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







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$endgroup$








  • 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
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$begingroup$

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












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$begingroup$

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


















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