If $x in A$ why does it not follow that $x in A-B$?
I understand that if $x in A-B$ then $x in A$ because $x notin B$ but why doesn't the reverse hold true?
elementary-set-theory
asked Nov 28 at 23:52
cap
103
add a comment |
I understand that if $x in A-B$ then $x in A$ because $x notin B$ but why doesn't the reverse hold true?
elementary-set-theory
asked Nov 28 at 23:52
cap
103
2
Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
– amWhy
Nov 28 at 23:55
1
Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
– JonathanZ
Nov 29 at 0:02
So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
– cap
Nov 29 at 0:11
add a comment |
I understand that if $x in A-B$ then $x in A$ because $x notin B$ but why doesn't the reverse hold true?
elementary-set-theory
asked Nov 28 at 23:52
cap
103
I understand that if $x in A-B$ then $x in A$ because $x notin B$ but why doesn't the reverse hold true?
elementary-set-theory
elementary-set-theory
asked Nov 28 at 23:52
cap
103
asked Nov 28 at 23:52
cap
103
asked Nov 28 at 23:52
cap
103
asked Nov 28 at 23:52
cap
103
asked Nov 28 at 23:52
cap
103
103
2
Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
– amWhy
Nov 28 at 23:55
1
Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
– JonathanZ
Nov 29 at 0:02
So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
– cap
Nov 29 at 0:11
add a comment |
2
Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
– amWhy
Nov 28 at 23:55
1
Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
– JonathanZ
Nov 29 at 0:02
So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
– cap
Nov 29 at 0:11
2
2
Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
– amWhy
Nov 28 at 23:55
Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
– amWhy
Nov 28 at 23:55
1
1
Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
– JonathanZ
Nov 29 at 0:02
Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
– JonathanZ
Nov 29 at 0:02
So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
– cap
Nov 29 at 0:11
So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
– cap
Nov 29 at 0:11
add a comment |
1 Answer
1
active
oldest
votes
Because $Asetminus B$ only includes elements in $A$ which are not also in $B$. Put differently, $A$ can be partitioned as
$$
A = (Asetminus B )cup (Acap B)
$$
and if $xin Acap B$, then $xin A$ but $xnotin Asetminus B$.
For instance, take $A=B={1}$ for a trivial counterexample. Then $Asetminus B = emptyset$: so $1in A$, but $1notin Asetminus B$.
answered Nov 28 at 23:56
Clement C.
49.4k33785
add a comment |
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Because $Asetminus B$ only includes elements in $A$ which are not also in $B$. Put differently, $A$ can be partitioned as
$$
A = (Asetminus B )cup (Acap B)
$$
and if $xin Acap B$, then $xin A$ but $xnotin Asetminus B$.
For instance, take $A=B={1}$ for a trivial counterexample. Then $Asetminus B = emptyset$: so $1in A$, but $1notin Asetminus B$.
answered Nov 28 at 23:56
Clement C.
49.4k33785
add a comment |
Because $Asetminus B$ only includes elements in $A$ which are not also in $B$. Put differently, $A$ can be partitioned as
$$
A = (Asetminus B )cup (Acap B)
$$
and if $xin Acap B$, then $xin A$ but $xnotin Asetminus B$.
For instance, take $A=B={1}$ for a trivial counterexample. Then $Asetminus B = emptyset$: so $1in A$, but $1notin Asetminus B$.
answered Nov 28 at 23:56
Clement C.
49.4k33785
add a comment |
Because $Asetminus B$ only includes elements in $A$ which are not also in $B$. Put differently, $A$ can be partitioned as
$$
A = (Asetminus B )cup (Acap B)
$$
and if $xin Acap B$, then $xin A$ but $xnotin Asetminus B$.
For instance, take $A=B={1}$ for a trivial counterexample. Then $Asetminus B = emptyset$: so $1in A$, but $1notin Asetminus B$.
answered Nov 28 at 23:56
Clement C.
49.4k33785
Because $Asetminus B$ only includes elements in $A$ which are not also in $B$. Put differently, $A$ can be partitioned as
$$
A = (Asetminus B )cup (Acap B)
$$
and if $xin Acap B$, then $xin A$ but $xnotin Asetminus B$.
For instance, take $A=B={1}$ for a trivial counterexample. Then $Asetminus B = emptyset$: so $1in A$, but $1notin Asetminus B$.
answered Nov 28 at 23:56
Clement C.
49.4k33785
answered Nov 28 at 23:56
Clement C.
49.4k33785
answered Nov 28 at 23:56
Clement C.
49.4k33785
answered Nov 28 at 23:56
Clement C.
49.4k33785
49.4k33785
add a comment |
add a comment |
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2
Suppose $A = {2, 3}, B = {1, 2, 3}.$ Then there is no $x in A$ that is in $A-B$. Similarly, if $A = {1, 2, 3}, B= {1, 2},$ we can't conclude that just because $xin A$ that $x in A-B$,
– amWhy
Nov 28 at 23:55
1
Because $A - B subseteq A$ always, but we can't generally say that $A subseteq A - B$.
– JonathanZ
Nov 29 at 0:02
So if $A$={1,2,3} and $B$={1,2} then there would be $x in A$ that's in $A-B$ namely {3}?
– cap
Nov 29 at 0:11