Cardinality of nested infinite subsets
$begingroup$
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
edited Dec 12 '18 at 2:09
Andrés E. Caicedo
65.8k8160252
asked Dec 11 '18 at 20:50
The FootprintThe Footprint
877
$endgroup$
add a comment |
$begingroup$
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
edited Dec 12 '18 at 2:09
Andrés E. Caicedo
65.8k8160252
asked Dec 11 '18 at 20:50
The FootprintThe Footprint
877
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 12 '18 at 2:09
Andrés E. Caicedo
65.8k8160252
asked Dec 11 '18 at 20:50
The FootprintThe Footprint
877
$endgroup$
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
edited Dec 12 '18 at 2:09
Andrés E. Caicedo
65.8k8160252
asked Dec 11 '18 at 20:50
The FootprintThe Footprint
877
edited Dec 12 '18 at 2:09
Andrés E. Caicedo
65.8k8160252
asked Dec 11 '18 at 20:50
The FootprintThe Footprint
877
edited Dec 12 '18 at 2:09
Andrés E. Caicedo
65.8k8160252
edited Dec 12 '18 at 2:09
Andrés E. Caicedo
65.8k8160252
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
asked Dec 11 '18 at 20:50
The FootprintThe Footprint
877
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $Ato B.$
answered Dec 12 '18 at 0:40
spaceisdarkgreenspaceisdarkgreen
33.8k21753
$endgroup$
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
add a comment |
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1 Answer
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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 $Ato B.$
answered Dec 12 '18 at 0:40
spaceisdarkgreenspaceisdarkgreen
33.8k21753
$endgroup$
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
add a comment |
$begingroup$
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 $Ato B.$
answered Dec 12 '18 at 0:40
spaceisdarkgreenspaceisdarkgreen
33.8k21753
$endgroup$
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
add a comment |
$begingroup$
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 $Ato B.$
answered Dec 12 '18 at 0:40
spaceisdarkgreenspaceisdarkgreen
33.8k21753
$endgroup$
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 $Ato B.$
answered Dec 12 '18 at 0:40
spaceisdarkgreenspaceisdarkgreen
33.8k21753
answered Dec 12 '18 at 0:40
spaceisdarkgreenspaceisdarkgreen
33.8k21753
answered Dec 12 '18 at 0:40
spaceisdarkgreenspaceisdarkgreen
33.8k21753
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
add a comment |
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
add a comment |
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$begingroup$
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