$ A = (Acap E^c) cup (Acap E)$? The 2019 Stack Overflow Developer Survey Results Are InIs $(X cup Y)cap Z subset X cup(Y cap Z)$?Does $(Acup B)cap(Ccup D)=(Acap C)cup(Bcap D)$?Proving $A cup (B cap C) = (A cup B) cap (A cup C)$.$A=(A cap B) cup(A cap B^mathsfc) $Prove that $ (A cup B) cap C subseteq A cup (B cap C)$Simple set theory: $Acap (Bcup C) subset (A cap B) cup C$Is $Acup B=Acup Bcap A^c$?Proving equalities with sets: $A cap (B cup C) = (A cap B) cup (A cap C)$ and $X setminus (A cap B) = (X setminus A) cup (X setminus B)$Proof verification: prove $Acup (Bcap C) = (Acup B) cap (A cup C)$.How to show equality $A = (Asetminus B) cup (Asetminus C) cup (A cap B cap C)$
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$ A = (Acap E^c) cup (Acap E)$?
The 2019 Stack Overflow Developer Survey Results Are InIs $(X cup Y)cap Z subset X cup(Y cap Z)$?Does $(Acup B)cap(Ccup D)=(Acap C)cup(Bcap D)$?Proving $A cup (B cap C) = (A cup B) cap (A cup C)$.$A=(A cap B) cup(A cap B^mathsfc) $Prove that $ (A cup B) cap C subseteq A cup (B cap C)$Simple set theory: $Acap (Bcup C) subset (A cap B) cup C$Is $Acup B=Acup Bcap A^c$?Proving equalities with sets: $A cap (B cup C) = (A cap B) cup (A cap C)$ and $X setminus (A cap B) = (X setminus A) cup (X setminus B)$Proof verification: prove $Acup (Bcap C) = (Acup B) cap (A cup C)$.How to show equality $A = (Asetminus B) cup (Asetminus C) cup (A cap B cap C)$
$begingroup$
$ A = (Acap E^c) cup (Acap E)$ ?
is that correct for any sets A and E?
Here is my proof:
Let $xin A$ since $Asubset Acap E$ $ xin Acap E cup (Acap E^c)$
now let $xin (Acap E^c) cup (Acap E)$ so either $x in (Acap E^c)$ or $x in cup (Acap E)$ it follows that $ x in A$ in both cases.
elementary-set-theory
asked Mar 23 at 17:21
José MarínJosé Marín
253211
$endgroup$
add a comment |
$begingroup$
$ A = (Acap E^c) cup (Acap E)$ ?
is that correct for any sets A and E?
Here is my proof:
Let $xin A$ since $Asubset Acap E$ $ xin Acap E cup (Acap E^c)$
now let $xin (Acap E^c) cup (Acap E)$ so either $x in (Acap E^c)$ or $x in cup (Acap E)$ it follows that $ x in A$ in both cases.
elementary-set-theory
asked Mar 23 at 17:21
José MarínJosé Marín
253211
$endgroup$
$begingroup$
$A subset A cap E$ <- this is false
$endgroup$
– Benitok
Mar 23 at 17:34
add a comment |
$begingroup$
$ A = (Acap E^c) cup (Acap E)$ ?
is that correct for any sets A and E?
Here is my proof:
Let $xin A$ since $Asubset Acap E$ $ xin Acap E cup (Acap E^c)$
now let $xin (Acap E^c) cup (Acap E)$ so either $x in (Acap E^c)$ or $x in cup (Acap E)$ it follows that $ x in A$ in both cases.
elementary-set-theory
asked Mar 23 at 17:21
José MarínJosé Marín
253211
$endgroup$
$ A = (Acap E^c) cup (Acap E)$ ?
is that correct for any sets A and E?
