Axiom of Choice in General TopologyIs the axiom of choice really all that important?Where do we need the axiom of choice in Riemannian geometry?Zorn's Lemma $equiv$ Axiom of ChoiceStatements equivalent to the Axiom of ChoiceHow to master general topology for analysis?Is the axiom of choice really all that important?General Topology on Complex field?Cardinality of the Euclidean topology and the axiom of choiceSoft Question: Why does the Axiom of Choice lead to the weirdest constructions?Intuition behind Covering AxiomsCovering characterization of Metrizability and Paracompactness
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Axiom of Choice in General Topology
Is the axiom of choice really all that important?Where do we need the axiom of choice in Riemannian geometry?Zorn's Lemma $equiv$ Axiom of ChoiceStatements equivalent to the Axiom of ChoiceHow to master general topology for analysis?Is the axiom of choice really all that important?General Topology on Complex field?Cardinality of the Euclidean topology and the axiom of choiceSoft Question: Why does the Axiom of Choice lead to the weirdest constructions?Intuition behind Covering AxiomsCovering characterization of Metrizability and Paracompactness
$begingroup$
I ask this question after having searched a bit the web and not having found much about the role of Axiom of Choice and General Topology.
I am trying to answer: where does the axiom of choice enters General Topology? Would it be possible to track its impact on the subject and explicit its most profound consequences? What would General Topology look without it?
Being only in my undergarduate studies I couldn't find a satisfactory answer.
The fact is: I studied proofs involving AOC, in its equivalent formulation of Zorn's Lemma, both in Algebra (mainly about rings ideals and, most famously, the existence of a Hammel Basis for a vector space) and in Analysis (extension theorems of linear maps, most importantly: the Hahn Banach theorem). In a hypothetical "foundational hierarchy" of Mathematics I would naively place General Topology somehow in the middle between Logic/Algebra and Analysis (I am obviously strongly stereotyping the concept! Don't misunderstand me) and so I would expect AOC to have a grip also in this field.
Coming to what I know about General Topoloogy, it seems to me one AOC enters "only" in coverings and their properties (paracompactness, types of refinements...) which in turn found all the theorems about metrizability, metrics spaces and so on (gauges, uniform spaces, completeness). But I mean, one should expet this: we are somehow going in the direction of Analysis. There is much gneral topology beyond these topics.
I hope you can help me, maybe also through articles/books.
general-topology soft-question axiom-of-choice
asked May 19 at 17:52
Francesco BilottaFrancesco Bilotta
1308
$endgroup$
|
show 3 more comments
$begingroup$
I ask this question after having searched a bit the web and not having found much about the role of Axiom of Choice and General Topology.
I am trying to answer: where does the axiom of choice enters General Topology? Would it be possible to track its impact on the subject and explicit its most profound consequences? What would General Topology look without it?
Being only in my undergarduate studies I couldn't find a satisfactory answer.
The fact is: I studied proofs involving AOC, in its equivalent formulation of Zorn's Lemma, both in Algebra (mainly about rings ideals and, most famously, the existence of a Hammel Basis for a vector space) and in Analysis (extension theorems of linear maps, most importantly: the Hahn Banach theorem). In a hypothetical "foundational hierarchy" of Mathematics I would naively place General Topology somehow in the middle between Logic/Algebra and Analysis (I am obviously strongly stereotyping the concept! Don't misunderstand me) and so I would expect AOC to have a grip also in this field.
Coming to what I know about General Topoloogy, it seems to me one AOC enters "only" in coverings and their properties (paracompactness, types of refinements...) which in turn found all the theorems about metrizability, metrics spaces and so on (gauges, uniform spaces, completeness). But I mean, one should expet this: we are somehow going in the direction of Analysis. There is much gneral topology beyond these topics.
I hope you can help me, maybe also through articles/books.
general-topology soft-question axiom-of-choice
asked May 19 at 17:52
Francesco BilottaFrancesco Bilotta
1308
$endgroup$
2
$begingroup$
I don't have the expertise to answer this fully, but as far as I know, Tychonoff's Theorem is equivalent to AOC.
