Find an appropriate set and a function such that neither is a subset of the other.
$begingroup$
I'm supposed to find a function $f:Xrightarrow Y$ and a set $Asubseteq X$ such that neither $f(A^{c})subseteq f(A)^{c}$ nor $f(A)^{c}subseteq f(A^{c})$.
I really don't know what to look for or start.
functions elementary-set-theory
edited Jan 19 at 14:20
Bernard
121k740116
asked Jan 19 at 13:54
Mathiaspilot123Mathiaspilot123
456
$endgroup$
add a comment |
$begingroup$
I'm supposed to find a function $f:Xrightarrow Y$ and a set $Asubseteq X$ such that neither $f(A^{c})subseteq f(A)^{c}$ nor $f(A)^{c}subseteq f(A^{c})$.
I really don't know what to look for or start.
functions elementary-set-theory
edited Jan 19 at 14:20
Bernard
121k740116
asked Jan 19 at 13:54
Mathiaspilot123Mathiaspilot123
456
$endgroup$
add a comment |
$begingroup$
I'm supposed to find a function $f:Xrightarrow Y$ and a set $Asubseteq X$ such that neither $f(A^{c})subseteq f(A)^{c}$ nor $f(A)^{c}subseteq f(A^{c})$.
I really don't know what to look for or start.
functions elementary-set-theory
edited Jan 19 at 14:20
Bernard
121k740116
asked Jan 19 at 13:54
Mathiaspilot123Mathiaspilot123
456
$endgroup$
I'm supposed to find a function $f:Xrightarrow Y$ and a set $Asubseteq X$ such that neither $f(A^{c})subseteq f(A)^{c}$ nor $f(A)^{c}subseteq f(A^{c})$.
I really don't know what to look for or start.
functions elementary-set-theory
functions elementary-set-theory
edited Jan 19 at 14:20
Bernard
121k740116
asked Jan 19 at 13:54
Mathiaspilot123Mathiaspilot123
456
edited Jan 19 at 14:20
Bernard
121k740116
asked Jan 19 at 13:54
Mathiaspilot123Mathiaspilot123
456
edited Jan 19 at 14:20
Bernard
121k740116
edited Jan 19 at 14:20
Bernard
121k740116
edited Jan 19 at 14:20
Bernard
121k740116
121k740116
asked Jan 19 at 13:54
Mathiaspilot123Mathiaspilot123
456
asked Jan 19 at 13:54
Mathiaspilot123Mathiaspilot123
456
asked Jan 19 at 13:54
Mathiaspilot123Mathiaspilot123
456
456
add a comment |
add a comment |
1 Answer
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$begingroup$
Always try some finite examples first:
$X=Y={1,2,3,4}$, $A={1,2}$, $f(1)=1, f(2)=2,f(3)=2, f(4)=3$.
Then $f[A]= {1,2}$, $f[A^c]={2,3}$, $f[A]^c = {3,4}$, so neither inclusion holds between $f[A^c]$ and $f[A]^c$. Found by some trial and error.
answered Jan 19 at 14:05
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Always try some finite examples first:
$X=Y={1,2,3,4}$, $A={1,2}$, $f(1)=1, f(2)=2,f(3)=2, f(4)=3$.
Then $f[A]= {1,2}$, $f[A^c]={2,3}$, $f[A]^c = {3,4}$, so neither inclusion holds between $f[A^c]$ and $f[A]^c$. Found by some trial and error.
answered Jan 19 at 14:05
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Always try some finite examples first:
$X=Y={1,2,3,4}$, $A={1,2}$, $f(1)=1, f(2)=2,f(3)=2, f(4)=3$.
Then $f[A]= {1,2}$, $f[A^c]={2,3}$, $f[A]^c = {3,4}$, so neither inclusion holds between $f[A^c]$ and $f[A]^c$. Found by some trial and error.
answered Jan 19 at 14:05
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Always try some finite examples first:
$X=Y={1,2,3,4}$, $A={1,2}$, $f(1)=1, f(2)=2,f(3)=2, f(4)=3$.
Then $f[A]= {1,2}$, $f[A^c]={2,3}$, $f[A]^c = {3,4}$, so neither inclusion holds between $f[A^c]$ and $f[A]^c$. Found by some trial and error.
answered Jan 19 at 14:05
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
Always try some finite examples first:
$X=Y={1,2,3,4}$, $A={1,2}$, $f(1)=1, f(2)=2,f(3)=2, f(4)=3$.
Then $f[A]= {1,2}$, $f[A^c]={2,3}$, $f[A]^c = {3,4}$, so neither inclusion holds between $f[A^c]$ and $f[A]^c$. Found by some trial and error.
answered Jan 19 at 14:05
Henno BrandsmaHenno Brandsma
110k347116
edited Jan 19 at 14:17
answered Jan 19 at 14:05
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 19 at 14:05
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 19 at 14:05
Henno BrandsmaHenno Brandsma
110k347116
110k347116
add a comment |
add a comment |
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