Mixing
Rigorously prove If $A setminus B = A$, then $A cap B = emptyset$
$begingroup$
Rigorously prove: If $Asetminus B = A$, then $A cap B = emptyset$.
Logically it makes sense i.e. since if $Asetminus B$ is equal to $A$, then $B$ must be the empty set hence $A cap B$ must be equal to the empty set since anything intersected with the empty set is the empty set.
My main question is how I can go about this to prove it formally, i.e 'mathematically/rigorously'
My attempt:
$left(A setminus B right) subseteq A$ (1)
$A$ is a subset of $A$ hence also a subset of $left(A setminus B right)$
So:
$A subseteq left(A setminus B right)$ (2)
Combining (1) and (2) we are allowed to conclude that:
$ left(A setminus B right) = A$
Hence since $ left(A setminus B right) = A$, then $B = emptyset$ then rtp that $A cap B = emptyset$
$A cap B$
$(x in A land x in B)$
$(x in A land x in emptyset)$
$x in emptyset$
Thanks in Advance
discrete-mathematics elementary-set-theory
edited Feb 2 at 10:34
Henno Brandsma
116k349127
asked Feb 1 at 12:23
xalalauxalalau
154
$endgroup$
add a comment |
$begingroup$
Rigorously prove: If $Asetminus B = A$, then $A cap B = emptyset$.
Logically it makes sense i.e. since if $Asetminus B$ is equal to $A$, then $B$ must be the empty set hence $A cap B$ must be equal to the empty set since anything intersected with the empty set is the empty set.
My main question is how I can go about this to prove it formally, i.e 'mathematically/rigorously'
My attempt:
$left(A setminus B right) subseteq A$ (1)
$A$ is a subset of $A$ hence also a subset of $left(A setminus B right)$
So:
$A subseteq left(A setminus B right)$ (2)
Combining (1) and (2) we are allowed to conclude that:
$ left(A setminus B right) = A$
Hence since $ left(A setminus B right) = A$, then $B = emptyset$ then rtp that $A cap B = emptyset$
$A cap B$
$(x in A land x in B)$
$(x in A land x in emptyset)$
$x in emptyset$
Thanks in Advance
discrete-mathematics elementary-set-theory
edited Feb 2 at 10:34
Henno Brandsma
116k349127
asked Feb 1 at 12:23
xalalauxalalau
154
$endgroup$
$begingroup$
Welcome to Maths SX! You mean $Asmallsetminus B$, not $A/B$, I suppose?
$endgroup$
– Bernard
Feb 1 at 12:29
$begingroup$
Apparently so (according to everyone on this forum) ... however this is how it is written on the textbook I'm using
$endgroup$
– xalalau
Feb 1 at 12:41
$begingroup$
What textbook is that?
$endgroup$
– bof
Feb 1 at 12:42
$begingroup$
Note that $A cap B$ and $A setminus B$ are two disjoint subsets of $A$.
$endgroup$
– Solomonoff's Secret
Feb 1 at 12:44
add a comment |
$begingroup$
Rigorously prove: If $Asetminus B = A$, then $A cap B = emptyset$.
Logically it makes sense i.e. since if $Asetminus B$ is equal to $A$, then $B$ must be the empty set hence $A cap B$ must be equal to the empty set since anything intersected with the empty set is the empty set.
My main question is how I can go about this to prove it formally, i.e 'mathematically/rigorously'
My attempt:
$left(A setminus B right) subseteq A$ (1)
$A$ is a subset of $A$ hence also a subset of $left(A setminus B right)$
So:
$A subseteq left(A setminus B right)$ (2)
Combining (1) and (2) we are allowed to conclude that:
$ left(A setminus B right) = A$
Hence since $ left(A setminus B right) = A$, then $B = emptyset$ then rtp that $A cap B = emptyset$
$A cap B$
$(x in A land x in B)$
$(x in A land x in emptyset)$
$x in emptyset$
Thanks in Advance
discrete-mathematics elementary-set-theory
edited Feb 2 at 10:34
Henno Brandsma
116k349127
asked Feb 1 at 12:23
xalalauxalalau
154
$endgroup$
Rigorously prove: If $Asetminus B = A$, then $A cap B = emptyset$.
Logically it makes sense i.e. since if $Asetminus B$ is equal to $A$, then $B$ must be the empty set hence $A cap B$ must be equal to the empty set since anything intersected with the empty set is the empty set.