$endgroup$
– Thorgott
May 19 at 17:55
2
$begingroup$
Here is an interesting paper on the subject: projecteuclid.org/download/pdf_1/euclid.ndjfl/1093870373
$endgroup$
– rubikscube09
May 19 at 17:55
$begingroup$
General topology is skewed towards analysis (it's why it was created!) so maybe that has something to do with your perception.
$endgroup$
– rubikscube09
May 19 at 17:56
1
$begingroup$
This answer focusses on consequences for the topology of $mathbb R$ which are not all that "analytic" in flavor. Having said that, your distinction between "general topology" and "analysis" is rather fuzzy...
$endgroup$
– Lee Mosher
May 19 at 18:30
1
$begingroup$
The difference between "any point has an open neighbourhood such that" and "there exists a covering with opens such that" is a subtle one, but it may appear. Other than that you have a bunch of properties relating to nets and filters that depend on some form of choice (just like for sequences in analysis). It often appears very subtly in places you wouldn't expect it to be. For instance, Tychonoff's theorem, if you insist that all spaces involved are Hausdorff, becomes equivalent to a strictly weaker principle than full AC (although still not provable in ZF)
$endgroup$
– Max
May 19 at 18:34
|
show 3 more comments
$begingroup$
I ask this question after having searched a bit the web and not having found much about the role of Axiom of Choice and General Topology.
I am trying to answer: where does the axiom of choice enters General Topology? Would it be possible to track its impact on the subject and explicit its most profound consequences? What would General Topology look without it?
Being only in my undergarduate studies I couldn't find a satisfactory answer.
The fact is: I studied proofs involving AOC, in its equivalent formulation of Zorn's Lemma, both in Algebra (mainly about rings ideals and, most famously, the existence of a Hammel Basis for a vector space) and in Analysis (extension theorems of linear maps, most importantly: the Hahn Banach theorem). In a hypothetical "foundational hierarchy" of Mathematics I would naively place General Topology somehow in the middle between Logic/Algebra and Analysis (I am obviously strongly stereotyping the concept! Don't misunderstand me) and so I would expect AOC to have a grip also in this field.
Coming to what I know about General Topoloogy, it seems to me one AOC enters "only" in coverings and their properties (paracompactness, types of refinements...) which in turn found all the theorems about metrizability, metrics spaces and so on (gauges, uniform spaces, completeness). But I mean, one should expet this: we are somehow going in the direction of Analysis. There is much gneral topology beyond these topics.
I hope you can help me, maybe also through articles/books.
general-topology soft-question axiom-of-choice
asked May 19 at 17:52
Francesco BilottaFrancesco Bilotta
1308
$endgroup$
I ask this question after having searched a bit the web and not having found much about the role of Axiom of Choice and General Topology.
I am trying to answer: where does the axiom of choice enters General Topology? Would it be possible to track its impact on the subject and explicit its most profound consequences? What would General Topology look without it?
Being only in my undergarduate studies I couldn't find a satisfactory answer.
The fact is: I studied proofs involving AOC, in its equivalent formulation of Zorn's Lemma, both in Algebra (mainly about rings ideals and, most famously, the existence of a Hammel Basis for a vector space) and in Analysis (extension theorems of linear maps, most importantly: the Hahn Banach theorem). In a hypothetical "foundational hierarchy" of Mathematics I would naively place General Topology somehow in the middle between Logic/Algebra and Analysis (I am obviously strongly stereotyping the concept! Don't misunderstand me) and so I would expect AOC to have a grip also in this field.
Coming to what I know about General Topoloogy, it seems to me one AOC enters "only" in coverings and their properties (paracompactness, types of refinements...) which in turn found all the theorems about metrizability, metrics spaces and so on (gauges, uniform spaces, completeness). But I mean, one should expet this: we are somehow going in the direction of Analysis. There is much gneral topology beyond these topics.
