Was there ever an axiom rendered a theorem?
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In the history of mathematics, are there notable examples of theorems which have been first considered axioms?
Alternatively, was there any statement first considered an axiom that later have been shown to be derived from other axiom(s), therefore rendering the statement a theorem?
logic math-history axioms
asked 12 hours ago
Eyal RothEyal Roth
1595
New contributor
Eyal Roth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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|
show 7 more comments
$begingroup$
In the history of mathematics, are there notable examples of theorems which have been first considered axioms?
Alternatively, was there any statement first considered an axiom that later have been shown to be derived from other axiom(s), therefore rendering the statement a theorem?
logic math-history axioms
asked 12 hours ago
Eyal RothEyal Roth
1595
New contributor
Eyal Roth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2
$begingroup$
All axioms are theorems, math.stackexchange.com/questions/1242021/… also of interest might be math.stackexchange.com/questions/258346/… and math.stackexchange.com/questions/1383457/… and math.stackexchange.com/questions/1131748/… might also be relevant.
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– Asaf Karagila♦
12 hours ago
1
$begingroup$
Eyal, the main point here is that "axiom" is a social agreement, rather than a mathematical definition.
$endgroup$
– Asaf Karagila♦
10 hours ago
4
$begingroup$
And indeed the Axiom of Choice is taken as an axiom and is reduced to a Theorem when assuming ZF+ZL, or or even to a false statement when assuming ZF+AD.
$endgroup$
– Asaf Karagila♦
10 hours ago
2
$begingroup$
I believe all the the axioms from Peano Arithmetic (PA) can be derived from ZF?
$endgroup$
– Bram28
9 hours ago
1
$begingroup$
@CliveNewstead: The axiom of infinity is ill-defined if you have not yet defined the empty set (because the empty set symbol appears in the axiom of infinity).
$endgroup$
– Kevin
8 hours ago
|
show 7 more comments
$begingroup$
In the history of mathematics, are there notable examples of theorems which have been first considered axioms?
Alternatively, was there any statement first considered an axiom that later have been shown to be derived from other axiom(s), therefore rendering the statement a theorem?
logic math-history axioms
asked 12 hours ago
Eyal RothEyal Roth
1595
New contributor
Eyal Roth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In the history of mathematics, are there notable examples of theorems which have been first considered axioms?
Alternatively, was there any statement first considered an axiom that later have been shown to be derived from other axiom(s), therefore rendering the statement a theorem?
logic math-history axioms
logic math-history axioms
asked 12 hours ago
Eyal RothEyal Roth
1595
New contributor
Eyal Roth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 12 hours ago
Eyal RothEyal Roth
1595
New contributor
Eyal Roth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 11 hours ago
Eyal Roth
asked 12 hours ago
Eyal RothEyal Roth
1595
New contributor
Eyal Roth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 12 hours ago
Eyal RothEyal Roth
1595
asked 12 hours ago
Eyal RothEyal Roth
1595
1595
New contributor
Eyal Roth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Eyal Roth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Eyal Roth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
All axioms are theorems, math.stackexchange.com/questions/1242021/… also of interest might be math.stackexchange.com/questions/258346/… and math.stackexchange.com/questions/1383457/… and math.stackexchange.com/questions/1131748/… might also be relevant.
$endgroup$
– Asaf Karagila♦
12 hours ago
1
$begingroup$
Eyal, the main point here is that "axiom" is a social agreement, rather than a mathematical definition.
$endgroup$
– Asaf Karagila♦
10 hours ago
4
$begingroup$
And indeed the Axiom of Choice is taken as an axiom and is reduced to a Theorem when assuming ZF+ZL, or or even to a false statement when assuming ZF+AD.
$endgroup$
– Asaf Karagila♦
10 hours ago
2
$begingroup$
I believe all the the axioms from Peano Arithmetic (PA) can be derived from ZF?
$endgroup$
– Bram28
9 hours ago
1
$begingroup$
@CliveNewstead: The axiom of infinity is ill-defined if you have not yet defined the empty set (because the empty set symbol appears in the axiom of infinity).
$endgroup$
– Kevin
8 hours ago
|
show 7 more comments
2
$begingroup$
All axioms are theorems, math.stackexchange.com/questions/1242021/… also of interest might be math.stackexchange.com/questions/258346/… and math.stackexchange.com/questions/1383457/… and math.stackexchange.com/questions/1131748/… might also be relevant.
$endgroup$
– Asaf Karagila♦
12 hours ago
1
$begingroup$
Eyal, the main point here is that "axiom" is a social agreement, rather than a mathematical definition.