My main question is how I can go about this to prove it formally, i.e 'mathematically/rigorously'
My attempt:
$left(A setminus B right) subseteq A$ (1)
$A$ is a subset of $A$ hence also a subset of $left(A setminus B right)$
So:
$A subseteq left(A setminus B right)$ (2)
Combining (1) and (2) we are allowed to conclude that:
$ left(A setminus B right) = A$
Hence since $ left(A setminus B right) = A$, then $B = emptyset$ then rtp that $A cap B = emptyset$
$A cap B$
$(x in A land x in B)$
$(x in A land x in emptyset)$
$x in emptyset$
Thanks in Advance
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
edited Feb 2 at 10:34
Henno Brandsma
116k349127
asked Feb 1 at 12:23
xalalauxalalau
154
edited Feb 2 at 10:34
Henno Brandsma
116k349127
asked Feb 1 at 12:23
xalalauxalalau
154
edited Feb 2 at 10:34
Henno Brandsma
116k349127
edited Feb 2 at 10:34
Henno Brandsma
116k349127
edited Feb 2 at 10:34
Henno Brandsma
116k349127
116k349127
asked Feb 1 at 12:23
xalalauxalalau
154
asked Feb 1 at 12:23
xalalauxalalau
154
asked Feb 1 at 12:23
xalalauxalalau
154
154
$begingroup$
Welcome to Maths SX! You mean $Asmallsetminus B$, not $A/B$, I suppose?
$endgroup$
– Bernard
Feb 1 at 12:29
$begingroup$
Apparently so (according to everyone on this forum) ... however this is how it is written on the textbook I'm using
$endgroup$
– xalalau
Feb 1 at 12:41
$begingroup$
What textbook is that?
$endgroup$
– bof
Feb 1 at 12:42
$begingroup$
Note that $A cap B$ and $A setminus B$ are two disjoint subsets of $A$.
$endgroup$
– Solomonoff's Secret
Feb 1 at 12:44
add a comment |
$begingroup$
Welcome to Maths SX! You mean $Asmallsetminus B$, not $A/B$, I suppose?
$endgroup$
– Bernard
Feb 1 at 12:29
$begingroup$
Apparently so (according to everyone on this forum) ... however this is how it is written on the textbook I'm using
$endgroup$
– xalalau
Feb 1 at 12:41
$begingroup$
What textbook is that?
$endgroup$
– bof
Feb 1 at 12:42
$begingroup$
Note that $A cap B$ and $A setminus B$ are two disjoint subsets of $A$.
$endgroup$
– Solomonoff's Secret
Feb 1 at 12:44
$begingroup$
Welcome to Maths SX! You mean $Asmallsetminus B$, not $A/B$, I suppose?
$endgroup$
– Bernard
Feb 1 at 12:29
$begingroup$
Welcome to Maths SX! You mean $Asmallsetminus B$, not $A/B$, I suppose?
$endgroup$
– Bernard
Feb 1 at 12:29
$begingroup$
Apparently so (according to everyone on this forum) ... however this is how it is written on the textbook I'm using
$endgroup$
– xalalau
Feb 1 at 12:41
$begingroup$
Apparently so (according to everyone on this forum) ... however this is how it is written on the textbook I'm using
$endgroup$
– xalalau
Feb 1 at 12:41
$begingroup$
What textbook is that?
$endgroup$
– bof
Feb 1 at 12:42
$begingroup$
What textbook is that?
$endgroup$
– bof
Feb 1 at 12:42
$begingroup$
Note that $A cap B$ and $A setminus B$ are two disjoint subsets of $A$.
$endgroup$
– Solomonoff's Secret
Feb 1 at 12:44
$begingroup$
Note that $A cap B$ and $A setminus B$ are two disjoint subsets of $A$.
$endgroup$
– Solomonoff's Secret
Feb 1 at 12:44
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
" Hence since $Asetminus B = A$, then $B = varnothing$ then rtp that $Acap B=varnothing$"
Nowhere we reach the conclusion that $B=varnothing$.
This is a way to do it:
Assume that $xin Acap B$.
Then $xin A=Asetminus B$ and $xin B$.
Then $xnotin B$ and $xin B$ so a contradiction is found.
We conclude that the assumption $xin Acap B$ is false, and this for every $x$.
That means exactly that $Acap B$ is empty.
answered Feb 1 at 12:34
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Note that it is always true that
$$A setminus B = A cap B^C$$
Hence, if $A setminus B =A$, then $A cap B^C =A$, and so:
$$Acap B= (A cap B^C) cap B= A cap (B^C cap B)=A cap emptyset=emptyset$$
answered Feb 1 at 12:54
Bram28Bram28
64.4k44793
$endgroup$
add a comment |
$begingroup$
A correct proof is sketched in the other answer. To answer your question about whether your reasoning is correct: Not really. You're trying to show that if $Asetminus B=A$ then $cap B=emptyset$. So $Asetminus B=A$ is given. Hence a line saying "we may conclude $Asetminus B=A$ makes very little sense.