I hope you can help me, maybe also through articles/books.
general-topology soft-question axiom-of-choice
general-topology soft-question axiom-of-choice
asked May 19 at 17:52
Francesco BilottaFrancesco Bilotta
1308
asked May 19 at 17:52
Francesco BilottaFrancesco Bilotta
1308
asked May 19 at 17:52
Francesco BilottaFrancesco Bilotta
1308
asked May 19 at 17:52
Francesco BilottaFrancesco Bilotta
1308
asked May 19 at 17:52
Francesco BilottaFrancesco Bilotta
1308
1308
2
$begingroup$
I don't have the expertise to answer this fully, but as far as I know, Tychonoff's Theorem is equivalent to AOC.
$endgroup$
– Thorgott
May 19 at 17:55
2
$begingroup$
Here is an interesting paper on the subject: projecteuclid.org/download/pdf_1/euclid.ndjfl/1093870373
$endgroup$
– rubikscube09
May 19 at 17:55
$begingroup$
General topology is skewed towards analysis (it's why it was created!) so maybe that has something to do with your perception.
$endgroup$
– rubikscube09
May 19 at 17:56
1
$begingroup$
This answer focusses on consequences for the topology of $mathbb R$ which are not all that "analytic" in flavor. Having said that, your distinction between "general topology" and "analysis" is rather fuzzy...
$endgroup$
– Lee Mosher
May 19 at 18:30
1
$begingroup$
The difference between "any point has an open neighbourhood such that" and "there exists a covering with opens such that" is a subtle one, but it may appear. Other than that you have a bunch of properties relating to nets and filters that depend on some form of choice (just like for sequences in analysis). It often appears very subtly in places you wouldn't expect it to be. For instance, Tychonoff's theorem, if you insist that all spaces involved are Hausdorff, becomes equivalent to a strictly weaker principle than full AC (although still not provable in ZF)
$endgroup$
– Max
May 19 at 18:34
|
show 3 more comments
2
$begingroup$
I don't have the expertise to answer this fully, but as far as I know, Tychonoff's Theorem is equivalent to AOC.
$endgroup$
– Thorgott
May 19 at 17:55
2
$begingroup$
Here is an interesting paper on the subject: projecteuclid.org/download/pdf_1/euclid.ndjfl/1093870373
$endgroup$
– rubikscube09
May 19 at 17:55
$begingroup$
General topology is skewed towards analysis (it's why it was created!) so maybe that has something to do with your perception.
$endgroup$
– rubikscube09
May 19 at 17:56
1
$begingroup$
This answer focusses on consequences for the topology of $mathbb R$ which are not all that "analytic" in flavor. Having said that, your distinction between "general topology" and "analysis" is rather fuzzy...
$endgroup$
– Lee Mosher
May 19 at 18:30
1
$begingroup$
The difference between "any point has an open neighbourhood such that" and "there exists a covering with opens such that" is a subtle one, but it may appear. Other than that you have a bunch of properties relating to nets and filters that depend on some form of choice (just like for sequences in analysis). It often appears very subtly in places you wouldn't expect it to be. For instance, Tychonoff's theorem, if you insist that all spaces involved are Hausdorff, becomes equivalent to a strictly weaker principle than full AC (although still not provable in ZF)
$endgroup$
– Max
May 19 at 18:34
2
2
$begingroup$
I don't have the expertise to answer this fully, but as far as I know, Tychonoff's Theorem is equivalent to AOC.
$endgroup$
– Thorgott
May 19 at 17:55
$begingroup$
I don't have the expertise to answer this fully, but as far as I know, Tychonoff's Theorem is equivalent to AOC.