$endgroup$
– Asaf Karagila♦
10 hours ago
4
$begingroup$
And indeed the Axiom of Choice is taken as an axiom and is reduced to a Theorem when assuming ZF+ZL, or or even to a false statement when assuming ZF+AD.
$endgroup$
– Asaf Karagila♦
10 hours ago
2
$begingroup$
I believe all the the axioms from Peano Arithmetic (PA) can be derived from ZF?
$endgroup$
– Bram28
9 hours ago
1
$begingroup$
@CliveNewstead: The axiom of infinity is ill-defined if you have not yet defined the empty set (because the empty set symbol appears in the axiom of infinity).
$endgroup$
– Kevin
8 hours ago
2
2
$begingroup$
All axioms are theorems, math.stackexchange.com/questions/1242021/… also of interest might be math.stackexchange.com/questions/258346/… and math.stackexchange.com/questions/1383457/… and math.stackexchange.com/questions/1131748/… might also be relevant.
$endgroup$
– Asaf Karagila♦
12 hours ago
$begingroup$
All axioms are theorems, math.stackexchange.com/questions/1242021/… also of interest might be math.stackexchange.com/questions/258346/… and math.stackexchange.com/questions/1383457/… and math.stackexchange.com/questions/1131748/… might also be relevant.
$endgroup$
– Asaf Karagila♦
12 hours ago
1
1
$begingroup$
Eyal, the main point here is that "axiom" is a social agreement, rather than a mathematical definition.
$endgroup$
– Asaf Karagila♦
10 hours ago
$begingroup$
Eyal, the main point here is that "axiom" is a social agreement, rather than a mathematical definition.
$endgroup$
– Asaf Karagila♦
10 hours ago
4
4
$begingroup$
And indeed the Axiom of Choice is taken as an axiom and is reduced to a Theorem when assuming ZF+ZL, or or even to a false statement when assuming ZF+AD.
$endgroup$
– Asaf Karagila♦
10 hours ago
$begingroup$
And indeed the Axiom of Choice is taken as an axiom and is reduced to a Theorem when assuming ZF+ZL, or or even to a false statement when assuming ZF+AD.
$endgroup$
– Asaf Karagila♦
10 hours ago
2
2
$begingroup$
I believe all the the axioms from Peano Arithmetic (PA) can be derived from ZF?
$endgroup$
– Bram28
9 hours ago
$begingroup$
I believe all the the axioms from Peano Arithmetic (PA) can be derived from ZF?
$endgroup$
– Bram28
9 hours ago
1
1
$begingroup$
@CliveNewstead: The axiom of infinity is ill-defined if you have not yet defined the empty set (because the empty set symbol appears in the axiom of infinity).
$endgroup$
– Kevin
8 hours ago
$begingroup$
@CliveNewstead: The axiom of infinity is ill-defined if you have not yet defined the empty set (because the empty set symbol appears in the axiom of infinity).
$endgroup$
– Kevin
8 hours ago
|
show 7 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Fraenkel introduced the axiom schema of replacement to set theory. This implied the axiom schema of comprehension, and allowed the empty set and unordered pair axioms to follow from the axiom of infinity. (Note Zermelo set theory includes the axiom of choice whereas ZF does not, so Zermelo+replacement is ZFC.) The "deleted" axioms are typically listed when describing ZF(C), partly so people realise they're in Zermelo set theory, partly for easier comparisons with other set theories of interest.
answered 10 hours ago
J.G.J.G.
33.1k23251
$endgroup$
add a comment |
$begingroup$
The most famous example I know is that of Hilbert's axiom II.4 for the linear ordering of points on a line, for Euclidean geometry, proven to be superfluous by E.H. Moore. See this wikipedia article, especially "Hilbert's discarded axiom". https://en.wikipedia.org/wiki/Hilbert%27s_axioms
In the article of Moore linked there, it is stated that also axiom I.4 is superfluous.
http://www.ams.org/journals/tran/1902-003-01/S0002-9947-1902-1500592-8/S0002-9947-1902-1500592-8.pdf
answered 7 hours ago
roy smithroy smith
999711
$endgroup$
add a comment |
$begingroup$
Yes, everywhere. What is an axiom from one theory can be a theorem in another.
Euclid's fifth postulate can be replaced by the statement that the angles on the inside of each triangle add up to $pi$ radians.
Another notable example is the axiom of choice, which is equivalent in some axiomatic systems to Zorn's Lemma.
Also, watch this Feynman clip.