Not that it matters, since you're given that $Asetminus B=A$, but your proof of this fact is also wrong. You state that $A$ is a subset of $Asetminus B$, but that's not true in general.
answered Feb 1 at 12:56
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
add a comment |
$begingroup$
Suppose $xin Acap B$. Then $xin A$ and $xin B$. But then $xnotin Asetminus B$. Since $Asetminus B=A$, this leads to a contradiction. Done.
answered Feb 1 at 13:02
WuestenfuxWuestenfux
5,5131513
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
" Hence since $Asetminus B = A$, then $B = varnothing$ then rtp that $Acap B=varnothing$"
Nowhere we reach the conclusion that $B=varnothing$.
This is a way to do it:
Assume that $xin Acap B$.
Then $xin A=Asetminus B$ and $xin B$.
Then $xnotin B$ and $xin B$ so a contradiction is found.
We conclude that the assumption $xin Acap B$ is false, and this for every $x$.
That means exactly that $Acap B$ is empty.
answered Feb 1 at 12:34
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
" Hence since $Asetminus B = A$, then $B = varnothing$ then rtp that $Acap B=varnothing$"
Nowhere we reach the conclusion that $B=varnothing$.
This is a way to do it:
Assume that $xin Acap B$.
Then $xin A=Asetminus B$ and $xin B$.
Then $xnotin B$ and $xin B$ so a contradiction is found.
We conclude that the assumption $xin Acap B$ is false, and this for every $x$.
That means exactly that $Acap B$ is empty.
answered Feb 1 at 12:34
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
" Hence since $Asetminus B = A$, then $B = varnothing$ then rtp that $Acap B=varnothing$"
Nowhere we reach the conclusion that $B=varnothing$.
This is a way to do it:
Assume that $xin Acap B$.
Then $xin A=Asetminus B$ and $xin B$.
Then $xnotin B$ and $xin B$ so a contradiction is found.
We conclude that the assumption $xin Acap B$ is false, and this for every $x$.
That means exactly that $Acap B$ is empty.
answered Feb 1 at 12:34
drhabdrhab
104k545136
$endgroup$
" Hence since $Asetminus B = A$, then $B = varnothing$ then rtp that $Acap B=varnothing$"
Nowhere we reach the conclusion that $B=varnothing$.
This is a way to do it:
Assume that $xin Acap B$.
Then $xin A=Asetminus B$ and $xin B$.
Then $xnotin B$ and $xin B$ so a contradiction is found.
We conclude that the assumption $xin Acap B$ is false, and this for every $x$.
That means exactly that $Acap B$ is empty.
answered Feb 1 at 12:34
drhabdrhab
104k545136
answered Feb 1 at 12:34
drhabdrhab
104k545136
answered Feb 1 at 12:34
drhabdrhab
104k545136
answered Feb 1 at 12:34
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
$begingroup$
Note that it is always true that
$$A setminus B = A cap B^C$$
Hence, if $A setminus B =A$, then $A cap B^C =A$, and so:
$$Acap B= (A cap B^C) cap B= A cap (B^C cap B)=A cap emptyset=emptyset$$
answered Feb 1 at 12:54
Bram28Bram28
64.4k44793
$endgroup$
add a comment |
$begingroup$
Note that it is always true that
$$A setminus B = A cap B^C$$
Hence, if $A setminus B =A$, then $A cap B^C =A$, and so:
$$Acap B= (A cap B^C) cap B= A cap (B^C cap B)=A cap emptyset=emptyset$$
answered Feb 1 at 12:54
Bram28Bram28
64.4k44793
$endgroup$
add a comment |
$begingroup$
Note that it is always true that
$$A setminus B = A cap B^C$$
Hence, if $A setminus B =A$, then $A cap B^C =A$, and so:
$$Acap B= (A cap B^C) cap B= A cap (B^C cap B)=A cap emptyset=emptyset$$
answered Feb 1 at 12:54
Bram28Bram28
64.4k44793
$endgroup$
Note that it is always true that
$$A setminus B = A cap B^C$$
Hence, if $A setminus B =A$, then $A cap B^C =A$, and so:
$$Acap B= (A cap B^C) cap B= A cap (B^C cap B)=A cap emptyset=emptyset$$
answered Feb 1 at 12:54
Bram28Bram28
64.4k44793
answered Feb 1 at 12:54
Bram28Bram28
64.4k44793
answered Feb 1 at 12:54
Bram28Bram28
64.4k44793
answered Feb 1 at 12:54
Bram28Bram28
64.4k44793
64.4k44793
add a comment |
add a comment |
$begingroup$
A correct proof is sketched in the other answer. To answer your question about whether your reasoning is correct: Not really. You're trying to show that if $Asetminus B=A$ then $cap B=emptyset$. So $Asetminus B=A$ is given. Hence a line saying "we may conclude $Asetminus B=A$ makes very little sense.