$endgroup$
– Thorgott
May 19 at 17:55
2
2
$begingroup$
Here is an interesting paper on the subject: projecteuclid.org/download/pdf_1/euclid.ndjfl/1093870373
$endgroup$
– rubikscube09
May 19 at 17:55
$begingroup$
Here is an interesting paper on the subject: projecteuclid.org/download/pdf_1/euclid.ndjfl/1093870373
$endgroup$
– rubikscube09
May 19 at 17:55
$begingroup$
General topology is skewed towards analysis (it's why it was created!) so maybe that has something to do with your perception.
$endgroup$
– rubikscube09
May 19 at 17:56
$begingroup$
General topology is skewed towards analysis (it's why it was created!) so maybe that has something to do with your perception.
$endgroup$
– rubikscube09
May 19 at 17:56
1
1
$begingroup$
This answer focusses on consequences for the topology of $mathbb R$ which are not all that "analytic" in flavor. Having said that, your distinction between "general topology" and "analysis" is rather fuzzy...
$endgroup$
– Lee Mosher
May 19 at 18:30
$begingroup$
This answer focusses on consequences for the topology of $mathbb R$ which are not all that "analytic" in flavor. Having said that, your distinction between "general topology" and "analysis" is rather fuzzy...
$endgroup$
– Lee Mosher
May 19 at 18:30
1
1
$begingroup$
The difference between "any point has an open neighbourhood such that" and "there exists a covering with opens such that" is a subtle one, but it may appear. Other than that you have a bunch of properties relating to nets and filters that depend on some form of choice (just like for sequences in analysis). It often appears very subtly in places you wouldn't expect it to be. For instance, Tychonoff's theorem, if you insist that all spaces involved are Hausdorff, becomes equivalent to a strictly weaker principle than full AC (although still not provable in ZF)
$endgroup$
– Max
May 19 at 18:34
$begingroup$
The difference between "any point has an open neighbourhood such that" and "there exists a covering with opens such that" is a subtle one, but it may appear. Other than that you have a bunch of properties relating to nets and filters that depend on some form of choice (just like for sequences in analysis). It often appears very subtly in places you wouldn't expect it to be. For instance, Tychonoff's theorem, if you insist that all spaces involved are Hausdorff, becomes equivalent to a strictly weaker principle than full AC (although still not provable in ZF)
$endgroup$
– Max
May 19 at 18:34
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
See the papers "Horrors of topology without AC, a non-normal orderable space" (van Douwen) and "continuing horrors of topology without choice" (Good and Tree)
Basic things that can go wrong in the absence of choice: $mathbbR$ can be the union of countably many countable sets (so is meagre in itself and Baire's theorem fails), a sequentially continuous function on a metric space need not be continuous, etc etc. It's used in very many places, especially its countable form. We cannot define compactifications in the usual way, a lot of dimension theory becomes invalid, no nice theory of ordered spaces etc. It's certainly not confined to just covering properties; it touches almost all parts of topology and analysis.
answered May 20 at 5:42
Henno BrandsmaHenno Brandsma
121k351134
$endgroup$
$begingroup$
Baire's theorem is provable in ZF for separable spaces. Even if $Bbb R$ is a countable union of countable sets, the countable intersection of dense open sets is still dense.
$endgroup$
– Asaf Karagila♦
May 20 at 10:21
add a comment |
$begingroup$
Tychonoff's theorem, which states that the product of any set of compact topological spaces is compact with respect to the product topology, is equivalent to the Axiom of Choice. And Tychonoff's theorem is one of the most important theorems in Topology.
answered May 19 at 17:56
José Carlos SantosJosé Carlos Santos
188k24145261
$endgroup$
$begingroup$
Yes, I knew about this, which I forgot to mention. Anyway, I think we could still proove that the product of a finite number of compact is compact, without axiom of choice., hence we would still have some very weak result. And the Tychonoff Theorem is used mainly (I think) to proove a number of central results in Analysis, mostly. What I mean is: seeing general topology as a "self-contained" field, would Tychonoff theorem retain all of its importance?
