Axiom of countable choice: a question about the domain. The 2019 Stack Overflow Developer Survey Results Are InAxiom of choice, non-measurable sets, countable unions3 Statements of axiom of choice are equivalentCountable infinity and the axiom of choiceAxiom of choice and the number of choice functionsQuestion about proof in Jech's The axiom of choice ($AC_omega$)Confused about Axiom of ChoiceAxiom of Choice (Naive Set Theory, Halmos)Confusion Regarding the Axiom of Countable ChoiceA proof that a countable product of countable sets is non-empty that does not use the axiom of choice. Is the proof correct?Axiom of Choice iff Every set has a choice function
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Axiom of countable choice: a question about the domain.
The 2019 Stack Overflow Developer Survey Results Are InAxiom of choice, non-measurable sets, countable unions3 Statements of axiom of choice are equivalentCountable infinity and the axiom of choiceAxiom of choice and the number of choice functionsQuestion about proof in Jech's The axiom of choice ($AC_omega$)Confused about Axiom of ChoiceAxiom of Choice (Naive Set Theory, Halmos)Confusion Regarding the Axiom of Countable ChoiceA proof that a countable product of countable sets is non-empty that does not use the axiom of choice. Is the proof correct?Axiom of Choice iff Every set has a choice function
$begingroup$
From Wikipedia: The axiom of countable choice or axiom of denumerable choice, denoted $AComega$, is an axiom of set theory that states that every countable collection of non-empty sets must have a choice function. I.e., given a function $A$ with domain $mathbbN$ (where $mathbbN$ denotes the set of natural numbers) such that $A(n)$ is a non-empty set for every $ninmathbbN$, then there exists a function $f$ with domain $mathbbN$ such that $f(n)in A(n)$ for every $ninmathbbN$.
Question Does function $A$ necessarily need to be defined on $mathbbN$ or can it also be defined on a set $M$ where $|M|=aleph_0$ or $|M|<+infty$?
Thanks!
set-theory axioms
asked Mar 30 at 21:59
Jack J.Jack J.
3592419
$endgroup$
add a comment |
$begingroup$
From Wikipedia: The axiom of countable choice or axiom of denumerable choice, denoted $AComega$, is an axiom of set theory that states that every countable collection of non-empty sets must have a choice function. I.e., given a function $A$ with domain $mathbbN$ (where $mathbbN$ denotes the set of natural numbers) such that $A(n)$ is a non-empty set for every $ninmathbbN$, then there exists a function $f$ with domain $mathbbN$ such that $f(n)in A(n)$ for every $ninmathbbN$.
Question Does function $A$ necessarily need to be defined on $mathbbN$ or can it also be defined on a set $M$ where $|M|=aleph_0$ or $|M|<+infty$?
Thanks!
set-theory axioms
asked Mar 30 at 21:59
Jack J.Jack J.
3592419
$endgroup$
1
$begingroup$
This is not really a question about the axiom of choice. It's about whether or not you can define a function from $Bbb N$ as function from a different countable set...
$endgroup$
– Asaf Karagila♦
Mar 31 at 0:30
add a comment |
$begingroup$
From Wikipedia: The axiom of countable choice or axiom of denumerable choice, denoted $AComega$, is an axiom of set theory that states that every countable collection of non-empty sets must have a choice function. I.e., given a function $A$ with domain $mathbbN$ (where $mathbbN$ denotes the set of natural numbers) such that $A(n)$ is a non-empty set for every $ninmathbbN$, then there exists a function $f$ with domain $mathbbN$ such that $f(n)in A(n)$ for every $ninmathbbN$.
Question Does function $A$ necessarily need to be defined on $mathbbN$ or can it also be defined on a set $M$ where $|M|=aleph_0$ or $|M|<+infty$?
Thanks!
set-theory axioms
asked Mar 30 at 21:59
Jack J.Jack J.