$endgroup$
$begingroup$
That is an interesting clip (and I love the accent). If I understand correctly, Feynman discusses axioms which have bi-directional relations; i.e, one can be deduced from the other and vice-versa; or perhaps, any two of three axioms can imply the third. I'm rather interested in cases of uni-directional axioms which have been discovered to be implied from another axiom or set of axioms.
$endgroup$
– Eyal Roth
11 hours ago
1
$begingroup$
These cases are considered to be alternative statements of the same axiom. You choose whichever one you want as an axiom and prove the other. If you have an axiom you suspect is redundant you might find a way to prove one of the statements, or you might find a statement that is obvious enough that people accept it as an axiom. In both of these cases, the axiom has been proven to be independent of the others. I think OP wants a case where a statement was thought to be independent of the other axioms of a subject and was shown to be a consequence of them.
$endgroup$
– Ross Millikan
5 hours ago
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
$begingroup$
Fraenkel introduced the axiom schema of replacement to set theory. This implied the axiom schema of comprehension, and allowed the empty set and unordered pair axioms to follow from the axiom of infinity. (Note Zermelo set theory includes the axiom of choice whereas ZF does not, so Zermelo+replacement is ZFC.) The "deleted" axioms are typically listed when describing ZF(C), partly so people realise they're in Zermelo set theory, partly for easier comparisons with other set theories of interest.
answered 10 hours ago
J.G.J.G.
33.1k23251
$endgroup$
add a comment |
$begingroup$
Fraenkel introduced the axiom schema of replacement to set theory. This implied the axiom schema of comprehension, and allowed the empty set and unordered pair axioms to follow from the axiom of infinity. (Note Zermelo set theory includes the axiom of choice whereas ZF does not, so Zermelo+replacement is ZFC.) The "deleted" axioms are typically listed when describing ZF(C), partly so people realise they're in Zermelo set theory, partly for easier comparisons with other set theories of interest.
answered 10 hours ago
J.G.J.G.
33.1k23251
$endgroup$
add a comment |
$begingroup$
Fraenkel introduced the axiom schema of replacement to set theory. This implied the axiom schema of comprehension, and allowed the empty set and unordered pair axioms to follow from the axiom of infinity. (Note Zermelo set theory includes the axiom of choice whereas ZF does not, so Zermelo+replacement is ZFC.) The "deleted" axioms are typically listed when describing ZF(C), partly so people realise they're in Zermelo set theory, partly for easier comparisons with other set theories of interest.
answered 10 hours ago
J.G.J.G.
33.1k23251
$endgroup$
Fraenkel introduced the axiom schema of replacement to set theory. This implied the axiom schema of comprehension, and allowed the empty set and unordered pair axioms to follow from the axiom of infinity. (Note Zermelo set theory includes the axiom of choice whereas ZF does not, so Zermelo+replacement is ZFC.) The "deleted" axioms are typically listed when describing ZF(C), partly so people realise they're in Zermelo set theory, partly for easier comparisons with other set theories of interest.
answered 10 hours ago
J.G.J.G.
33.1k23251
answered 10 hours ago
J.G.J.G.
33.1k23251
answered 10 hours ago
J.G.J.G.
33.1k23251
answered 10 hours ago
J.G.J.G.
33.1k23251
33.1k23251
add a comment |
add a comment |
$begingroup$
The most famous example I know is that of Hilbert's axiom II.4 for the linear ordering of points on a line, for Euclidean geometry, proven to be superfluous by E.H. Moore. See this wikipedia article, especially "Hilbert's discarded axiom". https://en.wikipedia.org/wiki/Hilbert%27s_axioms
In the article of Moore linked there, it is stated that also axiom I.4 is superfluous.
http://www.ams.org/journals/tran/1902-003-01/S0002-9947-1902-1500592-8/S0002-9947-1902-1500592-8.pdf
answered 7 hours ago
roy smithroy smith
999711
$endgroup$
add a comment |
$begingroup$
The most famous example I know is that of Hilbert's axiom II.4 for the linear ordering of points on a line, for Euclidean geometry, proven to be superfluous by E.H. Moore. See this wikipedia article, especially "Hilbert's discarded axiom". https://en.wikipedia.org/wiki/Hilbert%27s_axioms
In the article of Moore linked there, it is stated that also axiom I.4 is superfluous.