$endgroup$
– Francesco Bilotta
May 19 at 18:07
1
$begingroup$
Yes, for a finite number of compact spaces, the Axiom of Choice is not needed.
$endgroup$
– José Carlos Santos
May 19 at 18:08
add a comment |
$begingroup$
The proof of the Banach-Tarski paradox involves ideas from topology, geometry, measure theory, set theory and group theory. That, and a whole bunch of related results, are strongly dependent on the AOC. Stan Wagon's book offers comprehensive coverage of the subject. https://www.amazon.com/Banach-Tarski-Paradox-Encyclopedia-Mathematics-Applications/dp/0521457041/ref=sr_1_2?keywords=Banach-Tarski+paradox&qid=1558299855&s=books&sr=1-2
Len Wapner's book is a popularized discussion of the result. https://www.amazon.com/Pea-Sun-Mathematical-Paradox/dp/1568813279/ref=cm_cr_arp_d_product_sims?ie=UTF8#customerReviews
answered May 19 at 21:34
Richard GostanianRichard Gostanian
112
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
See the papers "Horrors of topology without AC, a non-normal orderable space" (van Douwen) and "continuing horrors of topology without choice" (Good and Tree)
Basic things that can go wrong in the absence of choice: $mathbbR$ can be the union of countably many countable sets (so is meagre in itself and Baire's theorem fails), a sequentially continuous function on a metric space need not be continuous, etc etc. It's used in very many places, especially its countable form. We cannot define compactifications in the usual way, a lot of dimension theory becomes invalid, no nice theory of ordered spaces etc. It's certainly not confined to just covering properties; it touches almost all parts of topology and analysis.
answered May 20 at 5:42
Henno BrandsmaHenno Brandsma
121k351134
$endgroup$
$begingroup$
Baire's theorem is provable in ZF for separable spaces. Even if $Bbb R$ is a countable union of countable sets, the countable intersection of dense open sets is still dense.
$endgroup$
– Asaf Karagila♦
May 20 at 10:21
add a comment |
$begingroup$
See the papers "Horrors of topology without AC, a non-normal orderable space" (van Douwen) and "continuing horrors of topology without choice" (Good and Tree)
Basic things that can go wrong in the absence of choice: $mathbbR$ can be the union of countably many countable sets (so is meagre in itself and Baire's theorem fails), a sequentially continuous function on a metric space need not be continuous, etc etc. It's used in very many places, especially its countable form. We cannot define compactifications in the usual way, a lot of dimension theory becomes invalid, no nice theory of ordered spaces etc. It's certainly not confined to just covering properties; it touches almost all parts of topology and analysis.
answered May 20 at 5:42
Henno BrandsmaHenno Brandsma
121k351134
$endgroup$
$begingroup$
Baire's theorem is provable in ZF for separable spaces. Even if $Bbb R$ is a countable union of countable sets, the countable intersection of dense open sets is still dense.
$endgroup$
– Asaf Karagila♦
May 20 at 10:21
add a comment |
$begingroup$
See the papers "Horrors of topology without AC, a non-normal orderable space" (van Douwen) and "continuing horrors of topology without choice" (Good and Tree)
Basic things that can go wrong in the absence of choice: $mathbbR$ can be the union of countably many countable sets (so is meagre in itself and Baire's theorem fails), a sequentially continuous function on a metric space need not be continuous, etc etc. It's used in very many places, especially its countable form. We cannot define compactifications in the usual way, a lot of dimension theory becomes invalid, no nice theory of ordered spaces etc. It's certainly not confined to just covering properties; it touches almost all parts of topology and analysis.
answered May 20 at 5:42
Henno BrandsmaHenno Brandsma
121k351134
$endgroup$
See the papers "Horrors of topology without AC, a non-normal orderable space" (van Douwen) and "continuing horrors of topology without choice" (Good and Tree)
Basic things that can go wrong in the absence of choice: $mathbbR$ can be the union of countably many countable sets (so is meagre in itself and Baire's theorem fails), a sequentially continuous function on a metric space need not be continuous, etc etc. It's used in very many places, especially its countable form. We cannot define compactifications in the usual way, a lot of dimension theory becomes invalid, no nice theory of ordered spaces etc. It's certainly not confined to just covering properties; it touches almost all parts of topology and analysis.