3592419
$endgroup$
From Wikipedia: The axiom of countable choice or axiom of denumerable choice, denoted $AComega$, is an axiom of set theory that states that every countable collection of non-empty sets must have a choice function. I.e., given a function $A$ with domain $mathbbN$ (where $mathbbN$ denotes the set of natural numbers) such that $A(n)$ is a non-empty set for every $ninmathbbN$, then there exists a function $f$ with domain $mathbbN$ such that $f(n)in A(n)$ for every $ninmathbbN$.
Question Does function $A$ necessarily need to be defined on $mathbbN$ or can it also be defined on a set $M$ where $|M|=aleph_0$ or $|M|<+infty$?
Thanks!
set-theory axioms
set-theory axioms
asked Mar 30 at 21:59
Jack J.Jack J.
3592419
asked Mar 30 at 21:59
Jack J.Jack J.
3592419
asked Mar 30 at 21:59
Jack J.Jack J.
3592419
asked Mar 30 at 21:59
Jack J.Jack J.
3592419
asked Mar 30 at 21:59
Jack J.Jack J.
3592419
3592419
1
$begingroup$
This is not really a question about the axiom of choice. It's about whether or not you can define a function from $Bbb N$ as function from a different countable set...
$endgroup$
– Asaf Karagila♦
Mar 31 at 0:30
add a comment |
1
$begingroup$
This is not really a question about the axiom of choice. It's about whether or not you can define a function from $Bbb N$ as function from a different countable set...
$endgroup$
– Asaf Karagila♦
Mar 31 at 0:30
1
1
$begingroup$
This is not really a question about the axiom of choice. It's about whether or not you can define a function from $Bbb N$ as function from a different countable set...
$endgroup$
– Asaf Karagila♦
Mar 31 at 0:30
$begingroup$
This is not really a question about the axiom of choice. It's about whether or not you can define a function from $Bbb N$ as function from a different countable set...
$endgroup$
– Asaf Karagila♦
Mar 31 at 0:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, the restriction that the index set (= domain of $A$) be $mathbbN$ is inessential: from the axiom of countable choice as stated we can deduce the apparently-more-general (arbitrary countably-infinite-or-finite index set) form as a corollary.
Below, I'll use the inclusive definition of countability: $X$ is countable if there is an injection from $X$ into $mathbbN$ (this is equivalent to demanding that $X$ admit a surjection from $mathbbN$ or be empty). Suppose $A$ is a function with domain a countable set $I$ such that $A(i)not=emptyset$ for each $iin I$; I'll show that there is a choice function for $A$, that is, a function $f$ with domain $I$ such that $f(i)in A(i)$ for each $iin I$.
Since $I$ is countable, fix an injection $j:IrightarrowmathbbN$. Let $B(n)=A(j^-1(n))$ if $nin ran(j)$ and let $B(n)=0$ (say) otherwise. By AC$omega$ we get a choice function $g$ for $B$. Now define $f$ by setting $$f(i)=g(j(i)).$$
answered Mar 30 at 22:05
Noah SchweberNoah Schweber
128k10152294
$endgroup$
add a comment |
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$begingroup$
No, the restriction that the index set (= domain of $A$) be $mathbbN$ is inessential: from the axiom of countable choice as stated we can deduce the apparently-more-general (arbitrary countably-infinite-or-finite index set) form as a corollary.
Below, I'll use the inclusive definition of countability: $X$ is countable if there is an injection from $X$ into $mathbbN$ (this is equivalent to demanding that $X$ admit a surjection from $mathbbN$ or be empty). Suppose $A$ is a function with domain a countable set $I$ such that $A(i)not=emptyset$ for each $iin I$; I'll show that there is a choice function for $A$, that is, a function $f$ with domain $I$ such that $f(i)in A(i)$ for each $iin I$.
