup$
In "Infinitesimal Calculus" by Henle and Kleinberg, page $52$, the statement is made that, "[U]nder some strange hypotheses, all bounded sets can be assigned a numerical 'area'."
I haven't been able to locate anything about this by Googling. Can some elucidate the statement, or give me a reference?
Full statement: https://books.google.com/books?id=mMPCAgAAQBAJ&pg=PA52
Incidentally, it has also been shown that under some strange hypotheses, all bounded sets can be assigned a numerical "area." The universe under these hypotheses is a very odd and interesting one, but we are not interested in it here, for in that universe the hyperreal numbers do not exist!
calculus measure-theory
calculus measure-theory
edited 8 hours ago
Chris Culter
21.6k4 gold badges38 silver badges89 bronze badges
asked 8 hours ago
saulspatzsaulspatz
21.1k4 gold badges16 silver badges38 bronze badges
edited 8 hours ago
Chris Culter
21.6k4 gold badges38 silver badges89 bronze badges
asked 8 hours ago
saulspatzsaulspatz
21.1k4 gold badges16 silver badges38 bronze badges
edited 8 hours ago
Chris Culter
21.6k4 gold badges38 silver badges89 bronze badges
edited 8 hours ago
Chris Culter
21.6k4 gold badges38 silver badges89 bronze badges
edited 8 hours ago
Chris Culter
21.6k4 gold badges38 silver badges89 bronze badges
21.6k4 gold badges38 silver badges89 bronze badges
asked 8 hours ago
saulspatzsaulspatz
21.1k4 gold badges16 silver badges38 bronze badges
asked 8 hours ago
saulspatzsaulspatz
21.1k4 gold badges16 silver badges38 bronze badges
asked 8 hours ago
saulspatzsaulspatz
21.1k4 gold badges16 silver badges38 bronze badges
21.1k4 gold badges16 silver badges38 bronze badges
$begingroup$
What does 'bounded' mean in context (e.g. bounded as in a metric space, topological space, order, etc.)?
$endgroup$
– R. Burton
8 hours ago
$begingroup$
@R.Burton I don't know. The statement appears as an aside, in a historical discussion at the beginning of the integral calculus portion of the book, right after a mention of Lebesgue.
$endgroup$
– saulspatz
8 hours ago
add a comment |
$begingroup$
What does 'bounded' mean in context (e.g. bounded as in a metric space, topological space, order, etc.)?
$endgroup$
– R. Burton
8 hours ago
$begingroup$
@R.Burton I don't know. The statement appears as an aside, in a historical discussion at the beginning of the integral calculus portion of the book, right after a mention of Lebesgue.
$endgroup$
– saulspatz
8 hours ago
$begingroup$
What does 'bounded' mean in context (e.g. bounded as in a metric space, topological space, order, etc.)?
$endgroup$
– R. Burton
8 hours ago
$begingroup$
What does 'bounded' mean in context (e.g. bounded as in a metric space, topological space, order, etc.)?
$endgroup$
– R. Burton
8 hours ago
$begingroup$
@R.Burton I don't know. The statement appears as an aside, in a historical discussion at the beginning of the integral calculus portion of the book, right after a mention of Lebesgue.
$endgroup$
– saulspatz
8 hours ago
$begingroup$
@R.Burton I don't know. The statement appears as an aside, in a historical discussion at the beginning of the integral calculus portion of the book, right after a mention of Lebesgue.
$endgroup$
– saulspatz
8 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's hard to give an answer because the authors might have been thinking of something else, but if I were to bet some money I'd say the authors were talking about the Axiom of Determinacy.
It's a not-so-strange-looking axiom which (under some large cardinal hypotheses) is consistent with ZF+ Dependent Choice, implies that all subsets of $mathbb R$ are Lebesgue measurable, and also implies that there is no nonprincipal ultrafilter on $mathbb N$, thus forbidding at least the usual construction of the hyperreal numbers (in particular it is inconsistent with the full axiom of choice). All subsets are Lebesgue measurable in particular implies that the Lebesgue measure $lambda$ is defined on all subsets of $mathbb R$ : in particular the bounded ones have a finite measure, and so we can assign to any of them a "meaningful value" (call it area in dimension $2$, volume in higher dimensions)
Essentially this axiom says that a certain class of two-player perfect information games is determined, i.e. one of the two players has a winning strategy. For a wide class of such games (Borel games) it is a theorem (in ZFC) that they are determined, it's called Borel Determinacy. The Axiom of Determinacy is just saying that a wider class is also determined.