answered May 20 at 5:42
Henno BrandsmaHenno Brandsma
121k351134
answered May 20 at 5:42
Henno BrandsmaHenno Brandsma
121k351134
answered May 20 at 5:42
Henno BrandsmaHenno Brandsma
121k351134
answered May 20 at 5:42
Henno BrandsmaHenno Brandsma
121k351134
121k351134
$begingroup$
Baire's theorem is provable in ZF for separable spaces. Even if $Bbb R$ is a countable union of countable sets, the countable intersection of dense open sets is still dense.
$endgroup$
– Asaf Karagila♦
May 20 at 10:21
add a comment |
$begingroup$
Baire's theorem is provable in ZF for separable spaces. Even if $Bbb R$ is a countable union of countable sets, the countable intersection of dense open sets is still dense.
$endgroup$
– Asaf Karagila♦
May 20 at 10:21
$begingroup$
Baire's theorem is provable in ZF for separable spaces. Even if $Bbb R$ is a countable union of countable sets, the countable intersection of dense open sets is still dense.
$endgroup$
– Asaf Karagila♦
May 20 at 10:21
$begingroup$
Baire's theorem is provable in ZF for separable spaces. Even if $Bbb R$ is a countable union of countable sets, the countable intersection of dense open sets is still dense.
$endgroup$
– Asaf Karagila♦
May 20 at 10:21
add a comment |
$begingroup$
Tychonoff's theorem, which states that the product of any set of compact topological spaces is compact with respect to the product topology, is equivalent to the Axiom of Choice. And Tychonoff's theorem is one of the most important theorems in Topology.
answered May 19 at 17:56
José Carlos SantosJosé Carlos Santos
188k24145261
$endgroup$
$begingroup$
Yes, I knew about this, which I forgot to mention. Anyway, I think we could still proove that the product of a finite number of compact is compact, without axiom of choice., hence we would still have some very weak result. And the Tychonoff Theorem is used mainly (I think) to proove a number of central results in Analysis, mostly. What I mean is: seeing general topology as a "self-contained" field, would Tychonoff theorem retain all of its importance?
$endgroup$
– Francesco Bilotta
May 19 at 18:07
1
$begingroup$
Yes, for a finite number of compact spaces, the Axiom of Choice is not needed.
$endgroup$
– José Carlos Santos
May 19 at 18:08
add a comment |
$begingroup$
Tychonoff's theorem, which states that the product of any set of compact topological spaces is compact with respect to the product topology, is equivalent to the Axiom of Choice. And Tychonoff's theorem is one of the most important theorems in Topology.
answered May 19 at 17:56
José Carlos SantosJosé Carlos Santos
188k24145261
$endgroup$
$begingroup$
Yes, I knew about this, which I forgot to mention. Anyway, I think we could still proove that the product of a finite number of compact is compact, without axiom of choice., hence we would still have some very weak result. And the Tychonoff Theorem is used mainly (I think) to proove a number of central results in Analysis, mostly. What I mean is: seeing general topology as a "self-contained" field, would Tychonoff theorem retain all of its importance?
$endgroup$
– Francesco Bilotta
May 19 at 18:07
1
$begingroup$
Yes, for a finite number of compact spaces, the Axiom of Choice is not needed.
