How to deduce two sets are equal when one being subset of another is proven?
Clash Royale CLAN TAG#URR8PPP
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Below is a proof that $(A cap B ) cup (A setminus B) = A$:
Let's consider $x in (A cap B ) cup (A setminus B)$.
- $ x ∈ (A∩B)∨x ∈ (A setminus B) $
- $ x ∈ A∧x ∈ B)∨(x ∈ A∧ x notin B) $
- $ x ∈ A∧(x ∈ B∨x notin B) $
Which means $ x ∈ A $ and thus $ (A∩B)∪(A setminus B) = A. $
The proof evidently shows that $ (A∩B)∪(A setminus B) ⊂ A $ but why does it show the two are equal?
elementary-set-theory proof-explanation
edited Sep 5 at 10:37
Mauro ALLEGRANZA
61.4k446105
asked Sep 5 at 10:27
jasno1
1
add a comment |Â
up vote
0
down vote
favorite
Below is a proof that $(A cap B ) cup (A setminus B) = A$:
Let's consider $x in (A cap B ) cup (A setminus B)$.
- $ x ∈ (A∩B)∨x ∈ (A setminus B) $
- $ x ∈ A∧x ∈ B)∨(x ∈ A∧ x notin B) $
- $ x ∈ A∧(x ∈ B∨x notin B) $
Which means $ x ∈ A $ and thus $ (A∩B)∪(A setminus B) = A. $
The proof evidently shows that $ (A∩B)∪(A setminus B) ⊂ A $ but why does it show the two are equal?
elementary-set-theory proof-explanation
edited Sep 5 at 10:37
Mauro ALLEGRANZA
61.4k446105
asked Sep 5 at 10:27
jasno1
1
Two ways : either 1) each step is an iff ($leftrightarrow$), or 2) show the reverse inclusion : $A subset ldots$.
– Mauro ALLEGRANZA
Sep 5 at 10:36
@MauroALLEGRANZA Doesn't it matter that A is a single group that is present in the union on the left side? If I was to prove this, I would show that $ A⊂ .... $ but this proof doesn't have that.
– jasno1
Sep 5 at 10:40
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Below is a proof that $(A cap B ) cup (A setminus B) = A$:
Let's consider $x in (A cap B ) cup (A setminus B)$.
- $ x ∈ (A∩B)∨x ∈ (A setminus B) $
- $ x ∈ A∧x ∈ B)∨(x ∈ A∧ x notin B) $
- $ x ∈ A∧(x ∈ B∨x notin B) $
Which means $ x ∈ A $ and thus $ (A∩B)∪(A setminus B) = A. $
The proof evidently shows that $ (A∩B)∪(A setminus B) ⊂ A $ but why does it show the two are equal?
elementary-set-theory proof-explanation
edited Sep 5 at 10:37
Mauro ALLEGRANZA
61.4k446105
asked Sep 5 at 10:27
jasno1
1
Below is a proof that $(A cap B ) cup (A setminus B) = A$:
Let's consider $x in (A cap B ) cup (A setminus B)$.
- $ x ∈ (A∩B)∨x ∈ (A setminus B) $
- $ x ∈ A∧x ∈ B)∨(x ∈ A∧ x notin B) $
- $ x ∈ A∧(x ∈ B∨x notin B) $
Which means $ x ∈ A $ and thus $ (A∩B)∪(A setminus B) = A. $
The proof evidently shows that $ (A∩B)∪(A setminus B) ⊂ A $ but why does it show the two are equal?
elementary-set-theory proof-explanation
elementary-set-theory proof-explanation
edited Sep 5 at 10:37
Mauro ALLEGRANZA
61.4k446105
asked Sep 5 at 10:27
jasno1
1
edited Sep 5 at 10:37
Mauro ALLEGRANZA
61.4k446105
asked Sep 5 at 10:27
jasno1
1
edited Sep 5 at 10:37
Mauro ALLEGRANZA
61.4k446105
edited Sep 5 at 10:37
Mauro ALLEGRANZA
61.4k446105
edited Sep 5 at 10:37
Mauro ALLEGRANZA
61.4k446105
61.4k446105
asked Sep 5 at 10:27
jasno1
1
asked Sep 5 at 10:27
jasno1
1
asked Sep 5 at 10:27
jasno1
1
1
Two ways : either 1) each step is an iff ($leftrightarrow$), or 2) show the reverse inclusion : $A subset ldots$.