There are some inbetween axioms such as Projective Determinacy and Analytic Determinacy, where Projective or Analytic also refer to some class of games (actually some class of subsets of topological spaces but the games in question, Gale-Stewart games, are related to topology)
Key words are : Gale-Stewart games, Determinacy (Axiom of, Projective, Analytic)
answered 8 hours ago
MaxMax
20.7k1 gold badge12 silver badges46 bronze badges
$endgroup$
$begingroup$
Thank you. I feel sure this is it, because right after the sentence I quoted, they make the statement, "... in that universe, the hyperreal numbers do not exist!"
$endgroup$
– saulspatz
8 hours ago
add a comment |
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active
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$begingroup$
It's hard to give an answer because the authors might have been thinking of something else, but if I were to bet some money I'd say the authors were talking about the Axiom of Determinacy.
It's a not-so-strange-looking axiom which (under some large cardinal hypotheses) is consistent with ZF+ Dependent Choice, implies that all subsets of $mathbb R$ are Lebesgue measurable, and also implies that there is no nonprincipal ultrafilter on $mathbb N$, thus forbidding at least the usual construction of the hyperreal numbers (in particular it is inconsistent with the full axiom of choice). All subsets are Lebesgue measurable in particular implies that the Lebesgue measure $lambda$ is defined on all subsets of $mathbb R$ : in particular the bounded ones have a finite measure, and so we can assign to any of them a "meaningful value" (call it area in dimension $2$, volume in higher dimensions)
Essentially this axiom says that a certain class of two-player perfect information games is determined, i.e. one of the two players has a winning strategy. For a wide class of such games (Borel games) it is a theorem (in ZFC) that they are determined, it's called Borel Determinacy. The Axiom of Determinacy is just saying that a wider class is also determined.
There are some inbetween axioms such as Projective Determinacy and Analytic Determinacy, where Projective or Analytic also refer to some class of games (actually some class of subsets of topological spaces but the games in question, Gale-Stewart games, are related to topology)
Key words are : Gale-Stewart games, Determinacy (Axiom of, Projective, Analytic)
answered 8 hours ago
MaxMax
20.7k1 gold badge12 silver badges46 bronze badges
$endgroup$
$begingroup$
Thank you. I feel sure this is it, because right after the sentence I quoted, they make the statement, "... in that universe, the hyperreal numbers do not exist!"
$endgroup$
– saulspatz
8 hours ago
add a comment |
$begingroup$
It's hard to give an answer because the authors might have been thinking of something else, but if I were to bet some money I'd say the authors were talking about the Axiom of Determinacy.
It's a not-so-strange-looking axiom which (under some large cardinal hypotheses) is consistent with ZF+ Dependent Choice, implies that all subsets of $mathbb R$ are Lebesgue measurable, and also implies that there is no nonprincipal ultrafilter on $mathbb N$, thus forbidding at least the usual construction of the hyperreal numbers (in particular it is inconsistent with the full axiom of choice). All subsets are Lebesgue measurable in particular implies that the Lebesgue measure $lambda$ is defined on all subsets of $mathbb R$ : in particular the bounded ones have a finite measure, and so we can assign to any of them a "meaningful value" (call it area in dimension $2$, volume in higher dimensions)
Essentially this axiom says that a certain class of two-player perfect information games is determined, i.e. one of the two players has a winning strategy. For a wide class of such games (Borel games) it is a theorem (in ZFC) that they are determined, it's called Borel Determinacy. The Axiom of Determinacy is just saying that a wider class is also determined.
There are some inbetween axioms such as Projective Determinacy and Analytic Determinacy, where Projective or Analytic also refer to some class of games (actually some class of subsets of topological spaces but the games in question, Gale-Stewart games, are related to topology)
Key words are : Gale-Stewart games, Determinacy (Axiom of, Projective, Analytic)
answered 8 hours ago
MaxMax
20.7k1 gold badge12 silver badges46 bronze badges
$endgroup$
$begingroup$
Thank you. I feel sure this is it, because right after the sentence I quoted, they make the statement, "... in that universe, the hyperreal numbers do not exist!"
$endgroup$
– saulspatz
8 hours ago
add a comment |
$begingroup$
It's hard to give an answer because the authors might have been thinking of something else, but if I were to bet some money I'd say the authors were talking about the Axiom of Determinacy.
