Do we need the Axiom of Choice to guarantee surjectiveness of projections?











up vote
1
down vote

favorite












Given a collection ${X_{alpha}}_{alphainOmega}$ of non-empty sets, do we need the Axiom of Choice to ensure that the projections
$$pi_{gamma}:prod_{alpha}X_{alpha}longrightarrow X_{gamma}$$
are surjective??



Given $Asubseteq X_{gamma}$ its preimage $pi_{gamma}^{-1}(A)$ is $Atimesleft(prod_{alphanegamma}X_{alpha}right)$.



When $Omega$ is finite we can choose elements $x_{alpha}$ in each one of the $X_{alpha}$ ($alphanegamma$) to get an $n$-tuple such that $x_{gamma}in X_{gamma}$ but what if $Omega$ isn't finite or even countable infinite? The fact that $pi_gamma$ is surjective would mean we can choose elements from the collection (using the AC).



Is it valid to say "for all" $x_{alpha}in X_{alpha}$ with $alphanegamma$ and some $x_gammain X_gamma$ then $f:Omegatobigcup_{alpha}X_{alpha}$ such that $f(alpha)=x_alpha$ for every $alphainOmega$ is a member of $prod_alpha X_alpha$ ?



Thanks






set-theory axiom-of-choice projection







share|cite|improve this question


















  • 2




    If $prod_alpha X_alpha$ is empty, then $pi_gamma$ is not surjective. That all such projections are nonempty is just a reformulation of the statement that product are nonempty, which is precisely the axiom of choice. I do not understand your last paragraph.
    – Andrés E. Caicedo
    Nov 22 at 19:38










  • @AndrésE.Caicedo the last paragraph is maybe a (probably wrong ) way to avoid the axiom of choice.
    – Pedro
    Nov 22 at 19:41















up vote
1
down vote

favorite












Given a collection ${X_{alpha}}_{alphainOmega}$ of non-empty sets, do we need the Axiom of Choice to ensure that the projections
$$pi_{gamma}:prod_{alpha}X_{alpha}longrightarrow X_{gamma}$$
are surjective??



Given $Asubseteq X_{gamma}$ its preimage $pi_{gamma}^{-1}(A)$ is $Atimesleft(prod_{alphanegamma}X_{alpha}right)$.



When $Omega$ is finite we can choose elements $x_{alpha}$ in each one of the $X_{alpha}$ ($alphanegamma$) to get an $n$-tuple such that $x_{gamma}in X_{gamma}$ but what if $Omega$ isn't finite or even countable infinite? The fact that $pi_gamma$ is surjective would mean we can choose elements from the collection (using the AC).



Is it valid to say "for all" $x_{alpha}in X_{alpha}$ with $alphanegamma$ and some $x_gammain X_gamma$ then $f:Omegatobigcup_{alpha}X_{alpha}$ such that $f(alpha)=x_alpha$ for every $alphainOmega$ is a member of $prod_alpha X_alpha$ ?



Thanks






set-theory axiom-of-choice projection







share|cite|improve this question


















  • 2




    If $prod_alpha X_alpha$ is empty, then $pi_gamma$ is not surjective. That all such projections are nonempty is just a reformulation of the statement that product are nonempty, which is precisely the axiom of choice. I do not understand your last paragraph.
    – Andrés E. Caicedo
    Nov 22 at 19:38










  • @AndrésE.Caicedo the last paragraph is maybe a (probably wrong ) way to avoid the axiom of choice.
    – Pedro
    Nov 22 at 19:41













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Given a collection ${X_{alpha}}_{alphainOmega}$ of non-empty sets, do we need the Axiom of Choice to ensure that the projections
$$pi_{gamma}:prod_{alpha}X_{alpha}longrightarrow X_{gamma}$$
are surjective??



Given $Asubseteq X_{gamma}$ its preimage $pi_{gamma}^{-1}(A)$ is $Atimesleft(prod_{alphanegamma}X_{alpha}right)$.



When $Omega$ is finite we can choose elements $x_{alpha}$ in each one of the $X_{alpha}$ ($alphanegamma$) to get an $n$-tuple such that $x_{gamma}in X_{gamma}$ but what if $Omega$ isn't finite or even countable infinite? The fact that $pi_gamma$ is surjective would mean we can choose elements from the collection (using the AC).



Is it valid to say "for all" $x_{alpha}in X_{alpha}$ with $alphanegamma$ and some $x_gammain X_gamma$ then $f:Omegatobigcup_{alpha}X_{alpha}$ such that $f(alpha)=x_alpha$ for every $alphainOmega$ is a member of $prod_alpha X_alpha$ ?



Thanks






set-theory axiom-of-choice projection







share|cite|improve this question













Given a collection ${X_{alpha}}_{alphainOmega}$ of non-empty sets, do we need the Axiom of Choice to ensure that the projections
$$pi_{gamma}:prod_{alpha}X_{alpha}longrightarrow X_{gamma}$$
are surjective??



