Check if $ f(A cap f^{-1}(B)) $ is a subset of $f(A) cap B $.
$f : X rightarrow Y , A subset X, Bsubset Y $
Check if $ f(A cap f^{-1}(B)) $ is a subset of $f(A) cap B $.
$f(A) = {y: forall x in A y=f(x) }$ Edit: this is incorrect!
$f^{-1}(B) = {x: f(x) = B }$
$f(A cap f^{-1}(B)) = \{y: forall x in (A cap f^{-1}(B)) y=f(x)} = \{y: forall x in A land forall xin f^{-1}(B) y=f(x)} =\ {y: forall x in A land forall xin f(x)=B y=f(x)} = ? $
I don't know how to proceed further.
functions elementary-set-theory
asked Nov 27 '18 at 15:09
Student
334
add a comment |
$f : X rightarrow Y , A subset X, Bsubset Y $
Check if $ f(A cap f^{-1}(B)) $ is a subset of $f(A) cap B $.
$f(A) = {y: forall x in A y=f(x) }$ Edit: this is incorrect!
$f^{-1}(B) = {x: f(x) = B }$
$f(A cap f^{-1}(B)) = \{y: forall x in (A cap f^{-1}(B)) y=f(x)} = \{y: forall x in A land forall xin f^{-1}(B) y=f(x)} =\ {y: forall x in A land forall xin f(x)=B y=f(x)} = ? $
I don't know how to proceed further.
functions elementary-set-theory
asked Nov 27 '18 at 15:09
Student
334
Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
– Robert Israel
Nov 27 '18 at 15:15
@Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
– Yadati Kiran
Nov 27 '18 at 15:21
Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
– Student
Nov 27 '18 at 15:25
add a comment |
$f : X rightarrow Y , A subset X, Bsubset Y $
Check if $ f(A cap f^{-1}(B)) $ is a subset of $f(A) cap B $.
$f(A) = {y: forall x in A y=f(x) }$ Edit: this is incorrect!
$f^{-1}(B) = {x: f(x) = B }$
$f(A cap f^{-1}(B)) = \{y: forall x in (A cap f^{-1}(B)) y=f(x)} = \{y: forall x in A land forall xin f^{-1}(B) y=f(x)} =\ {y: forall x in A land forall xin f(x)=B y=f(x)} = ? $
I don't know how to proceed further.
functions elementary-set-theory
asked Nov 27 '18 at 15:09
Student
334
$f : X rightarrow Y , A subset X, Bsubset Y $
Check if $ f(A cap f^{-1}(B)) $ is a subset of $f(A) cap B $.
$f(A) = {y: forall x in A y=f(x) }$ Edit: this is incorrect!
$f^{-1}(B) = {x: f(x) = B }$
$f(A cap f^{-1}(B)) = \{y: forall x in (A cap f^{-1}(B)) y=f(x)} = \{y: forall x in A land forall xin f^{-1}(B) y=f(x)} =\ {y: forall x in A land forall xin f(x)=B y=f(x)} = ? $
I don't know how to proceed further.
functions elementary-set-theory
functions elementary-set-theory
asked Nov 27 '18 at 15:09
Student
334
asked Nov 27 '18 at 15:09
Student
334
edited Nov 27 '18 at 15:26
asked Nov 27 '18 at 15:09
Student
334
asked Nov 27 '18 at 15:09
Student
334
asked Nov 27 '18 at 15:09
Student
334
334
Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
– Robert Israel
Nov 27 '18 at 15:15
@Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
– Yadati Kiran
Nov 27 '18 at 15:21
Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
– Student
Nov 27 '18 at 15:25
add a comment |
Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
– Robert Israel
Nov 27 '18 at 15:15
@Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
– Yadati Kiran
Nov 27 '18 at 15:21
Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
– Student
Nov 27 '18 at 15:25
Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
– Robert Israel
Nov 27 '18 at 15:15
Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
– Robert Israel
Nov 27 '18 at 15:15
@Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
– Yadati Kiran
Nov 27 '18 at 15:21
@Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
– Yadati Kiran
Nov 27 '18 at 15:21
Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
– Student
Nov 27 '18 at 15:25
Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
– Student
Nov 27 '18 at 15:25
add a comment |
2 Answers
2
active
oldest
votes
By definition $f(A)={f(a)mid ain A}$ and $f^{-1}(B)={xin Xmid f(x)in B}$.
The last definition states that $xin f^{-1}(B)iff f(x)in B$.
We can conclude directly that $f(f^{-1}(B))subseteq B$.
Now observe that:
$Acap f^{-1}(B)subseteq A$ so that $f(Acap f^{-1}(B))subseteq f(A)$
$Acap f^{-1}(B)subseteq f^{-1}(B)$ so that $f(Acap f^{-1}(B))subseteq f(f^{-1}(B)subseteq B$
Then consequently: $$f(Acap f^{-1}(B))subseteq f(A)cap B$$
answered Nov 27 '18 at 15:24
drhab
98k544129
add a comment |
It is a subset of $f(A)$ (why?). It is a subset of $B$ (why?). Therefore ...
answered Nov 27 '18 at 15:14
Robert Israel
318k23208457
Wow very good answer
– Akash Roy
Nov 27 '18 at 15:37
add a comment |
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2 Answers
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2 Answers
2
active
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votes
By definition $f(A)={f(a)mid ain A}$ and $f^{-1}(B)={xin Xmid f(x)in B}$.