– Mauro ALLEGRANZA
Sep 5 at 10:36
@MauroALLEGRANZA Doesn't it matter that A is a single group that is present in the union on the left side? If I was to prove this, I would show that $ A⊂ .... $ but this proof doesn't have that.
– jasno1
Sep 5 at 10:40
add a comment |Â
Two ways : either 1) each step is an iff ($leftrightarrow$), or 2) show the reverse inclusion : $A subset ldots$.
– Mauro ALLEGRANZA
Sep 5 at 10:36
@MauroALLEGRANZA Doesn't it matter that A is a single group that is present in the union on the left side? If I was to prove this, I would show that $ A⊂ .... $ but this proof doesn't have that.
– jasno1
Sep 5 at 10:40
Two ways : either 1) each step is an iff ($leftrightarrow$), or 2) show the reverse inclusion : $A subset ldots$.
– Mauro ALLEGRANZA
Sep 5 at 10:36
Two ways : either 1) each step is an iff ($leftrightarrow$), or 2) show the reverse inclusion : $A subset ldots$.
– Mauro ALLEGRANZA
Sep 5 at 10:36
@MauroALLEGRANZA Doesn't it matter that A is a single group that is present in the union on the left side? If I was to prove this, I would show that $ A⊂ .... $ but this proof doesn't have that.
– jasno1
Sep 5 at 10:40
@MauroALLEGRANZA Doesn't it matter that A is a single group that is present in the union on the left side? If I was to prove this, I would show that $ A⊂ .... $ but this proof doesn't have that.
– jasno1
Sep 5 at 10:40
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
You only need to rewrite the $setminus$ as complement and use a distributive law:
$
(A cap B ) cup (A setminus B)
= (A cap B ) cup (A cap B^C)
= A cap (B cup B^C)
= A
$
answered Sep 5 at 11:01
pH 74
93
Using the notation $B^mathrm C$ requires a universe $U$ so that $B^mathrm C= Usetminus B$. The distributive law does not need this change in notation though: $Asetminus B= A cap (Asetminus B)$ achieves the same.
– Christoph
Sep 5 at 11:20
add a comment |Â
up vote
0
down vote
The proof should look more like this:
beginalign*
& x in (Acap B)cup (Asetminus B) \
Longleftrightarrowquad& left( xin (Acap B)right) vee left(xin (Asetminus B)right) \
Longleftrightarrowquad& left( xin A wedge xin Bright) vee left(xin A wedge X notin Bright) \
Longleftrightarrowquad& xin A wedge left(xin B vee X notin Bright) \
Longleftrightarrowquad& xin A wedge mathrmtrue \
Longleftrightarrowquad& xin A.
endalign*
Since each step is an equivalence, you can read it top to bottom to get $subset$ and bottom to top to get $supset$.
answered Sep 5 at 11:19
Christoph
11k1240
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You only need to rewrite the $setminus$ as complement and use a distributive law:
$
(A cap B ) cup (A setminus B)
= (A cap B ) cup (A cap B^C)
= A cap (B cup B^C)
= A
$
answered Sep 5 at 11:01
pH 74
93
Using the notation $B^mathrm C$ requires a universe $U$ so that $B^mathrm C= Usetminus B$. The distributive law does not need this change in notation though: $Asetminus B= A cap (Asetminus B)$ achieves the same.
– Christoph
Sep 5 at 11:20
add a comment |Â
up vote
0
down vote
You only need to rewrite the $setminus$ as complement and use a distributive law:
$
(A cap B ) cup (A setminus B)
= (A cap B ) cup (A cap B^C)
= A cap (B cup B^C)
= A
$
answered Sep 5 at 11:01
pH 74
93
Using the notation $B^mathrm C$ requires a universe $U$ so that $B^mathrm C= Usetminus B$. The distributive law does not need this change in notation though: $Asetminus B= A cap (Asetminus B)$ achieves the same.
– Christoph
Sep 5 at 11:20
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You only need to rewrite the $setminus$ as complement and use a distributive law:
$
(A cap B ) cup (A setminus B)
= (A cap B ) cup (A cap B^C)
= A cap (B cup B^C)
= A
$
answered Sep 5 at 11:01
pH 74
93
You only need to rewrite the $setminus$ as complement and use a distributive law:
$
(A cap B ) cup (A setminus B)
= (A cap B ) cup (A cap B^C)
= A cap (B cup B^C)
= A
$
answered Sep 5 at 11:01
pH 74
93
answered Sep 5 at 11:01
pH 74
93
answered Sep 5 at 11:01
pH 74
93
answered Sep 5 at 11:01
pH 74
93
93
Using the notation $B^mathrm C$ requires a universe $U$ so that $B^mathrm C= Usetminus B$. The distributive law does not need this change in notation though: $Asetminus B= A cap (Asetminus B)$ achieves the same.