Here is my proof:
Let $xin A$ since $Asubset Acap E$ $ xin Acap E cup (Acap E^c)$
now let $xin (Acap E^c) cup (Acap E)$ so either $x in (Acap E^c)$ or $x in cup (Acap E)$ it follows that $ x in A$ in both cases.
elementary-set-theory
elementary-set-theory
asked Mar 23 at 17:21
José MarínJosé Marín
253211
asked Mar 23 at 17:21
José MarínJosé Marín
253211
asked Mar 23 at 17:21
José MarínJosé Marín
253211
asked Mar 23 at 17:21
José MarínJosé Marín
253211
asked Mar 23 at 17:21
José MarínJosé Marín
253211
253211
$begingroup$
$A subset A cap E$ <- this is false
$endgroup$
– Benitok
Mar 23 at 17:34
add a comment |
$begingroup$
$A subset A cap E$ <- this is false
$endgroup$
– Benitok
Mar 23 at 17:34
$begingroup$
$A subset A cap E$ <- this is false
$endgroup$
– Benitok
Mar 23 at 17:34
$begingroup$
$A subset A cap E$ <- this is false
$endgroup$
– Benitok
Mar 23 at 17:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is a flaw in your proof (if I understood it well). It is not true that $A subset A cap E$; for example, if $E = 1,2$ and $A = 0,1$, then $A cap E = 1$.
I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A subset X$.
Let $x in A$. If $x notin E$, then $x in E^c$, and hence $x in A cap E^c$. Hence $A subset (A cap E)cup(A cap E^c)$.
To prove the other inclusion, note that both $(A cap E)$ and $(A cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.
If $A notsubset X$ then you have $A cap X = (A cap E)cup(A cap E^c)$, since now $A cap X subset X$.
answered Mar 23 at 17:36
AspiringMathematicianAspiringMathematician
1,878621
$endgroup$
$begingroup$
That's true. Thank you.
$endgroup$
– José Marín
Mar 23 at 18:03
add a comment |
$begingroup$
That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.
$$
beginalign
A &= A cap U hskip1cm text (Identity law - $U$ is the universe)\
&= A cap ( E cup E^c) hskip1cm text (Complement law)\
&= (A cap E) cup (A cap E^c) hskip1cm text (Distributive property)\\
endalign
$$
answered Mar 23 at 17:36
leonbloyleonbloy
42.2k647108
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
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oldest
votes
$begingroup$
There is a flaw in your proof (if I understood it well). It is not true that $A subset A cap E$; for example, if $E = 1,2$ and $A = 0,1$, then $A cap E = 1$.
I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A subset X$.
Let $x in A$. If $x notin E$, then $x in E^c$, and hence $x in A cap E^c$. Hence $A subset (A cap E)cup(A cap E^c)$.
To prove the other inclusion, note that both $(A cap E)$ and $(A cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.
If $A notsubset X$ then you have $A cap X = (A cap E)cup(A cap E^c)$, since now $A cap X subset X$.
answered Mar 23 at 17:36
AspiringMathematicianAspiringMathematician
1,878621
$endgroup$
$begingroup$
That's true. Thank you.
$endgroup$
– José Marín
Mar 23 at 18:03
add a comment |
$begingroup$
There is a flaw in your proof (if I understood it well). It is not true that $A subset A cap E$; for example, if $E = 1,2$ and $A = 0,1$, then $A cap E = 1$.
I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A subset X$.
Let $x in A$. If $x notin E$, then $x in E^c$, and hence $x in A cap E^c$. Hence $A subset (A cap E)cup(A cap E^c)$.
To prove the other inclusion, note that both $(A cap E)$ and $(A cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.
If $A notsubset X$ then you have $A cap X = (A cap E)cup(A cap E^c)$, since now $A cap X subset X$.
answered Mar 23 at 17:36
AspiringMathematicianAspiringMathematician
1,878621
$endgroup$
$begingroup$
That's true. Thank you.
$endgroup$
– José Marín
Mar 23 at 18:03
add a comment |
$begingroup$
There is a flaw in your proof (if I understood it well). It is not true that $A subset A cap E$; for example, if $E = 1,2$ and $A = 0,1$, then $A cap E = 1$.
I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A subset X$.
Let $x in A$. If $x notin E$, then $x in E^c$, and hence $x in A cap E^c$. Hence $A subset (A cap E)cup(A cap E^c)$.