The last definition states that $xin f^{-1}(B)iff f(x)in B$.
We can conclude directly that $f(f^{-1}(B))subseteq B$.
Now observe that:
$Acap f^{-1}(B)subseteq A$ so that $f(Acap f^{-1}(B))subseteq f(A)$
$Acap f^{-1}(B)subseteq f^{-1}(B)$ so that $f(Acap f^{-1}(B))subseteq f(f^{-1}(B)subseteq B$
Then consequently: $$f(Acap f^{-1}(B))subseteq f(A)cap B$$
answered Nov 27 '18 at 15:24
drhab
98k544129
add a comment |
By definition $f(A)={f(a)mid ain A}$ and $f^{-1}(B)={xin Xmid f(x)in B}$.
The last definition states that $xin f^{-1}(B)iff f(x)in B$.
We can conclude directly that $f(f^{-1}(B))subseteq B$.
Now observe that:
$Acap f^{-1}(B)subseteq A$ so that $f(Acap f^{-1}(B))subseteq f(A)$
$Acap f^{-1}(B)subseteq f^{-1}(B)$ so that $f(Acap f^{-1}(B))subseteq f(f^{-1}(B)subseteq B$
Then consequently: $$f(Acap f^{-1}(B))subseteq f(A)cap B$$
answered Nov 27 '18 at 15:24
drhab
98k544129
add a comment |
By definition $f(A)={f(a)mid ain A}$ and $f^{-1}(B)={xin Xmid f(x)in B}$.
The last definition states that $xin f^{-1}(B)iff f(x)in B$.
We can conclude directly that $f(f^{-1}(B))subseteq B$.
Now observe that:
$Acap f^{-1}(B)subseteq A$ so that $f(Acap f^{-1}(B))subseteq f(A)$
$Acap f^{-1}(B)subseteq f^{-1}(B)$ so that $f(Acap f^{-1}(B))subseteq f(f^{-1}(B)subseteq B$
Then consequently: $$f(Acap f^{-1}(B))subseteq f(A)cap B$$
answered Nov 27 '18 at 15:24
drhab
98k544129
By definition $f(A)={f(a)mid ain A}$ and $f^{-1}(B)={xin Xmid f(x)in B}$.
The last definition states that $xin f^{-1}(B)iff f(x)in B$.
We can conclude directly that $f(f^{-1}(B))subseteq B$.
Now observe that:
$Acap f^{-1}(B)subseteq A$ so that $f(Acap f^{-1}(B))subseteq f(A)$
$Acap f^{-1}(B)subseteq f^{-1}(B)$ so that $f(Acap f^{-1}(B))subseteq f(f^{-1}(B)subseteq B$
Then consequently: $$f(Acap f^{-1}(B))subseteq f(A)cap B$$
answered Nov 27 '18 at 15:24
drhab
98k544129
answered Nov 27 '18 at 15:24
drhab
98k544129
answered Nov 27 '18 at 15:24
drhab
98k544129
answered Nov 27 '18 at 15:24
drhab
98k544129
98k544129
add a comment |
add a comment |
It is a subset of $f(A)$ (why?). It is a subset of $B$ (why?). Therefore ...
answered Nov 27 '18 at 15:14
Robert Israel
318k23208457
Wow very good answer
– Akash Roy
Nov 27 '18 at 15:37
add a comment |
It is a subset of $f(A)$ (why?). It is a subset of $B$ (why?). Therefore ...
answered Nov 27 '18 at 15:14
Robert Israel
318k23208457
Wow very good answer
– Akash Roy
Nov 27 '18 at 15:37
add a comment |
It is a subset of $f(A)$ (why?). It is a subset of $B$ (why?). Therefore ...
answered Nov 27 '18 at 15:14
Robert Israel
318k23208457
It is a subset of $f(A)$ (why?). It is a subset of $B$ (why?). Therefore ...
answered Nov 27 '18 at 15:14
Robert Israel
318k23208457
answered Nov 27 '18 at 15:14
Robert Israel
318k23208457
answered Nov 27 '18 at 15:14
Robert Israel
318k23208457
answered Nov 27 '18 at 15:14
Robert Israel
318k23208457
318k23208457
Wow very good answer
– Akash Roy
Nov 27 '18 at 15:37
add a comment |
Wow very good answer
– Akash Roy
Nov 27 '18 at 15:37
Wow very good answer
– Akash Roy
Nov 27 '18 at 15:37
Wow very good answer
– Akash Roy
Nov 27 '18 at 15:37
add a comment |
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Your statements $f(A)=ldots$ and $f^{-1}(B) = ldots$ are wrong.
– Robert Israel
Nov 27 '18 at 15:15
@Student: $f(A)={f(x):::xin A}$ and $f^{-1}(B)={xin X:::f(x)in B}$
– Yadati Kiran
Nov 27 '18 at 15:21
Oh the book I was using had some other quantifier notation and made a mistake reading which was which.
– Student
Nov 27 '18 at 15:25