Since $I$ is countable, fix an injection $j:IrightarrowmathbbN$. Let $B(n)=A(j^-1(n))$ if $nin ran(j)$ and let $B(n)=0$ (say) otherwise. By AC$omega$ we get a choice function $g$ for $B$. Now define $f$ by setting $$f(i)=g(j(i)).$$
answered Mar 30 at 22:05
Noah SchweberNoah Schweber
128k10152294
$endgroup$
add a comment |
$begingroup$
No, the restriction that the index set (= domain of $A$) be $mathbbN$ is inessential: from the axiom of countable choice as stated we can deduce the apparently-more-general (arbitrary countably-infinite-or-finite index set) form as a corollary.
Below, I'll use the inclusive definition of countability: $X$ is countable if there is an injection from $X$ into $mathbbN$ (this is equivalent to demanding that $X$ admit a surjection from $mathbbN$ or be empty). Suppose $A$ is a function with domain a countable set $I$ such that $A(i)not=emptyset$ for each $iin I$; I'll show that there is a choice function for $A$, that is, a function $f$ with domain $I$ such that $f(i)in A(i)$ for each $iin I$.
Since $I$ is countable, fix an injection $j:IrightarrowmathbbN$. Let $B(n)=A(j^-1(n))$ if $nin ran(j)$ and let $B(n)=0$ (say) otherwise. By AC$omega$ we get a choice function $g$ for $B$. Now define $f$ by setting $$f(i)=g(j(i)).$$
answered Mar 30 at 22:05
Noah SchweberNoah Schweber
128k10152294
$endgroup$
add a comment |
$begingroup$
No, the restriction that the index set (= domain of $A$) be $mathbbN$ is inessential: from the axiom of countable choice as stated we can deduce the apparently-more-general (arbitrary countably-infinite-or-finite index set) form as a corollary.
Below, I'll use the inclusive definition of countability: $X$ is countable if there is an injection from $X$ into $mathbbN$ (this is equivalent to demanding that $X$ admit a surjection from $mathbbN$ or be empty). Suppose $A$ is a function with domain a countable set $I$ such that $A(i)not=emptyset$ for each $iin I$; I'll show that there is a choice function for $A$, that is, a function $f$ with domain $I$ such that $f(i)in A(i)$ for each $iin I$.
Since $I$ is countable, fix an injection $j:IrightarrowmathbbN$. Let $B(n)=A(j^-1(n))$ if $nin ran(j)$ and let $B(n)=0$ (say) otherwise. By AC$omega$ we get a choice function $g$ for $B$. Now define $f$ by setting $$f(i)=g(j(i)).$$
answered Mar 30 at 22:05
Noah SchweberNoah Schweber
128k10152294
$endgroup$
No, the restriction that the index set (= domain of $A$) be $mathbbN$ is inessential: from the axiom of countable choice as stated we can deduce the apparently-more-general (arbitrary countably-infinite-or-finite index set) form as a corollary.
Below, I'll use the inclusive definition of countability: $X$ is countable if there is an injection from $X$ into $mathbbN$ (this is equivalent to demanding that $X$ admit a surjection from $mathbbN$ or be empty). Suppose $A$ is a function with domain a countable set $I$ such that $A(i)not=emptyset$ for each $iin I$; I'll show that there is a choice function for $A$, that is, a function $f$ with domain $I$ such that $f(i)in A(i)$ for each $iin I$.
Since $I$ is countable, fix an injection $j:IrightarrowmathbbN$. Let $B(n)=A(j^-1(n))$ if $nin ran(j)$ and let $B(n)=0$ (say) otherwise. By AC$omega$ we get a choice function $g$ for $B$. Now define $f$ by setting $$f(i)=g(j(i)).$$
answered Mar 30 at 22:05
Noah SchweberNoah Schweber
128k10152294
answered Mar 30 at 22:05
Noah SchweberNoah Schweber
128k10152294
answered Mar 30 at 22:05
Noah SchweberNoah Schweber
128k10152294
answered Mar 30 at 22:05
Noah SchweberNoah Schweber
128k10152294
128k10152294
add a comment |
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This is not really a question about the axiom of choice. It's about whether or not you can define a function from $Bbb N$ as function from a different countable set...
$endgroup$
– Asaf Karagila♦
Mar 31 at 0:30