http://www.ams.org/journals/tran/1902-003-01/S0002-9947-1902-1500592-8/S0002-9947-1902-1500592-8.pdf
answered 7 hours ago
roy smithroy smith
999711
$endgroup$
add a comment |
$begingroup$
The most famous example I know is that of Hilbert's axiom II.4 for the linear ordering of points on a line, for Euclidean geometry, proven to be superfluous by E.H. Moore. See this wikipedia article, especially "Hilbert's discarded axiom". https://en.wikipedia.org/wiki/Hilbert%27s_axioms
In the article of Moore linked there, it is stated that also axiom I.4 is superfluous.
http://www.ams.org/journals/tran/1902-003-01/S0002-9947-1902-1500592-8/S0002-9947-1902-1500592-8.pdf
answered 7 hours ago
roy smithroy smith
999711
$endgroup$
The most famous example I know is that of Hilbert's axiom II.4 for the linear ordering of points on a line, for Euclidean geometry, proven to be superfluous by E.H. Moore. See this wikipedia article, especially "Hilbert's discarded axiom". https://en.wikipedia.org/wiki/Hilbert%27s_axioms
In the article of Moore linked there, it is stated that also axiom I.4 is superfluous.
http://www.ams.org/journals/tran/1902-003-01/S0002-9947-1902-1500592-8/S0002-9947-1902-1500592-8.pdf
answered 7 hours ago
roy smithroy smith
999711
answered 7 hours ago
roy smithroy smith
999711
answered 7 hours ago
roy smithroy smith
999711
answered 7 hours ago
roy smithroy smith
999711
999711
add a comment |
add a comment |
$begingroup$
Yes, everywhere. What is an axiom from one theory can be a theorem in another.
Euclid's fifth postulate can be replaced by the statement that the angles on the inside of each triangle add up to $pi$ radians.
Another notable example is the axiom of choice, which is equivalent in some axiomatic systems to Zorn's Lemma.
Also, watch this Feynman clip.
$endgroup$
$begingroup$
That is an interesting clip (and I love the accent). If I understand correctly, Feynman discusses axioms which have bi-directional relations; i.e, one can be deduced from the other and vice-versa; or perhaps, any two of three axioms can imply the third. I'm rather interested in cases of uni-directional axioms which have been discovered to be implied from another axiom or set of axioms.
$endgroup$
– Eyal Roth
11 hours ago
1
$begingroup$
These cases are considered to be alternative statements of the same axiom. You choose whichever one you want as an axiom and prove the other. If you have an axiom you suspect is redundant you might find a way to prove one of the statements, or you might find a statement that is obvious enough that people accept it as an axiom. In both of these cases, the axiom has been proven to be independent of the others. I think OP wants a case where a statement was thought to be independent of the other axioms of a subject and was shown to be a consequence of them.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
$begingroup$
Yes, everywhere. What is an axiom from one theory can be a theorem in another.
Euclid's fifth postulate can be replaced by the statement that the angles on the inside of each triangle add up to $pi$ radians.
Another notable example is the axiom of choice, which is equivalent in some axiomatic systems to Zorn's Lemma.
Also, watch this Feynman clip.
$endgroup$
$begingroup$
That is an interesting clip (and I love the accent). If I understand correctly, Feynman discusses axioms which have bi-directional relations; i.e, one can be deduced from the other and vice-versa; or perhaps, any two of three axioms can imply the third. I'm rather interested in cases of uni-directional axioms which have been discovered to be implied from another axiom or set of axioms.
$endgroup$
– Eyal Roth
11 hours ago
1
$begingroup$
These cases are considered to be alternative statements of the same axiom. You choose whichever one you want as an axiom and prove the other. If you have an axiom you suspect is redundant you might find a way to prove one of the statements, or you might find a statement that is obvious enough that people accept it as an axiom. In both of these cases, the axiom has been proven to be independent of the others. I think OP wants a case where a statement was thought to be independent of the other axioms of a subject and was shown to be a consequence of them.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
$begingroup$
Yes, everywhere. What is an axiom from one theory can be a theorem in another.
Euclid's fifth postulate can be replaced by the statement that the angles on the inside of each triangle add up to $pi$ radians.
Another notable example is the axiom of choice, which is equivalent in some axiomatic systems to Zorn's Lemma.
Also, watch this Feynman clip.
$endgroup$
Yes, everywhere. What is an axiom from one theory can be a theorem in another.
Euclid's fifth postulate can be replaced by the statement that the angles on the inside of each triangle add up to $pi$ radians.
Another notable example is the axiom of choice, which is equivalent in some axiomatic systems to Zorn's Lemma.
Also, watch this Feynman clip.
edited 12 hours ago
community wiki
2 revs
Shaun
$begingroup$
That is an interesting clip (and I love the accent). If I understand correctly, Feynman discusses axioms which have bi-directional relations; i.e, one can be deduced from the other and vice-versa; or perhaps, any two of three axioms can imply the third. I'm rather interested in cases of uni-directional axioms which have been discovered to be implied from another axiom or set of axioms.