It's a not-so-strange-looking axiom which (under some large cardinal hypotheses) is consistent with ZF+ Dependent Choice, implies that all subsets of $mathbb R$ are Lebesgue measurable, and also implies that there is no nonprincipal ultrafilter on $mathbb N$, thus forbidding at least the usual construction of the hyperreal numbers (in particular it is inconsistent with the full axiom of choice). All subsets are Lebesgue measurable in particular implies that the Lebesgue measure $lambda$ is defined on all subsets of $mathbb R$ : in particular the bounded ones have a finite measure, and so we can assign to any of them a "meaningful value" (call it area in dimension $2$, volume in higher dimensions)
Essentially this axiom says that a certain class of two-player perfect information games is determined, i.e. one of the two players has a winning strategy. For a wide class of such games (Borel games) it is a theorem (in ZFC) that they are determined, it's called Borel Determinacy. The Axiom of Determinacy is just saying that a wider class is also determined.
There are some inbetween axioms such as Projective Determinacy and Analytic Determinacy, where Projective or Analytic also refer to some class of games (actually some class of subsets of topological spaces but the games in question, Gale-Stewart games, are related to topology)
Key words are : Gale-Stewart games, Determinacy (Axiom of, Projective, Analytic)
answered 8 hours ago
MaxMax
20.7k1 gold badge12 silver badges46 bronze badges
$endgroup$
It's hard to give an answer because the authors might have been thinking of something else, but if I were to bet some money I'd say the authors were talking about the Axiom of Determinacy.
It's a not-so-strange-looking axiom which (under some large cardinal hypotheses) is consistent with ZF+ Dependent Choice, implies that all subsets of $mathbb R$ are Lebesgue measurable, and also implies that there is no nonprincipal ultrafilter on $mathbb N$, thus forbidding at least the usual construction of the hyperreal numbers (in particular it is inconsistent with the full axiom of choice). All subsets are Lebesgue measurable in particular implies that the Lebesgue measure $lambda$ is defined on all subsets of $mathbb R$ : in particular the bounded ones have a finite measure, and so we can assign to any of them a "meaningful value" (call it area in dimension $2$, volume in higher dimensions)
Essentially this axiom says that a certain class of two-player perfect information games is determined, i.e. one of the two players has a winning strategy. For a wide class of such games (Borel games) it is a theorem (in ZFC) that they are determined, it's called Borel Determinacy. The Axiom of Determinacy is just saying that a wider class is also determined.
There are some inbetween axioms such as Projective Determinacy and Analytic Determinacy, where Projective or Analytic also refer to some class of games (actually some class of subsets of topological spaces but the games in question, Gale-Stewart games, are related to topology)
Key words are : Gale-Stewart games, Determinacy (Axiom of, Projective, Analytic)
answered 8 hours ago
MaxMax
20.7k1 gold badge12 silver badges46 bronze badges
answered 8 hours ago
MaxMax
20.7k1 gold badge12 silver badges46 bronze badges
answered 8 hours ago
MaxMax
20.7k1 gold badge12 silver badges46 bronze badges
answered 8 hours ago
MaxMax
20.7k1 gold badge12 silver badges46 bronze badges
20.7k1 gold badge12 silver badges46 bronze badges
$begingroup$
Thank you. I feel sure this is it, because right after the sentence I quoted, they make the statement, "... in that universe, the hyperreal numbers do not exist!"
$endgroup$
– saulspatz
8 hours ago
add a comment |
$begingroup$
Thank you. I feel sure this is it, because right after the sentence I quoted, they make the statement, "... in that universe, the hyperreal numbers do not exist!"
$endgroup$
– saulspatz
8 hours ago
$begingroup$
Thank you. I feel sure this is it, because right after the sentence I quoted, they make the statement, "... in that universe, the hyperreal numbers do not exist!"
$endgroup$
– saulspatz
8 hours ago
$begingroup$
Thank you. I feel sure this is it, because right after the sentence I quoted, they make the statement, "... in that universe, the hyperreal numbers do not exist!"
$endgroup$
– saulspatz
8 hours ago
add a comment |
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$begingroup$
What does 'bounded' mean in context (e.g. bounded as in a metric space, topological space, order, etc.)?
$endgroup$
– R. Burton
8 hours ago
$begingroup$
@R.Burton I don't know. The statement appears as an aside, in a historical discussion at the beginning of the integral calculus portion of the book, right after a mention of Lebesgue.
$endgroup$
– saulspatz
8 hours ago