Cardinality of nested infinite subsets












0












$begingroup$


Let $A$ and $C$ be infinite sets, with $C subset A$, and suppose $|A|=|C|$. Now suppose there exists a set $B$ such that $C subset B subset A$.



Intuitively, $A$ and $B$ should have the same cardinality. I'm sure the proof is ridiculously easy but I haven't been able to find it. Can anyone help me, please?






elementary-set-theory cardinals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem ?
    $endgroup$
    – Lord Shark the Unknown
    Dec 11 '18 at 20:52






  • 2




    $begingroup$
    why is nested in quotes? the sets are nested
    $endgroup$
    – mathworker21
    Dec 11 '18 at 20:53












  • $begingroup$
    @LordSharktheUnknown thank you. Maybe it wasn't so trivial after all?
    $endgroup$
    – The Footprint
    Dec 11 '18 at 20:58










  • $begingroup$
    @LordSharktheUnknown You should give basically that as an answer.
    $endgroup$
    – Noah Schweber
    Dec 11 '18 at 23:06
















0












$begingroup$


Let $A$ and $C$ be infinite sets, with $C subset A$, and suppose $|A|=|C|$. Now suppose there exists a set $B$ such that $C subset B subset A$.



Intuitively, $A$ and $B$ should have the same cardinality. I'm sure the proof is ridiculously easy but I haven't been able to find it. Can anyone help me, please?






elementary-set-theory cardinals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem ?
    $endgroup$
    – Lord Shark the Unknown
    Dec 11 '18 at 20:52






  • 2




    $begingroup$
    why is nested in quotes? the sets are nested
    $endgroup$
    – mathworker21
    Dec 11 '18 at 20:53












  • $begingroup$
    @LordSharktheUnknown thank you. Maybe it wasn't so trivial after all?
    $endgroup$
    – The Footprint
    Dec 11 '18 at 20:58










  • $begingroup$
    @LordSharktheUnknown You should give basically that as an answer.
    $endgroup$
    – Noah Schweber
    Dec 11 '18 at 23:06














0












0








0





$begingroup$


Let $A$ and $C$ be infinite sets, with $C subset A$, and suppose $|A|=|C|$. Now suppose there exists a set $B$ such that $C subset B subset A$.



Intuitively, $A$ and $B$ should have the same cardinality. I'm sure the proof is ridiculously easy but I haven't been able to find it. Can anyone help me, please?






elementary-set-theory cardinals







share|cite|improve this question











$endgroup$




Let $A$ and $C$ be infinite sets, with $C subset A$, and suppose $|A|=|C|$. Now suppose there exists a set $B$ such that $C subset B subset A$.



Intuitively, $A$ and $B$ should have the same cardinality. I'm sure the proof is ridiculously easy but I haven't been able to find it. Can anyone help me, please?






elementary-set-theory cardinals




elementary-set-theory cardinals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 2:09









Andrés E. Caicedo

65.8k8160252




65.8k8160252










asked Dec 11 '18 at 20:50









The FootprintThe Footprint

877




877








  • 2




    $begingroup$
    en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem ?
    $endgroup$
    – Lord Shark the Unknown
    Dec 11 '18 at 20:52






  • 2




    $begingroup$
    why is nested in quotes? the sets are nested
    $endgroup$
    – mathworker21
    Dec 11 '18 at 20:53












  • $begingroup$
    @LordSharktheUnknown thank you. Maybe it wasn't so trivial after all?
    $endgroup$
    – The Footprint
    Dec 11 '18 at 20:58










  • $begingroup$
    @LordSharktheUnknown You should give basically that as an answer.
    $endgroup$
    – Noah Schweber
    Dec 11 '18 at 23:06














  • 2




    $begingroup$
    en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem ?
    $endgroup$
    – Lord Shark the Unknown
    Dec 11 '18 at 20:52






  • 2




    $begingroup$
    why is nested in quotes? the sets are nested
    $endgroup$
    – mathworker21
    Dec 11 '18 at 20:53












  • $begingroup$
    @LordSharktheUnknown thank you. Maybe it wasn't so trivial after all?
    $endgroup$
    – The Footprint
    Dec 11 '18 at 20:58










  • $begingroup$
    @LordSharktheUnknown You should give basically that as an answer.
    $endgroup$
    – Noah Schweber
    Dec 11 '18 at 23:06








2




2




$begingroup$
en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem ?
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 20:52




$begingroup$
en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem ?
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 20:52




2




2




$begingroup$
why is nested in quotes? the sets are nested
$endgroup$
– mathworker21
Dec 11 '18 at 20:53






$begingroup$
why is nested in quotes? the sets are nested
$endgroup$
– mathworker21
Dec 11 '18 at 20:53














$begingroup$
@LordSharktheUnknown thank you. Maybe it wasn't so trivial after all?
$endgroup$
– The Footprint
Dec 11 '18 at 20:58




$begingroup$
@LordSharktheUnknown thank you. Maybe it wasn't so trivial after all?
$endgroup$
– The Footprint
Dec 11 '18 at 20:58












$begingroup$
@LordSharktheUnknown You should give basically that as an answer.
$endgroup$
– Noah Schweber
Dec 11 '18 at 23:06




$begingroup$
@LordSharktheUnknown You should give basically that as an answer.
$endgroup$
– Noah Schweber
Dec 11 '18 at 23:06