To prove the other inclusion, note that both $(A cap E)$ and $(A cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.
If $A notsubset X$ then you have $A cap X = (A cap E)cup(A cap E^c)$, since now $A cap X subset X$.
answered Mar 23 at 17:36
AspiringMathematicianAspiringMathematician
1,878621
$endgroup$
There is a flaw in your proof (if I understood it well). It is not true that $A subset A cap E$; for example, if $E = 1,2$ and $A = 0,1$, then $A cap E = 1$.
I assume $E^c$ means the complement of $E$ in some other set $X$. That statement is true if $A subset X$.
Let $x in A$. If $x notin E$, then $x in E^c$, and hence $x in A cap E^c$. Hence $A subset (A cap E)cup(A cap E^c)$.
To prove the other inclusion, note that both $(A cap E)$ and $(A cap E^c)$ are subsets of $A$, so their union also is a subset of $A$.
If $A notsubset X$ then you have $A cap X = (A cap E)cup(A cap E^c)$, since now $A cap X subset X$.
answered Mar 23 at 17:36
AspiringMathematicianAspiringMathematician
1,878621
answered Mar 23 at 17:36
AspiringMathematicianAspiringMathematician
1,878621
answered Mar 23 at 17:36
AspiringMathematicianAspiringMathematician
1,878621
answered Mar 23 at 17:36
AspiringMathematicianAspiringMathematician
1,878621
1,878621
$begingroup$
That's true. Thank you.
$endgroup$
– José Marín
Mar 23 at 18:03
add a comment |
$begingroup$
That's true. Thank you.
$endgroup$
– José Marín
Mar 23 at 18:03
$begingroup$
That's true. Thank you.
$endgroup$
– José Marín
Mar 23 at 18:03
$begingroup$
That's true. Thank you.
$endgroup$
– José Marín
Mar 23 at 18:03
add a comment |
$begingroup$
That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.
$$
beginalign
A &= A cap U hskip1cm text (Identity law - $U$ is the universe)\
&= A cap ( E cup E^c) hskip1cm text (Complement law)\
&= (A cap E) cup (A cap E^c) hskip1cm text (Distributive property)\\
endalign
$$
answered Mar 23 at 17:36
leonbloyleonbloy
42.2k647108
$endgroup$
add a comment |
$begingroup$
That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.
$$
beginalign
A &= A cap U hskip1cm text (Identity law - $U$ is the universe)\
&= A cap ( E cup E^c) hskip1cm text (Complement law)\
&= (A cap E) cup (A cap E^c) hskip1cm text (Distributive property)\\
endalign
$$
answered Mar 23 at 17:36
leonbloyleonbloy
42.2k647108
$endgroup$
add a comment |
$begingroup$
That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.
$$
beginalign
A &= A cap U hskip1cm text (Identity law - $U$ is the universe)\
&= A cap ( E cup E^c) hskip1cm text (Complement law)\
&= (A cap E) cup (A cap E^c) hskip1cm text (Distributive property)\\
endalign
$$
answered Mar 23 at 17:36
leonbloyleonbloy
42.2k647108
$endgroup$
That's right. It can also be proved by (assuming as known) the elementary properties of algebra of sets.
$$
beginalign
A &= A cap U hskip1cm text (Identity law - $U$ is the universe)\
&= A cap ( E cup E^c) hskip1cm text (Complement law)\
&= (A cap E) cup (A cap E^c) hskip1cm text (Distributive property)\\
endalign
$$
answered Mar 23 at 17:36
leonbloyleonbloy
42.2k647108
answered Mar 23 at 17:36
leonbloyleonbloy
42.2k647108
answered Mar 23 at 17:36
leonbloyleonbloy
42.2k647108
answered Mar 23 at 17:36
leonbloyleonbloy
42.2k647108
42.2k647108
add a comment |
add a comment |
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$begingroup$
$A subset A cap E$ <- this is false
$endgroup$
– Benitok
Mar 23 at 17:34