$endgroup$
– Eyal Roth
11 hours ago
1
$begingroup$
These cases are considered to be alternative statements of the same axiom. You choose whichever one you want as an axiom and prove the other. If you have an axiom you suspect is redundant you might find a way to prove one of the statements, or you might find a statement that is obvious enough that people accept it as an axiom. In both of these cases, the axiom has been proven to be independent of the others. I think OP wants a case where a statement was thought to be independent of the other axioms of a subject and was shown to be a consequence of them.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
$begingroup$
That is an interesting clip (and I love the accent). If I understand correctly, Feynman discusses axioms which have bi-directional relations; i.e, one can be deduced from the other and vice-versa; or perhaps, any two of three axioms can imply the third. I'm rather interested in cases of uni-directional axioms which have been discovered to be implied from another axiom or set of axioms.
$endgroup$
– Eyal Roth
11 hours ago
1
$begingroup$
These cases are considered to be alternative statements of the same axiom. You choose whichever one you want as an axiom and prove the other. If you have an axiom you suspect is redundant you might find a way to prove one of the statements, or you might find a statement that is obvious enough that people accept it as an axiom. In both of these cases, the axiom has been proven to be independent of the others. I think OP wants a case where a statement was thought to be independent of the other axioms of a subject and was shown to be a consequence of them.
$endgroup$
– Ross Millikan
5 hours ago
$begingroup$
That is an interesting clip (and I love the accent). If I understand correctly, Feynman discusses axioms which have bi-directional relations; i.e, one can be deduced from the other and vice-versa; or perhaps, any two of three axioms can imply the third. I'm rather interested in cases of uni-directional axioms which have been discovered to be implied from another axiom or set of axioms.
$endgroup$
– Eyal Roth
11 hours ago
$begingroup$
That is an interesting clip (and I love the accent). If I understand correctly, Feynman discusses axioms which have bi-directional relations; i.e, one can be deduced from the other and vice-versa; or perhaps, any two of three axioms can imply the third. I'm rather interested in cases of uni-directional axioms which have been discovered to be implied from another axiom or set of axioms.
$endgroup$
– Eyal Roth
11 hours ago
1
1
$begingroup$
These cases are considered to be alternative statements of the same axiom. You choose whichever one you want as an axiom and prove the other. If you have an axiom you suspect is redundant you might find a way to prove one of the statements, or you might find a statement that is obvious enough that people accept it as an axiom. In both of these cases, the axiom has been proven to be independent of the others. I think OP wants a case where a statement was thought to be independent of the other axioms of a subject and was shown to be a consequence of them.
$endgroup$
– Ross Millikan
5 hours ago
$begingroup$
These cases are considered to be alternative statements of the same axiom. You choose whichever one you want as an axiom and prove the other. If you have an axiom you suspect is redundant you might find a way to prove one of the statements, or you might find a statement that is obvious enough that people accept it as an axiom. In both of these cases, the axiom has been proven to be independent of the others. I think OP wants a case where a statement was thought to be independent of the other axioms of a subject and was shown to be a consequence of them.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
Eyal Roth is a new contributor. Be nice, and check out our Code of Conduct.
Eyal Roth is a new contributor. Be nice, and check out our Code of Conduct.
Eyal Roth is a new contributor. Be nice, and check out our Code of Conduct.
Eyal Roth is a new contributor. Be nice, and check out our Code of Conduct.
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All axioms are theorems, math.stackexchange.com/questions/1242021/… also of interest might be math.stackexchange.com/questions/258346/… and math.stackexchange.com/questions/1383457/… and math.stackexchange.com/questions/1131748/… might also be relevant.
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– Asaf Karagila♦
12 hours ago
1
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Eyal, the main point here is that "axiom" is a social agreement, rather than a mathematical definition.
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– Asaf Karagila♦
10 hours ago
4
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And indeed the Axiom of Choice is taken as an axiom and is reduced to a Theorem when assuming ZF+ZL, or or even to a false statement when assuming ZF+AD.
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– Asaf Karagila♦
10 hours ago
2
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I believe all the the axioms from Peano Arithmetic (PA) can be derived from ZF?
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– Bram28
9 hours ago
1
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@CliveNewstead: The axiom of infinity is ill-defined if you have not yet defined the empty set (because the empty set symbol appears in the axiom of infinity).
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– Kevin
8 hours ago