Let $f:Xto Y$, and $Asubseteq X$, $Bsubseteq Y$. Show that $f[A]cap Bsubseteq f[Acap f^-1[B]]$

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Let $f:Xto Y$, and $Asubseteq X$, $Bsubseteq Y$. Show that $f[A]cap Bsubseteq f[Acap f^-1[B]]$

I've tried to introduce some elements as the preimage for $B$ but I can't seem to make sense of it.

edited Sep 10 at 7:43

Asaf Karagilaâ™¦

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asked Sep 10 at 7:37

Peter Celinski

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up vote
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down vote

favorite

Let $f:Xto Y$, and $Asubseteq X$, $Bsubseteq Y$. Show that $f[A]cap Bsubseteq f[Acap f^-1[B]]$

I've tried to introduce some elements as the preimage for $B$ but I can't seem to make sense of it.

edited Sep 10 at 7:43

Asaf Karagilaâ™¦

295k32410738

asked Sep 10 at 7:37

Peter Celinski

423

add a commentÂ |Â

up vote
0
down vote

favorite

Let $f:Xto Y$, and $Asubseteq X$, $Bsubseteq Y$. Show that $f[A]cap Bsubseteq f[Acap f^-1[B]]$

I've tried to introduce some elements as the preimage for $B$ but I can't seem to make sense of it.

edited Sep 10 at 7:43

Asaf Karagilaâ™¦

295k32410738

asked Sep 10 at 7:37

Peter Celinski

423

Let $f:Xto Y$, and $Asubseteq X$, $Bsubseteq Y$. Show that $f[A]cap Bsubseteq f[Acap f^-1[B]]$

I've tried to introduce some elements as the preimage for $B$ but I can't seem to make sense of it.

functions elementary-set-theory

edited Sep 10 at 7:43

Asaf Karagilaâ™¦

295k32410738

asked Sep 10 at 7:37

Peter Celinski

423

edited Sep 10 at 7:43

Asaf Karagilaâ™¦

295k32410738

asked Sep 10 at 7:37

Peter Celinski

423

edited Sep 10 at 7:43

Asaf Karagilaâ™¦

295k32410738

edited Sep 10 at 7:43

Asaf Karagilaâ™¦

295k32410738

edited Sep 10 at 7:43

Asaf Karagilaâ™¦

295k32410738

asked Sep 10 at 7:37

Peter Celinski

423

asked Sep 10 at 7:37

Peter Celinski

423

asked Sep 10 at 7:37

Peter Celinski

423

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
1
down vote

accepted

The sets are even the same.

$$yin f[A]cap Biffexists xin A; [y=f(x)wedge yin B]iffexists xin A; [y=f(x)wedge f(x)in B]tag1$$
$$yin f[Acap f^-1(B)]iffexists xin A; [y=f(x)wedge xin f^-1(B)]tag2$$

and here: $$f(x)in Biff xin f^-1(B)$$ so that $yin f[A]cap B$ and $yin f[Acap f^-1(B)]$ are equivalent.

answered Sep 10 at 8:03

drhab

89.4k541123

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3
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Let $y in f(A)cap B$. Then $y =f(a)$ for some $a in A$. Now $a in A cap f^-1 (B)$ because $a in A$ and $f(a)=y in B$. Hence $y =f(a) in f(Acap f^-1(B))$.

answered Sep 10 at 7:44

Kavi Rama Murthy

27.1k31439

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0
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Let $y in f(A) cap B$,

Then $y in f(A)$ and $y in B$, i.e there is a

$a in A$ such that $y=f(a)=b in B$,

then $a in f^-1(b) subset f^-1(B)$;

Since $a in A$ and $a in f^-1(B)$, we have

$a in Acap f^-1B$, and finally

$y=f(a) in f(Acap f^-1(B))$.

edited Sep 10 at 8:39

answered Sep 10 at 8:33

Peter Szilas

8,4452617

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
1
down vote

accepted

The sets are even the same.

$$yin f[A]cap Biffexists xin A; [y=f(x)wedge yin B]iffexists xin A; [y=f(x)wedge f(x)in B]tag1$$
$$yin f[Acap f^-1(B)]iffexists xin A; [y=f(x)wedge xin f^-1(B)]tag2$$

and here: $$f(x)in Biff xin f^-1(B)$$ so that $yin f[A]cap B$ and $yin f[Acap f^-1(B)]$ are equivalent.

answered Sep 10 at 8:03

drhab

89.4k541123

add a commentÂ |Â

up vote
1
down vote

accepted

The sets are even the same.

$$yin f[A]cap Biffexists xin A; [y=f(x)wedge yin B]iffexists xin A; [y=f(x)wedge f(x)in B]tag1$$
$$yin f[Acap f^-1(B)]iffexists xin A; [y=f(x)wedge xin f^-1(B)]tag2$$

and here: $$f(x)in Biff xin f^-1(B)$$ so that $yin f[A]cap B$ and $yin f[Acap f^-1(B)]$ are equivalent.

answered Sep 10 at 8:03

drhab

89.4k541123

add a commentÂ |Â

up vote
1
down vote

accepted

The sets are even the same.

