Mixing
If $a notin A setminus B$ and $a in A$ , show that $a in B $.
$begingroup$
If $a notin A setminus B$ and $a in A$ , show that $a in B $.
hi. i am studying the basic set theory.
I want to prove that statement. but precisely i don't know the proof.
elementary-set-theory
asked Feb 2 at 23:53
y2noy2no
204
$endgroup$
add a comment |
$begingroup$
If $a notin A setminus B$ and $a in A$ , show that $a in B $.
hi. i am studying the basic set theory.
I want to prove that statement. but precisely i don't know the proof.
elementary-set-theory
asked Feb 2 at 23:53
y2noy2no
204
$endgroup$
$begingroup$
$Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
$endgroup$
– babemcnuggets
Feb 3 at 0:05
add a comment |
$begingroup$
If $a notin A setminus B$ and $a in A$ , show that $a in B $.
hi. i am studying the basic set theory.
I want to prove that statement. but precisely i don't know the proof.
elementary-set-theory
asked Feb 2 at 23:53
y2noy2no
204
$endgroup$
If $a notin A setminus B$ and $a in A$ , show that $a in B $.
hi. i am studying the basic set theory.
I want to prove that statement. but precisely i don't know the proof.
elementary-set-theory
elementary-set-theory
asked Feb 2 at 23:53
y2noy2no
204
asked Feb 2 at 23:53
y2noy2no
204
asked Feb 2 at 23:53
y2noy2no
204
asked Feb 2 at 23:53
y2noy2no
204
asked Feb 2 at 23:53
y2noy2no
204
204
$begingroup$
$Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
$endgroup$
– babemcnuggets
Feb 3 at 0:05
add a comment |
$begingroup$
$Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
$endgroup$
– babemcnuggets
Feb 3 at 0:05
$begingroup$
$Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
$endgroup$
– babemcnuggets
Feb 3 at 0:05
$begingroup$
$Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
$endgroup$
– babemcnuggets
Feb 3 at 0:05
add a comment |
3 Answers
3
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oldest
votes
$begingroup$
Using De Morgan's law and the identity $A setminus B = A cap B^complement$, we have
$$begin{align} a notin A setminus B &iff a in (A setminus B)^complement \&iff a in (A cap B^complement)^complement \&iff a in A^complement cup B end{align}$$
So at least one of $a in A^complement$ or $a in B$ must be true. But $a in A implies a notin A^complement$, so we must have $a in B.$
answered Feb 3 at 0:10
greeliousgreelious
472112
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add a comment |
$begingroup$
The key here is that $A = (A setminus B) cup (A cap B)$. [If you haven't proved that, or perhaps defined $A setminus B$ through this relation, then you should do that).
But given that, you have that any $a$ is either in $A setminus B$, or $a in A cap B$. Since your prescribed $a$ is not in $A setminus B$, we have that $a in A cap B$, and thus $a in A$ and $a in B$.
answered Feb 3 at 0:07
davidlowryduda♦davidlowryduda
75.2k7121256
$endgroup$
add a comment |
$begingroup$
$A - B = {a : a in A land anotin B}$.
Let's say $a in A$. Then $a notin A - B$ implies directly:
$a$ is s.t $neg(a in A land a notin B)$. Which is equivalent to saying $(a notin A lor a in B)$. Since we have $a in A$ you get $a in B$..
answered Feb 3 at 0:33
MariahMariah
2,1471718
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add a comment |
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3 Answers
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3 Answers
3
active
oldest
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$begingroup$
Using De Morgan's law and the identity $A setminus B = A cap B^complement$, we have
$$begin{align} a notin A setminus B &iff a in (A setminus B)^complement \&iff a in (A cap B^complement)^complement \&iff a in A^complement cup B end{align}$$
So at least one of $a in A^complement$ or $a in B$ must be true. But $a in A implies a notin A^complement$, so we must have $a in B.$
answered Feb 3 at 0:10
greeliousgreelious
472112
$endgroup$
add a comment |
$begingroup$
Using De Morgan's law and the identity $A setminus B = A cap B^complement$, we have
$$begin{align} a notin A setminus B &iff a in (A setminus B)^complement \&iff a in (A cap B^complement)^complement \&iff a in A^complement cup B end{align}$$
So at least one of $a in A^complement$ or $a in B$ must be true. But $a in A implies a notin A^complement$, so we must have $a in B.$
answered Feb 3 at 0:10
greeliousgreelious
472112
$endgroup$
add a comment |
$begingroup$
Using De Morgan's law and the identity $A setminus B = A cap B^complement$, we have
$$begin{align} a notin A setminus B &iff a in (A setminus B)^complement \&iff a in (A cap B^complement)^complement \&iff a in A^complement cup B end{align}$$
So at least one of $a in A^complement$ or $a in B$ must be true. But $a in A implies a notin A^complement$, so we must have $a in B.$
answered Feb 3 at 0:10
greeliousgreelious
472112
$endgroup$
Using De Morgan's law and the identity $A setminus B = A cap B^complement$, we have
$$begin{align} a notin A setminus B &iff a in (A setminus B)^complement \&iff a in (A cap B^complement)^complement \&iff a in A^complement cup B end{align}$$
So at least one of $a in A^complement$ or $a in B$ must be true. But $a in A implies a notin A^complement$, so we must have $a in B.$
answered Feb 3 at 0:10
greeliousgreelious
472112
answered Feb 3 at 0:10
greeliousgreelious
472112
answered Feb 3 at 0:10
greeliousgreelious
472112
answered Feb 3 at 0:10
greeliousgreelious
472112
472112
add a comment |
add a comment |
$begingroup$
The key here is that $A = (A setminus B) cup (A cap B)$. [If you haven't proved that, or perhaps defined $A setminus B$ through this relation, then you should do that).
