Prove that if $A preceq B$ and $B approx C$ then $A preceq C$
$begingroup$
Could anyone check my working please?
Prove that if $A preceq B$ and $B approx C$ then $A preceq C$
Assume $A preceq B$ (ie. there is a $f:Ato B$ such that it is one-one) and $B approx C$ (ie. there is a $g:B to C$ that is one-one and onto), we need to show that there is a $h:Ato C$ that is one-one.
Let this $h$ be $gcirc f$, and show that it is one-one. ie. If $g(f(a))=g(f(a'))$ then $f(a)=f(a')$.
We know that $g(f(a))=g(f(a'))$, but since we know that $g$ is one-one, ie. if $g(b)=g(b')$ then $b=b'$, therefore $f(a)=f(a')$.
proof-verification elementary-set-theory
edited Dec 30 '18 at 20:29
Andrés E. Caicedo
66.1k8160252
asked Dec 30 '18 at 16:14
Daniel MakDaniel Mak
520517
$endgroup$
add a comment |
$begingroup$
Could anyone check my working please?
Prove that if $A preceq B$ and $B approx C$ then $A preceq C$
Assume $A preceq B$ (ie. there is a $f:Ato B$ such that it is one-one) and $B approx C$ (ie. there is a $g:B to C$ that is one-one and onto), we need to show that there is a $h:Ato C$ that is one-one.
Let this $h$ be $gcirc f$, and show that it is one-one. ie. If $g(f(a))=g(f(a'))$ then $f(a)=f(a')$.
We know that $g(f(a))=g(f(a'))$, but since we know that $g$ is one-one, ie. if $g(b)=g(b')$ then $b=b'$, therefore $f(a)=f(a')$.
proof-verification elementary-set-theory
edited Dec 30 '18 at 20:29
Andrés E. Caicedo
66.1k8160252
asked Dec 30 '18 at 16:14
Daniel MakDaniel Mak
520517
$endgroup$
add a comment |
$begingroup$
Could anyone check my working please?
Prove that if $A preceq B$ and $B approx C$ then $A preceq C$
Assume $A preceq B$ (ie. there is a $f:Ato B$ such that it is one-one) and $B approx C$ (ie. there is a $g:B to C$ that is one-one and onto), we need to show that there is a $h:Ato C$ that is one-one.
Let this $h$ be $gcirc f$, and show that it is one-one. ie. If $g(f(a))=g(f(a'))$ then $f(a)=f(a')$.
We know that $g(f(a))=g(f(a'))$, but since we know that $g$ is one-one, ie. if $g(b)=g(b')$ then $b=b'$, therefore $f(a)=f(a')$.
proof-verification elementary-set-theory
edited Dec 30 '18 at 20:29
Andrés E. Caicedo
66.1k8160252
asked Dec 30 '18 at 16:14
Daniel MakDaniel Mak
520517
$endgroup$
Could anyone check my working please?
Prove that if $A preceq B$ and $B approx C$ then $A preceq C$
Assume $A preceq B$ (ie. there is a $f:Ato B$ such that it is one-one) and $B approx C$ (ie. there is a $g:B to C$ that is one-one and onto), we need to show that there is a $h:Ato C$ that is one-one.
Let this $h$ be $gcirc f$, and show that it is one-one. ie. If $g(f(a))=g(f(a'))$ then $f(a)=f(a')$.
We know that $g(f(a))=g(f(a'))$, but since we know that $g$ is one-one, ie. if $g(b)=g(b')$ then $b=b'$, therefore $f(a)=f(a')$.
proof-verification elementary-set-theory
proof-verification elementary-set-theory
edited Dec 30 '18 at 20:29
Andrés E. Caicedo
66.1k8160252
asked Dec 30 '18 at 16:14
Daniel MakDaniel Mak
520517
edited Dec 30 '18 at 20:29
Andrés E. Caicedo
66.1k8160252
asked Dec 30 '18 at 16:14
Daniel MakDaniel Mak
520517
edited Dec 30 '18 at 20:29
Andrés E. Caicedo
66.1k8160252
edited Dec 30 '18 at 20:29
Andrés E. Caicedo
66.1k8160252
edited Dec 30 '18 at 20:29
Andrés E. Caicedo
66.1k8160252
66.1k8160252
asked Dec 30 '18 at 16:14
Daniel MakDaniel Mak
520517
asked Dec 30 '18 at 16:14
Daniel MakDaniel Mak
520517
asked Dec 30 '18 at 16:14
Daniel MakDaniel Mak
520517
520517
add a comment |
add a comment |
1 Answer
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$begingroup$
If you want to show $g circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.
Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.
answered Dec 30 '18 at 17:01
angryavianangryavian
42.7k23481
$endgroup$
add a comment |
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$begingroup$
If you want to show $g circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.
Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.
answered Dec 30 '18 at 17:01
angryavianangryavian
42.7k23481
$endgroup$
add a comment |
$begingroup$
If you want to show $g circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.
Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.
answered Dec 30 '18 at 17:01
angryavianangryavian
42.7k23481
$endgroup$
add a comment |
$begingroup$
If you want to show $g circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.
Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.
answered Dec 30 '18 at 17:01
angryavianangryavian
42.7k23481
$endgroup$
If you want to show $g circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.
Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.
answered Dec 30 '18 at 17:01
angryavianangryavian
42.7k23481
answered Dec 30 '18 at 17:01
angryavianangryavian
42.7k23481
answered Dec 30 '18 at 17:01
angryavianangryavian
42.7k23481
answered Dec 30 '18 at 17:01
angryavianangryavian
42.7k23481
42.7k23481
add a comment |
add a comment |
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