Not that it matters, since you're given that $Asetminus B=A$, but your proof of this fact is also wrong. You state that $A$ is a subset of $Asetminus B$, but that's not true in general.
answered Feb 1 at 12:56
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
add a comment |
$begingroup$
A correct proof is sketched in the other answer. To answer your question about whether your reasoning is correct: Not really. You're trying to show that if $Asetminus B=A$ then $cap B=emptyset$. So $Asetminus B=A$ is given. Hence a line saying "we may conclude $Asetminus B=A$ makes very little sense.
Not that it matters, since you're given that $Asetminus B=A$, but your proof of this fact is also wrong. You state that $A$ is a subset of $Asetminus B$, but that's not true in general.
answered Feb 1 at 12:56
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
add a comment |
$begingroup$
A correct proof is sketched in the other answer. To answer your question about whether your reasoning is correct: Not really. You're trying to show that if $Asetminus B=A$ then $cap B=emptyset$. So $Asetminus B=A$ is given. Hence a line saying "we may conclude $Asetminus B=A$ makes very little sense.
Not that it matters, since you're given that $Asetminus B=A$, but your proof of this fact is also wrong. You state that $A$ is a subset of $Asetminus B$, but that's not true in general.
answered Feb 1 at 12:56
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
A correct proof is sketched in the other answer. To answer your question about whether your reasoning is correct: Not really. You're trying to show that if $Asetminus B=A$ then $cap B=emptyset$. So $Asetminus B=A$ is given. Hence a line saying "we may conclude $Asetminus B=A$ makes very little sense.
Not that it matters, since you're given that $Asetminus B=A$, but your proof of this fact is also wrong. You state that $A$ is a subset of $Asetminus B$, but that's not true in general.
answered Feb 1 at 12:56
David C. UllrichDavid C. Ullrich
61.7k44095
answered Feb 1 at 12:56
David C. UllrichDavid C. Ullrich
61.7k44095
answered Feb 1 at 12:56
David C. UllrichDavid C. Ullrich
61.7k44095
answered Feb 1 at 12:56
David C. UllrichDavid C. Ullrich
61.7k44095
61.7k44095
add a comment |
add a comment |
$begingroup$
Suppose $xin Acap B$. Then $xin A$ and $xin B$. But then $xnotin Asetminus B$. Since $Asetminus B=A$, this leads to a contradiction. Done.
answered Feb 1 at 13:02
WuestenfuxWuestenfux
5,5131513
$endgroup$
add a comment |
$begingroup$
Suppose $xin Acap B$. Then $xin A$ and $xin B$. But then $xnotin Asetminus B$. Since $Asetminus B=A$, this leads to a contradiction. Done.
answered Feb 1 at 13:02
WuestenfuxWuestenfux
5,5131513
$endgroup$
add a comment |
$begingroup$
Suppose $xin Acap B$. Then $xin A$ and $xin B$. But then $xnotin Asetminus B$. Since $Asetminus B=A$, this leads to a contradiction. Done.
answered Feb 1 at 13:02
WuestenfuxWuestenfux
5,5131513
$endgroup$
Suppose $xin Acap B$. Then $xin A$ and $xin B$. But then $xnotin Asetminus B$. Since $Asetminus B=A$, this leads to a contradiction. Done.
answered Feb 1 at 13:02
WuestenfuxWuestenfux
5,5131513
answered Feb 1 at 13:02
WuestenfuxWuestenfux
5,5131513
answered Feb 1 at 13:02
WuestenfuxWuestenfux
5,5131513
answered Feb 1 at 13:02
WuestenfuxWuestenfux
5,5131513
5,5131513
add a comment |
add a comment |
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$begingroup$
Welcome to Maths SX! You mean $Asmallsetminus B$, not $A/B$, I suppose?
$endgroup$
– Bernard
Feb 1 at 12:29
$begingroup$
Apparently so (according to everyone on this forum) ... however this is how it is written on the textbook I'm using
$endgroup$
– xalalau
Feb 1 at 12:41
$begingroup$
What textbook is that?
$endgroup$
– bof
Feb 1 at 12:42
$begingroup$
Note that $A cap B$ and $A setminus B$ are two disjoint subsets of $A$.
$endgroup$
– Solomonoff's Secret
Feb 1 at 12:44