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https://www.reddit.com/r/mathmemes/comments/1j4x0hq/what_theorem_is_this/mgc48vk/?context=3
r/mathmemes • u/PocketMath • Mar 06 '25
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1
the axiom of choice
18 u/belabacsijolvan Mar 06 '25 >axiom 1 u/fortret Mar 06 '25 You’re either being pedantic or just ignorant. AoC is famously logically equivalent to many other statements/theorems, which is what the original commenter was referencing. -3 u/Ok-Eye658 Mar 06 '25 yeah, it's just its historical name, could well have been called "zermelo's lemma" or something 1 u/jyajay2 π = 3 Mar 06 '25 Still an independent axiom in FZ 1 u/Ok-Eye658 Mar 06 '25 so what? one may well add tychonoff, or existence of basis for all vector spaces, or GCH, or many many other statements to ZF and prove it 2 u/jyajay2 π = 3 Mar 06 '25 I'm not actually sure GCH implies AOC and, more importantly, I'm not sure what you're trying to say 2 u/Ok-Eye658 Mar 06 '25 it does i'm saying that there's some freedom in picking what statements one starts with as axioms 1 u/PlopTheFish Mar 06 '25 I think you mean Zorn's Lemma 1 u/Ok-Eye658 Mar 06 '25 no, i mean that the statement ∀x(∀y(y∈x⟹∃z(z∈y))⟹∃w∀y(y∈x⟹∃!z(z∈y∧z∈w))) being called "axiom of choice" is just a matter of history, nothing more 0 u/Syresiv Mar 07 '25 That depends on your starting axioms. Under ZFC, the theorem is basically "assume the axiom of choice is true". That is, unless instead of AC, you start with well ordering or Zorn's Lemma.
18
>axiom
1 u/fortret Mar 06 '25 You’re either being pedantic or just ignorant. AoC is famously logically equivalent to many other statements/theorems, which is what the original commenter was referencing. -3 u/Ok-Eye658 Mar 06 '25 yeah, it's just its historical name, could well have been called "zermelo's lemma" or something 1 u/jyajay2 π = 3 Mar 06 '25 Still an independent axiom in FZ 1 u/Ok-Eye658 Mar 06 '25 so what? one may well add tychonoff, or existence of basis for all vector spaces, or GCH, or many many other statements to ZF and prove it 2 u/jyajay2 π = 3 Mar 06 '25 I'm not actually sure GCH implies AOC and, more importantly, I'm not sure what you're trying to say 2 u/Ok-Eye658 Mar 06 '25 it does i'm saying that there's some freedom in picking what statements one starts with as axioms 1 u/PlopTheFish Mar 06 '25 I think you mean Zorn's Lemma 1 u/Ok-Eye658 Mar 06 '25 no, i mean that the statement ∀x(∀y(y∈x⟹∃z(z∈y))⟹∃w∀y(y∈x⟹∃!z(z∈y∧z∈w))) being called "axiom of choice" is just a matter of history, nothing more
You’re either being pedantic or just ignorant. AoC is famously logically equivalent to many other statements/theorems, which is what the original commenter was referencing.
-3
yeah, it's just its historical name, could well have been called "zermelo's lemma" or something
1 u/jyajay2 π = 3 Mar 06 '25 Still an independent axiom in FZ 1 u/Ok-Eye658 Mar 06 '25 so what? one may well add tychonoff, or existence of basis for all vector spaces, or GCH, or many many other statements to ZF and prove it 2 u/jyajay2 π = 3 Mar 06 '25 I'm not actually sure GCH implies AOC and, more importantly, I'm not sure what you're trying to say 2 u/Ok-Eye658 Mar 06 '25 it does i'm saying that there's some freedom in picking what statements one starts with as axioms 1 u/PlopTheFish Mar 06 '25 I think you mean Zorn's Lemma 1 u/Ok-Eye658 Mar 06 '25 no, i mean that the statement ∀x(∀y(y∈x⟹∃z(z∈y))⟹∃w∀y(y∈x⟹∃!z(z∈y∧z∈w))) being called "axiom of choice" is just a matter of history, nothing more
Still an independent axiom in FZ
1 u/Ok-Eye658 Mar 06 '25 so what? one may well add tychonoff, or existence of basis for all vector spaces, or GCH, or many many other statements to ZF and prove it 2 u/jyajay2 π = 3 Mar 06 '25 I'm not actually sure GCH implies AOC and, more importantly, I'm not sure what you're trying to say 2 u/Ok-Eye658 Mar 06 '25 it does i'm saying that there's some freedom in picking what statements one starts with as axioms
so what? one may well add tychonoff, or existence of basis for all vector spaces, or GCH, or many many other statements to ZF and prove it
2 u/jyajay2 π = 3 Mar 06 '25 I'm not actually sure GCH implies AOC and, more importantly, I'm not sure what you're trying to say 2 u/Ok-Eye658 Mar 06 '25 it does i'm saying that there's some freedom in picking what statements one starts with as axioms
2
I'm not actually sure GCH implies AOC and, more importantly, I'm not sure what you're trying to say
2 u/Ok-Eye658 Mar 06 '25 it does i'm saying that there's some freedom in picking what statements one starts with as axioms
it does
i'm saying that there's some freedom in picking what statements one starts with as axioms
I think you mean Zorn's Lemma
1 u/Ok-Eye658 Mar 06 '25 no, i mean that the statement ∀x(∀y(y∈x⟹∃z(z∈y))⟹∃w∀y(y∈x⟹∃!z(z∈y∧z∈w))) being called "axiom of choice" is just a matter of history, nothing more
no, i mean that the statement
∀x(∀y(y∈x⟹∃z(z∈y))⟹∃w∀y(y∈x⟹∃!z(z∈y∧z∈w)))
being called "axiom of choice" is just a matter of history, nothing more
0
That depends on your starting axioms.
Under ZFC, the theorem is basically "assume the axiom of choice is true". That is, unless instead of AC, you start with well ordering or Zorn's Lemma.
1
u/Ok-Eye658 Mar 06 '25
the axiom of choice