Given $Asubseteq X_{gamma}$ its preimage $pi_{gamma}^{-1}(A)$ is $Atimesleft(prod_{alphanegamma}X_{alpha}right)$.



When $Omega$ is finite we can choose elements $x_{alpha}$ in each one of the $X_{alpha}$ ($alphanegamma$) to get an $n$-tuple such that $x_{gamma}in X_{gamma}$ but what if $Omega$ isn't finite or even countable infinite? The fact that $pi_gamma$ is surjective would mean we can choose elements from the collection (using the AC).



Is it valid to say "for all" $x_{alpha}in X_{alpha}$ with $alphanegamma$ and some $x_gammain X_gamma$ then $f:Omegatobigcup_{alpha}X_{alpha}$ such that $f(alpha)=x_alpha$ for every $alphainOmega$ is a member of $prod_alpha X_alpha$ ?



Thanks






set-theory axiom-of-choice projection




set-theory axiom-of-choice projection






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 19:32









Pedro

512212




512212








  • 2




    If $prod_alpha X_alpha$ is empty, then $pi_gamma$ is not surjective. That all such projections are nonempty is just a reformulation of the statement that product are nonempty, which is precisely the axiom of choice. I do not understand your last paragraph.
    – Andrés E. Caicedo
    Nov 22 at 19:38










  • @AndrésE.Caicedo the last paragraph is maybe a (probably wrong ) way to avoid the axiom of choice.
    – Pedro
    Nov 22 at 19:41














  • 2




    If $prod_alpha X_alpha$ is empty, then $pi_gamma$ is not surjective. That all such projections are nonempty is just a reformulation of the statement that product are nonempty, which is precisely the axiom of choice. I do not understand your last paragraph.
    – Andrés E. Caicedo
    Nov 22 at 19:38










  • @AndrésE.Caicedo the last paragraph is maybe a (probably wrong ) way to avoid the axiom of choice.
    – Pedro
    Nov 22 at 19:41








2




2




If $prod_alpha X_alpha$ is empty, then $pi_gamma$ is not surjective. That all such projections are nonempty is just a reformulation of the statement that product are nonempty, which is precisely the axiom of choice. I do not understand your last paragraph.
– Andrés E. Caicedo
Nov 22 at 19:38




If $prod_alpha X_alpha$ is empty, then $pi_gamma$ is not surjective. That all such projections are nonempty is just a reformulation of the statement that product are nonempty, which is precisely the axiom of choice. I do not understand your last paragraph.
– Andrés E. Caicedo
Nov 22 at 19:38












@AndrésE.Caicedo the last paragraph is maybe a (probably wrong ) way to avoid the axiom of choice.
– Pedro
Nov 22 at 19:41




@AndrésE.Caicedo the last paragraph is maybe a (probably wrong ) way to avoid the axiom of choice.
– Pedro
Nov 22 at 19:41










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










Yes, choice is needed.



Without choice, $prod_{alphain Omega}X_alpha$ could be empty, even if each $X_alpha$ is nonempty - indeed, "every product of nonempty sets is nonempty" is just the axiom of choice itself! (A choice function for a family of disjoint nonempty sets is exactly an element of their product.)



Re: your last paragraph, the $f$ you're trying to define there need not exist without choice - note that you're beginning right away by picking $x_alphain X_alpha$. So that's why it doesn't get around AC.




What is true is that if $prod_{alphainOmega}X_alpha$ is nonempty, then each $pi_gamma$ is surjective.



Proof: For simplicity, assume the $X_alpha$s are disjoint. Let $hinprod_{alphainOmega}X_alpha$, and fix $gammainOmega$. For each $xin X_gamma$, let $h_x$ be gotten from $h$ by replacing $h(gamma)$ with $x$: $$h_x(alpha)=h(alpha)mbox{ for $alphainOmegasetminus{gamma}$}, quad h_x(gamma)=x.$$ Then $h_xinprod_{alphainOmega}X_alpha$ and $pi_gamma(h_x)=x$.



Note that this means that in fact "the $pi_gamma$s are always surjective" is equivalent to the axiom of choice.






share|cite|improve this answer





















  • Thanks! Nice answer
    – Pedro
    Nov 22 at 19:47











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009550%2fdo-we-need-the-axiom-of-choice-to-guarantee-surjectiveness-of-projections%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Yes, choice is needed.



Without choice, $prod_{alphain Omega}X_alpha$ could be empty, even if each $X_alpha$ is nonempty - indeed, "every product of nonempty sets is nonempty" is just the axiom of choice itself! (A choice function for a family of disjoint nonempty sets is exactly an element of their product.)



Re: your last paragraph, the $f$ you're trying to define there need not exist without choice - note that you're beginning right away by picking $x_alphain X_alpha$. So that's why it doesn't get around AC.




What is true is that if $prod_{alphainOmega}X_alpha$ is nonempty, then each $pi_gamma$ is surjective.