$endgroup$
– José Carlos Santos
May 19 at 18:08
add a comment |
$begingroup$
Tychonoff's theorem, which states that the product of any set of compact topological spaces is compact with respect to the product topology, is equivalent to the Axiom of Choice. And Tychonoff's theorem is one of the most important theorems in Topology.
answered May 19 at 17:56
José Carlos SantosJosé Carlos Santos
188k24145261
$endgroup$
Tychonoff's theorem, which states that the product of any set of compact topological spaces is compact with respect to the product topology, is equivalent to the Axiom of Choice. And Tychonoff's theorem is one of the most important theorems in Topology.
answered May 19 at 17:56
José Carlos SantosJosé Carlos Santos
188k24145261
answered May 19 at 17:56
José Carlos SantosJosé Carlos Santos
188k24145261
answered May 19 at 17:56
José Carlos SantosJosé Carlos Santos
188k24145261
answered May 19 at 17:56
José Carlos SantosJosé Carlos Santos
188k24145261
188k24145261
$begingroup$
Yes, I knew about this, which I forgot to mention. Anyway, I think we could still proove that the product of a finite number of compact is compact, without axiom of choice., hence we would still have some very weak result. And the Tychonoff Theorem is used mainly (I think) to proove a number of central results in Analysis, mostly. What I mean is: seeing general topology as a "self-contained" field, would Tychonoff theorem retain all of its importance?
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– Francesco Bilotta
May 19 at 18:07
1
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Yes, for a finite number of compact spaces, the Axiom of Choice is not needed.
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– José Carlos Santos
May 19 at 18:08
add a comment |
$begingroup$
Yes, I knew about this, which I forgot to mention. Anyway, I think we could still proove that the product of a finite number of compact is compact, without axiom of choice., hence we would still have some very weak result. And the Tychonoff Theorem is used mainly (I think) to proove a number of central results in Analysis, mostly. What I mean is: seeing general topology as a "self-contained" field, would Tychonoff theorem retain all of its importance?
$endgroup$
– Francesco Bilotta
May 19 at 18:07
1
$begingroup$
Yes, for a finite number of compact spaces, the Axiom of Choice is not needed.
$endgroup$
– José Carlos Santos
May 19 at 18:08
$begingroup$
Yes, I knew about this, which I forgot to mention. Anyway, I think we could still proove that the product of a finite number of compact is compact, without axiom of choice., hence we would still have some very weak result. And the Tychonoff Theorem is used mainly (I think) to proove a number of central results in Analysis, mostly. What I mean is: seeing general topology as a "self-contained" field, would Tychonoff theorem retain all of its importance?
$endgroup$
– Francesco Bilotta
May 19 at 18:07
$begingroup$
Yes, I knew about this, which I forgot to mention. Anyway, I think we could still proove that the product of a finite number of compact is compact, without axiom of choice., hence we would still have some very weak result. And the Tychonoff Theorem is used mainly (I think) to proove a number of central results in Analysis, mostly. What I mean is: seeing general topology as a "self-contained" field, would Tychonoff theorem retain all of its importance?
$endgroup$
– Francesco Bilotta
May 19 at 18:07
1
1
$begingroup$
Yes, for a finite number of compact spaces, the Axiom of Choice is not needed.
$endgroup$
– José Carlos Santos
May 19 at 18:08
$begingroup$
Yes, for a finite number of compact spaces, the Axiom of Choice is not needed.