1 Answer
1






active

oldest

votes


















1












$begingroup$

Since nobody's answered this one yet, as Lord Shark suggests, the Schroeder-Bernstein theorem makes short work of this. Inclusion is an injective map $Bto A,$ whereas composing the inclusion $Cto B$ with the bijection $Ato C$ that exists by assumption gives an injective map $Ato B.$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    It might be of some interest that this fact is equivalent to the Schröder-Bernstein theorem. To prove the converse, suppose we have injective maps $f:Xto Y$ and $g:Yto X$. Then consider the nested sets $Xsupseteq g(Y)supseteq g(f(X))$, note that we have a bijection $gf$ from$X$ to $g(f(X))$, infer that there's a bijection from $X$ to $g(Y)$, and compose that with $g^{-1}$ to get a bijection from $X$ to $Y$.
    $endgroup$
    – Andreas Blass
    Dec 12 '18 at 2:51












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',
autoActivateHeartbeat: false,
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%2f3035796%2fcardinality-of-nested-infinite-subsets%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









1












$begingroup$

Since nobody's answered this one yet, as Lord Shark suggests, the Schroeder-Bernstein theorem makes short work of this. Inclusion is an injective map $Bto A,$ whereas composing the inclusion $Cto B$ with the bijection $Ato C$ that exists by assumption gives an injective map $Ato B.$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    It might be of some interest that this fact is equivalent to the Schröder-Bernstein theorem. To prove the converse, suppose we have injective maps $f:Xto Y$ and $g:Yto X$. Then consider the nested sets $Xsupseteq g(Y)supseteq g(f(X))$, note that we have a bijection $gf$ from$X$ to $g(f(X))$, infer that there's a bijection from $X$ to $g(Y)$, and compose that with $g^{-1}$ to get a bijection from $X$ to $Y$.
    $endgroup$
    – Andreas Blass
    Dec 12 '18 at 2:51
















1












$begingroup$

Since nobody's answered this one yet, as Lord Shark suggests, the Schroeder-Bernstein theorem makes short work of this. Inclusion is an injective map $Bto A,$ whereas composing the inclusion $Cto B$ with the bijection $Ato C$ that exists by assumption gives an injective map $Ato B.$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    It might be of some interest that this fact is equivalent to the Schröder-Bernstein theorem. To prove the converse, suppose we have injective maps $f:Xto Y$ and $g:Yto X$. Then consider the nested sets $Xsupseteq g(Y)supseteq g(f(X))$, note that we have a bijection $gf$ from$X$ to $g(f(X))$, infer that there's a bijection from $X$ to $g(Y)$, and compose that with $g^{-1}$ to get a bijection from $X$ to $Y$.
    $endgroup$
    – Andreas Blass
    Dec 12 '18 at 2:51














1












1








1





$begingroup$

Since nobody's answered this one yet, as Lord Shark suggests, the Schroeder-Bernstein theorem makes short work of this. Inclusion is an injective map $Bto A,$ whereas composing the inclusion $Cto B$ with the bijection $Ato C$ that exists by assumption gives an injective map $Ato B.$






share|cite|improve this answer









$endgroup$



Since nobody's answered this one yet, as Lord Shark suggests, the Schroeder-Bernstein theorem makes short work of this. Inclusion is an injective map $Bto A,$ whereas composing the inclusion $Cto B$ with the bijection $Ato C$ that exists by assumption gives an injective map $Ato B.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 0:40









spaceisdarkgreenspaceisdarkgreen

33.8k21753




33.8k21753








  • 3




    $begingroup$
    It might be of some interest that this fact is equivalent to the Schröder-Bernstein theorem. To prove the converse, suppose we have injective maps $f:Xto Y$ and $g:Yto X$. Then consider the nested sets $Xsupseteq g(Y)supseteq g(f(X))$, note that we have a bijection $gf$ from$X$ to $g(f(X))$, infer that there's a bijection from $X$ to $g(Y)$, and compose that with $g^{-1}$ to get a bijection from $X$ to $Y$.
    $endgroup$
    – Andreas Blass
    Dec 12 '18 at 2:51














  • 3




    $begingroup$
    It might be of some interest that this fact is equivalent to the Schröder-Bernstein theorem. To prove the converse, suppose we have injective maps $f:Xto Y$ and $g:Yto X$. Then consider the nested sets $Xsupseteq g(Y)supseteq g(f(X))$, note that we have a bijection $gf$ from$X$ to $g(f(X))$, infer that there's a bijection from $X$ to $g(Y)$, and compose that with $g^{-1}$ to get a bijection from $X$ to $Y$.
    $endgroup$
    – Andreas Blass
    Dec 12 '18 at 2:51








3




3




$begingroup$
It might be of some interest that this fact is equivalent to the Schröder-Bernstein theorem. To prove the converse, suppose we have injective maps $f:Xto Y$ and $g:Yto X$. Then consider the nested sets $Xsupseteq g(Y)supseteq g(f(X))$, note that we have a bijection $gf$ from$X$ to $g(f(X))$, infer that there's a bijection from $X$ to $g(Y)$, and compose that with $g^{-1}$ to get a bijection from $X$ to $Y$.
$endgroup$
– Andreas Blass
Dec 12 '18 at 2:51




$begingroup$
It might be of some interest that this fact is equivalent to the Schröder-Bernstein theorem. To prove the converse, suppose we have injective maps $f:Xto Y$ and $g:Yto X$. Then consider the nested sets $Xsupseteq g(Y)supseteq g(f(X))$, note that we have a bijection $gf$ from$X$ to $g(f(X))$, infer that there's a bijection from $X$ to $g(Y)$, and compose that with $g^{-1}$ to get a bijection from $X$ to $Y$.
$endgroup$
– Andreas Blass
Dec 12 '18 at 2:51


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035796%2fcardinality-of-nested-infinite-subsets%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

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more