Proof: For simplicity, assume the $X_alpha$s are disjoint. Let $hinprod_{alphainOmega}X_alpha$, and fix $gammainOmega$. For each $xin X_gamma$, let $h_x$ be gotten from $h$ by replacing $h(gamma)$ with $x$: $$h_x(alpha)=h(alpha)mbox{ for $alphainOmegasetminus{gamma}$}, quad h_x(gamma)=x.$$ Then $h_xinprod_{alphainOmega}X_alpha$ and $pi_gamma(h_x)=x$.



Note that this means that in fact "the $pi_gamma$s are always surjective" is equivalent to the axiom of choice.






share|cite|improve this answer





















  • Thanks! Nice answer
    – Pedro
    Nov 22 at 19:47















up vote
2
down vote



accepted










Yes, choice is needed.



Without choice, $prod_{alphain Omega}X_alpha$ could be empty, even if each $X_alpha$ is nonempty - indeed, "every product of nonempty sets is nonempty" is just the axiom of choice itself! (A choice function for a family of disjoint nonempty sets is exactly an element of their product.)



Re: your last paragraph, the $f$ you're trying to define there need not exist without choice - note that you're beginning right away by picking $x_alphain X_alpha$. So that's why it doesn't get around AC.




What is true is that if $prod_{alphainOmega}X_alpha$ is nonempty, then each $pi_gamma$ is surjective.



Proof: For simplicity, assume the $X_alpha$s are disjoint. Let $hinprod_{alphainOmega}X_alpha$, and fix $gammainOmega$. For each $xin X_gamma$, let $h_x$ be gotten from $h$ by replacing $h(gamma)$ with $x$: $$h_x(alpha)=h(alpha)mbox{ for $alphainOmegasetminus{gamma}$}, quad h_x(gamma)=x.$$ Then $h_xinprod_{alphainOmega}X_alpha$ and $pi_gamma(h_x)=x$.



Note that this means that in fact "the $pi_gamma$s are always surjective" is equivalent to the axiom of choice.






share|cite|improve this answer





















  • Thanks! Nice answer
    – Pedro
    Nov 22 at 19:47













up vote
2
down vote



accepted







up vote
2
down vote



accepted






Yes, choice is needed.



Without choice, $prod_{alphain Omega}X_alpha$ could be empty, even if each $X_alpha$ is nonempty - indeed, "every product of nonempty sets is nonempty" is just the axiom of choice itself! (A choice function for a family of disjoint nonempty sets is exactly an element of their product.)



Re: your last paragraph, the $f$ you're trying to define there need not exist without choice - note that you're beginning right away by picking $x_alphain X_alpha$. So that's why it doesn't get around AC.




What is true is that if $prod_{alphainOmega}X_alpha$ is nonempty, then each $pi_gamma$ is surjective.



Proof: For simplicity, assume the $X_alpha$s are disjoint. Let $hinprod_{alphainOmega}X_alpha$, and fix $gammainOmega$. For each $xin X_gamma$, let $h_x$ be gotten from $h$ by replacing $h(gamma)$ with $x$: $$h_x(alpha)=h(alpha)mbox{ for $alphainOmegasetminus{gamma}$}, quad h_x(gamma)=x.$$ Then $h_xinprod_{alphainOmega}X_alpha$ and $pi_gamma(h_x)=x$.



Note that this means that in fact "the $pi_gamma$s are always surjective" is equivalent to the axiom of choice.






share|cite|improve this answer












Yes, choice is needed.



Without choice, $prod_{alphain Omega}X_alpha$ could be empty, even if each $X_alpha$ is nonempty - indeed, "every product of nonempty sets is nonempty" is just the axiom of choice itself! (A choice function for a family of disjoint nonempty sets is exactly an element of their product.)



Re: your last paragraph, the $f$ you're trying to define there need not exist without choice - note that you're beginning right away by picking $x_alphain X_alpha$. So that's why it doesn't get around AC.




What is true is that if $prod_{alphainOmega}X_alpha$ is nonempty, then each $pi_gamma$ is surjective.



Proof: For simplicity, assume the $X_alpha$s are disjoint. Let $hinprod_{alphainOmega}X_alpha$, and fix $gammainOmega$. For each $xin X_gamma$, let $h_x$ be gotten from $h$ by replacing $h(gamma)$ with $x$: $$h_x(alpha)=h(alpha)mbox{ for $alphainOmegasetminus{gamma}$}, quad h_x(gamma)=x.$$ Then $h_xinprod_{alphainOmega}X_alpha$ and $pi_gamma(h_x)=x$.



Note that this means that in fact "the $pi_gamma$s are always surjective" is equivalent to the axiom of choice.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 19:42









Noah Schweber

120k10146278




120k10146278












  • Thanks! Nice answer
    – Pedro
    Nov 22 at 19:47


















  • Thanks! Nice answer
    – Pedro
    Nov 22 at 19:47
















Thanks! Nice answer
– Pedro
Nov 22 at 19:47




Thanks! Nice answer
– Pedro
Nov 22 at 19:47


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009550%2fdo-we-need-the-axiom-of-choice-to-guarantee-surjectiveness-of-projections%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more