$$yin f[A]cap Biffexists xin A; [y=f(x)wedge yin B]iffexists xin A; [y=f(x)wedge f(x)in B]tag1$$
$$yin f[Acap f^-1(B)]iffexists xin A; [y=f(x)wedge xin f^-1(B)]tag2$$

and here: $$f(x)in Biff xin f^-1(B)$$ so that $yin f[A]cap B$ and $yin f[Acap f^-1(B)]$ are equivalent.

answered Sep 10 at 8:03

drhab

89.4k541123

The sets are even the same.

$$yin f[A]cap Biffexists xin A; [y=f(x)wedge yin B]iffexists xin A; [y=f(x)wedge f(x)in B]tag1$$
$$yin f[Acap f^-1(B)]iffexists xin A; [y=f(x)wedge xin f^-1(B)]tag2$$

and here: $$f(x)in Biff xin f^-1(B)$$ so that $yin f[A]cap B$ and $yin f[Acap f^-1(B)]$ are equivalent.

answered Sep 10 at 8:03

drhab

89.4k541123

answered Sep 10 at 8:03

drhab

89.4k541123

answered Sep 10 at 8:03

drhab

89.4k541123

answered Sep 10 at 8:03

drhab

89.4k541123

add a commentÂ |Â

up vote
3
down vote

Let $y in f(A)cap B$. Then $y =f(a)$ for some $a in A$. Now $a in A cap f^-1 (B)$ because $a in A$ and $f(a)=y in B$. Hence $y =f(a) in f(Acap f^-1(B))$.

answered Sep 10 at 7:44

Kavi Rama Murthy

27.1k31439

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up vote
3
down vote

Let $y in f(A)cap B$. Then $y =f(a)$ for some $a in A$. Now $a in A cap f^-1 (B)$ because $a in A$ and $f(a)=y in B$. Hence $y =f(a) in f(Acap f^-1(B))$.

answered Sep 10 at 7:44

Kavi Rama Murthy

27.1k31439

add a commentÂ |Â

up vote
3
down vote

Let $y in f(A)cap B$. Then $y =f(a)$ for some $a in A$. Now $a in A cap f^-1 (B)$ because $a in A$ and $f(a)=y in B$. Hence $y =f(a) in f(Acap f^-1(B))$.

answered Sep 10 at 7:44

Kavi Rama Murthy

27.1k31439

Let $y in f(A)cap B$. Then $y =f(a)$ for some $a in A$. Now $a in A cap f^-1 (B)$ because $a in A$ and $f(a)=y in B$. Hence $y =f(a) in f(Acap f^-1(B))$.

answered Sep 10 at 7:44

Kavi Rama Murthy

27.1k31439

answered Sep 10 at 7:44

Kavi Rama Murthy

27.1k31439

answered Sep 10 at 7:44

Kavi Rama Murthy

27.1k31439

answered Sep 10 at 7:44

Kavi Rama Murthy

27.1k31439

add a commentÂ |Â

up vote
0
down vote

Let $y in f(A) cap B$,

Then $y in f(A)$ and $y in B$, i.e there is a

$a in A$ such that $y=f(a)=b in B$,

then $a in f^-1(b) subset f^-1(B)$;

Since $a in A$ and $a in f^-1(B)$, we have

$a in Acap f^-1B$, and finally

$y=f(a) in f(Acap f^-1(B))$.

edited Sep 10 at 8:39

answered Sep 10 at 8:33

Peter Szilas

8,4452617

add a commentÂ |Â

up vote
0
down vote

Let $y in f(A) cap B$,

Then $y in f(A)$ and $y in B$, i.e there is a

$a in A$ such that $y=f(a)=b in B$,

then $a in f^-1(b) subset f^-1(B)$;

Since $a in A$ and $a in f^-1(B)$, we have

$a in Acap f^-1B$, and finally

$y=f(a) in f(Acap f^-1(B))$.

edited Sep 10 at 8:39

answered Sep 10 at 8:33

Peter Szilas

8,4452617

add a commentÂ |Â

up vote
0
down vote

Let $y in f(A) cap B$,

Then $y in f(A)$ and $y in B$, i.e there is a

$a in A$ such that $y=f(a)=b in B$,

then $a in f^-1(b) subset f^-1(B)$;

Since $a in A$ and $a in f^-1(B)$, we have

$a in Acap f^-1B$, and finally

$y=f(a) in f(Acap f^-1(B))$.

edited Sep 10 at 8:39

answered Sep 10 at 8:33

Peter Szilas

8,4452617

Let $y in f(A) cap B$,

Then $y in f(A)$ and $y in B$, i.e there is a

$a in A$ such that $y=f(a)=b in B$,

then $a in f^-1(b) subset f^-1(B)$;

Since $a in A$ and $a in f^-1(B)$, we have

$a in Acap f^-1B$, and finally

$y=f(a) in f(Acap f^-1(B))$.

edited Sep 10 at 8:39

answered Sep 10 at 8:33

Peter Szilas

8,4452617

edited Sep 10 at 8:39

answered Sep 10 at 8:33

Peter Szilas

8,4452617

answered Sep 10 at 8:33

Peter Szilas

8,4452617

answered Sep 10 at 8:33

Peter Szilas

8,4452617

add a commentÂ |Â

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