But given that, you have that any $a$ is either in $A setminus B$, or $a in A cap B$. Since your prescribed $a$ is not in $A setminus B$, we have that $a in A cap B$, and thus $a in A$ and $a in B$.
answered Feb 3 at 0:07
davidlowryduda♦davidlowryduda
75.2k7121256
$endgroup$
add a comment |
$begingroup$
The key here is that $A = (A setminus B) cup (A cap B)$. [If you haven't proved that, or perhaps defined $A setminus B$ through this relation, then you should do that).
But given that, you have that any $a$ is either in $A setminus B$, or $a in A cap B$. Since your prescribed $a$ is not in $A setminus B$, we have that $a in A cap B$, and thus $a in A$ and $a in B$.
answered Feb 3 at 0:07
davidlowryduda♦davidlowryduda
75.2k7121256
$endgroup$
add a comment |
$begingroup$
The key here is that $A = (A setminus B) cup (A cap B)$. [If you haven't proved that, or perhaps defined $A setminus B$ through this relation, then you should do that).
But given that, you have that any $a$ is either in $A setminus B$, or $a in A cap B$. Since your prescribed $a$ is not in $A setminus B$, we have that $a in A cap B$, and thus $a in A$ and $a in B$.
answered Feb 3 at 0:07
davidlowryduda♦davidlowryduda
75.2k7121256
$endgroup$
The key here is that $A = (A setminus B) cup (A cap B)$. [If you haven't proved that, or perhaps defined $A setminus B$ through this relation, then you should do that).
But given that, you have that any $a$ is either in $A setminus B$, or $a in A cap B$. Since your prescribed $a$ is not in $A setminus B$, we have that $a in A cap B$, and thus $a in A$ and $a in B$.
answered Feb 3 at 0:07
davidlowryduda♦davidlowryduda
75.2k7121256
answered Feb 3 at 0:07
davidlowryduda♦davidlowryduda
75.2k7121256
answered Feb 3 at 0:07
davidlowryduda♦davidlowryduda
75.2k7121256
answered Feb 3 at 0:07
davidlowryduda♦davidlowryduda
75.2k7121256
75.2k7121256
add a comment |
add a comment |
$begingroup$
$A - B = {a : a in A land anotin B}$.
Let's say $a in A$. Then $a notin A - B$ implies directly:
$a$ is s.t $neg(a in A land a notin B)$. Which is equivalent to saying $(a notin A lor a in B)$. Since we have $a in A$ you get $a in B$..
answered Feb 3 at 0:33
MariahMariah
2,1471718
$endgroup$
add a comment |
$begingroup$
$A - B = {a : a in A land anotin B}$.
Let's say $a in A$. Then $a notin A - B$ implies directly:
$a$ is s.t $neg(a in A land a notin B)$. Which is equivalent to saying $(a notin A lor a in B)$. Since we have $a in A$ you get $a in B$..
answered Feb 3 at 0:33
MariahMariah
2,1471718
$endgroup$
add a comment |
$begingroup$
$A - B = {a : a in A land anotin B}$.
Let's say $a in A$. Then $a notin A - B$ implies directly:
$a$ is s.t $neg(a in A land a notin B)$. Which is equivalent to saying $(a notin A lor a in B)$. Since we have $a in A$ you get $a in B$..
answered Feb 3 at 0:33
MariahMariah
2,1471718
$endgroup$
$A - B = {a : a in A land anotin B}$.
Let's say $a in A$. Then $a notin A - B$ implies directly:
$a$ is s.t $neg(a in A land a notin B)$. Which is equivalent to saying $(a notin A lor a in B)$. Since we have $a in A$ you get $a in B$..
answered Feb 3 at 0:33
MariahMariah
2,1471718
answered Feb 3 at 0:33
MariahMariah
2,1471718
answered Feb 3 at 0:33
MariahMariah
2,1471718
answered Feb 3 at 0:33
MariahMariah
2,1471718
2,1471718
add a comment |
add a comment |
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$begingroup$
$Asetminus B$={$ain A | a notin B$}. What means $a notin Asetminus B$?
$endgroup$
– babemcnuggets
Feb 3 at 0:05