– Christoph
Sep 5 at 11:20
add a comment |Â
Using the notation $B^mathrm C$ requires a universe $U$ so that $B^mathrm C= Usetminus B$. The distributive law does not need this change in notation though: $Asetminus B= A cap (Asetminus B)$ achieves the same.
– Christoph
Sep 5 at 11:20
Using the notation $B^mathrm C$ requires a universe $U$ so that $B^mathrm C= Usetminus B$. The distributive law does not need this change in notation though: $Asetminus B= A cap (Asetminus B)$ achieves the same.
– Christoph
Sep 5 at 11:20
Using the notation $B^mathrm C$ requires a universe $U$ so that $B^mathrm C= Usetminus B$. The distributive law does not need this change in notation though: $Asetminus B= A cap (Asetminus B)$ achieves the same.
– Christoph
Sep 5 at 11:20
add a comment |Â
up vote
0
down vote
The proof should look more like this:
beginalign*
& x in (Acap B)cup (Asetminus B) \
Longleftrightarrowquad& left( xin (Acap B)right) vee left(xin (Asetminus B)right) \
Longleftrightarrowquad& left( xin A wedge xin Bright) vee left(xin A wedge X notin Bright) \
Longleftrightarrowquad& xin A wedge left(xin B vee X notin Bright) \
Longleftrightarrowquad& xin A wedge mathrmtrue \
Longleftrightarrowquad& xin A.
endalign*
Since each step is an equivalence, you can read it top to bottom to get $subset$ and bottom to top to get $supset$.
answered Sep 5 at 11:19
Christoph
11k1240
add a comment |Â
up vote
0
down vote
The proof should look more like this:
beginalign*
& x in (Acap B)cup (Asetminus B) \
Longleftrightarrowquad& left( xin (Acap B)right) vee left(xin (Asetminus B)right) \
Longleftrightarrowquad& left( xin A wedge xin Bright) vee left(xin A wedge X notin Bright) \
Longleftrightarrowquad& xin A wedge left(xin B vee X notin Bright) \
Longleftrightarrowquad& xin A wedge mathrmtrue \
Longleftrightarrowquad& xin A.
endalign*
Since each step is an equivalence, you can read it top to bottom to get $subset$ and bottom to top to get $supset$.
answered Sep 5 at 11:19
Christoph
11k1240
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The proof should look more like this:
beginalign*
& x in (Acap B)cup (Asetminus B) \
Longleftrightarrowquad& left( xin (Acap B)right) vee left(xin (Asetminus B)right) \
Longleftrightarrowquad& left( xin A wedge xin Bright) vee left(xin A wedge X notin Bright) \
Longleftrightarrowquad& xin A wedge left(xin B vee X notin Bright) \
Longleftrightarrowquad& xin A wedge mathrmtrue \
Longleftrightarrowquad& xin A.
endalign*
Since each step is an equivalence, you can read it top to bottom to get $subset$ and bottom to top to get $supset$.
answered Sep 5 at 11:19
Christoph
11k1240
The proof should look more like this:
beginalign*
& x in (Acap B)cup (Asetminus B) \
Longleftrightarrowquad& left( xin (Acap B)right) vee left(xin (Asetminus B)right) \
Longleftrightarrowquad& left( xin A wedge xin Bright) vee left(xin A wedge X notin Bright) \
Longleftrightarrowquad& xin A wedge left(xin B vee X notin Bright) \
Longleftrightarrowquad& xin A wedge mathrmtrue \
Longleftrightarrowquad& xin A.
endalign*
Since each step is an equivalence, you can read it top to bottom to get $subset$ and bottom to top to get $supset$.
answered Sep 5 at 11:19
Christoph
11k1240
answered Sep 5 at 11:19
Christoph
11k1240
answered Sep 5 at 11:19
Christoph
11k1240
answered Sep 5 at 11:19
Christoph
11k1240
11k1240
add a comment |Â
add a comment |Â
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Two ways : either 1) each step is an iff ($leftrightarrow$), or 2) show the reverse inclusion : $A subset ldots$.
– Mauro ALLEGRANZA
Sep 5 at 10:36
@MauroALLEGRANZA Doesn't it matter that A is a single group that is present in the union on the left side? If I was to prove this, I would show that $ A⊂ .... $ but this proof doesn't have that.
– jasno1
Sep 5 at 10:40