$endgroup$
– José Carlos Santos
May 19 at 18:08
add a comment |
$begingroup$
The proof of the Banach-Tarski paradox involves ideas from topology, geometry, measure theory, set theory and group theory. That, and a whole bunch of related results, are strongly dependent on the AOC. Stan Wagon's book offers comprehensive coverage of the subject. https://www.amazon.com/Banach-Tarski-Paradox-Encyclopedia-Mathematics-Applications/dp/0521457041/ref=sr_1_2?keywords=Banach-Tarski+paradox&qid=1558299855&s=books&sr=1-2
Len Wapner's book is a popularized discussion of the result. https://www.amazon.com/Pea-Sun-Mathematical-Paradox/dp/1568813279/ref=cm_cr_arp_d_product_sims?ie=UTF8#customerReviews
answered May 19 at 21:34
Richard GostanianRichard Gostanian
112
$endgroup$
add a comment |
$begingroup$
The proof of the Banach-Tarski paradox involves ideas from topology, geometry, measure theory, set theory and group theory. That, and a whole bunch of related results, are strongly dependent on the AOC. Stan Wagon's book offers comprehensive coverage of the subject. https://www.amazon.com/Banach-Tarski-Paradox-Encyclopedia-Mathematics-Applications/dp/0521457041/ref=sr_1_2?keywords=Banach-Tarski+paradox&qid=1558299855&s=books&sr=1-2
Len Wapner's book is a popularized discussion of the result. https://www.amazon.com/Pea-Sun-Mathematical-Paradox/dp/1568813279/ref=cm_cr_arp_d_product_sims?ie=UTF8#customerReviews
answered May 19 at 21:34
Richard GostanianRichard Gostanian
112
$endgroup$
add a comment |
$begingroup$
The proof of the Banach-Tarski paradox involves ideas from topology, geometry, measure theory, set theory and group theory. That, and a whole bunch of related results, are strongly dependent on the AOC. Stan Wagon's book offers comprehensive coverage of the subject. https://www.amazon.com/Banach-Tarski-Paradox-Encyclopedia-Mathematics-Applications/dp/0521457041/ref=sr_1_2?keywords=Banach-Tarski+paradox&qid=1558299855&s=books&sr=1-2
Len Wapner's book is a popularized discussion of the result. https://www.amazon.com/Pea-Sun-Mathematical-Paradox/dp/1568813279/ref=cm_cr_arp_d_product_sims?ie=UTF8#customerReviews
answered May 19 at 21:34
Richard GostanianRichard Gostanian
112
$endgroup$
The proof of the Banach-Tarski paradox involves ideas from topology, geometry, measure theory, set theory and group theory. That, and a whole bunch of related results, are strongly dependent on the AOC. Stan Wagon's book offers comprehensive coverage of the subject. https://www.amazon.com/Banach-Tarski-Paradox-Encyclopedia-Mathematics-Applications/dp/0521457041/ref=sr_1_2?keywords=Banach-Tarski+paradox&qid=1558299855&s=books&sr=1-2
Len Wapner's book is a popularized discussion of the result. https://www.amazon.com/Pea-Sun-Mathematical-Paradox/dp/1568813279/ref=cm_cr_arp_d_product_sims?ie=UTF8#customerReviews
answered May 19 at 21:34
Richard GostanianRichard Gostanian
112
edited May 20 at 1:26
answered May 19 at 21:34
Richard GostanianRichard Gostanian
112
answered May 19 at 21:34
Richard GostanianRichard Gostanian
112
answered May 19 at 21:34
Richard GostanianRichard Gostanian
112
112
add a comment |
add a comment |
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I don't have the expertise to answer this fully, but as far as I know, Tychonoff's Theorem is equivalent to AOC.
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– Thorgott
May 19 at 17:55
2
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Here is an interesting paper on the subject: projecteuclid.org/download/pdf_1/euclid.ndjfl/1093870373
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– rubikscube09
May 19 at 17:55
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General topology is skewed towards analysis (it's why it was created!) so maybe that has something to do with your perception.
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– rubikscube09
May 19 at 17:56
1
$begingroup$
This answer focusses on consequences for the topology of $mathbb R$ which are not all that "analytic" in flavor. Having said that, your distinction between "general topology" and "analysis" is rather fuzzy...
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– Lee Mosher
May 19 at 18:30
1
$begingroup$
The difference between "any point has an open neighbourhood such that" and "there exists a covering with opens such that" is a subtle one, but it may appear. Other than that you have a bunch of properties relating to nets and filters that depend on some form of choice (just like for sequences in analysis). It often appears very subtly in places you wouldn't expect it to be. For instance, Tychonoff's theorem, if you insist that all spaces involved are Hausdorff, becomes equivalent to a strictly weaker principle than full AC (although still not provable in ZF)
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– Max
May 19 at 18:34