Mixing
How do I prove set inequality: $(A-B)-C=(A-C)-(B-C)$
$begingroup$
We are letting A,B, and C be non-empty sets. My issue with this is that I do not know how to prove it formally. Intuitively I just say let $xin (A-B)-C$, so it is clear to me that $x$ is in $A$ but not in $B$ and then $x$ is in $A$ but not in $B$ and not in $C$. So I say $xin(A-C)$ and $xnotin(B-C)$ which makes me jump to the conclusion $xin(A-C)-(B-C)$ thus $(A-B)-Csubseteq(A-C)-(B-C)$.
The other way around is even more weird, I assume $xin(A-C)-(B-C)$ AND intuitively I know that if I break it down I get that $xin A,xnotin B,xnotin C$ but how do I get to this result I just end up writing the explanation. Is that perfectly fine? From there I say clearly $xnotin C$ and $xin(A-B)$ so $xin (A-B)-C$ and I have that they are both subsets of eachother hence equal.
elementary-set-theory
asked Jan 23 at 22:18
Albert DiazAlbert Diaz
1157
$endgroup$
add a comment |
$begingroup$
We are letting A,B, and C be non-empty sets. My issue with this is that I do not know how to prove it formally. Intuitively I just say let $xin (A-B)-C$, so it is clear to me that $x$ is in $A$ but not in $B$ and then $x$ is in $A$ but not in $B$ and not in $C$. So I say $xin(A-C)$ and $xnotin(B-C)$ which makes me jump to the conclusion $xin(A-C)-(B-C)$ thus $(A-B)-Csubseteq(A-C)-(B-C)$.
The other way around is even more weird, I assume $xin(A-C)-(B-C)$ AND intuitively I know that if I break it down I get that $xin A,xnotin B,xnotin C$ but how do I get to this result I just end up writing the explanation. Is that perfectly fine? From there I say clearly $xnotin C$ and $xin(A-B)$ so $xin (A-B)-C$ and I have that they are both subsets of eachother hence equal.
elementary-set-theory
asked Jan 23 at 22:18
Albert DiazAlbert Diaz
1157
$endgroup$
$begingroup$
You can use a venn diagram with 3 circles.
$endgroup$
– Math Lover
Jan 23 at 23:46
add a comment |
$begingroup$
We are letting A,B, and C be non-empty sets. My issue with this is that I do not know how to prove it formally. Intuitively I just say let $xin (A-B)-C$, so it is clear to me that $x$ is in $A$ but not in $B$ and then $x$ is in $A$ but not in $B$ and not in $C$. So I say $xin(A-C)$ and $xnotin(B-C)$ which makes me jump to the conclusion $xin(A-C)-(B-C)$ thus $(A-B)-Csubseteq(A-C)-(B-C)$.
The other way around is even more weird, I assume $xin(A-C)-(B-C)$ AND intuitively I know that if I break it down I get that $xin A,xnotin B,xnotin C$ but how do I get to this result I just end up writing the explanation. Is that perfectly fine? From there I say clearly $xnotin C$ and $xin(A-B)$ so $xin (A-B)-C$ and I have that they are both subsets of eachother hence equal.
elementary-set-theory
asked Jan 23 at 22:18
Albert DiazAlbert Diaz
1157
$endgroup$
We are letting A,B, and C be non-empty sets. My issue with this is that I do not know how to prove it formally. Intuitively I just say let $xin (A-B)-C$, so it is clear to me that $x$ is in $A$ but not in $B$ and then $x$ is in $A$ but not in $B$ and not in $C$. So I say $xin(A-C)$ and $xnotin(B-C)$ which makes me jump to the conclusion $xin(A-C)-(B-C)$ thus $(A-B)-Csubseteq(A-C)-(B-C)$.
The other way around is even more weird, I assume $xin(A-C)-(B-C)$ AND intuitively I know that if I break it down I get that $xin A,xnotin B,xnotin C$ but how do I get to this result I just end up writing the explanation. Is that perfectly fine? From there I say clearly $xnotin C$ and $xin(A-B)$ so $xin (A-B)-C$ and I have that they are both subsets of eachother hence equal.
elementary-set-theory
elementary-set-theory
asked Jan 23 at 22:18
Albert DiazAlbert Diaz
1157
asked Jan 23 at 22:18
Albert DiazAlbert Diaz
1157
asked Jan 23 at 22:18
Albert DiazAlbert Diaz
1157
asked Jan 23 at 22:18
Albert DiazAlbert Diaz
1157
asked Jan 23 at 22:18
Albert DiazAlbert Diaz
1157
1157
$begingroup$
You can use a venn diagram with 3 circles.
$endgroup$
– Math Lover
Jan 23 at 23:46
add a comment |
$begingroup$
You can use a venn diagram with 3 circles.
$endgroup$
– Math Lover
Jan 23 at 23:46
$begingroup$
You can use a venn diagram with 3 circles.
$endgroup$
– Math Lover
Jan 23 at 23:46
$begingroup$
You can use a venn diagram with 3 circles.
$endgroup$
– Math Lover
Jan 23 at 23:46
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Turn $A-B$ into $A capbar{B}$ and use De Morgan's rules.
answered Jan 23 at 22:26
lightxbulblightxbulb
1,125311
$endgroup$
add a comment |
$begingroup$
Element chase:
If $x in (A-B)-C$ then $x in A$ and $x not in C$ so $xin (A-C$.
And if $x in (A-B) -C$ then $x not in B$ so $x not in B-C subset B$. So $x in (A-C) -(B-C)$.
So $(A-B)-C subset (A-C) - (B-C)$.
Likewise if $y in (A-C)-(B-C)$ then $x in A$. $x not in B-C$ so if $x in B$ then $x in C$. But $x in A-C$ so $x not in C$. So $x not in B$. So $x in A-B$. And $x not in C$ so $x in (A-B) -C$.
So $(A-C) -(B-C) subset (A-B) -C$.
So $(A-B) - C = (A-C) -(B-C)$.
... Another way is to consider which elements are in which sets based on whether they are or are not in $A,B$ or $C$:
Take an element $x in (A-B) -C$.
Is $x in A$? Yes. Is $xin B$? No. Is $x in C$? No.
Take an element $y in (A-C) - (B-C)$.
Is $x in A$? Yes. Is $x in B$? Hmmm. Not if it's not in $C$ But it could be that it is in $B$ and $C$. Is $x in C$. No.... Oh, wait, if it's not $C$ then it can't be in both $B$ and $C$ and since it can't be in $B-C$, it can't be in $B$ after all.
So in both cases $(A-B)-C$ and $(A-C)-(B-C)$ are both precisely the elements in $A$ that aren't in $B$ or $C$.
...
And if we want to do it formally:
$(A-B) - C = (Acap B^c) cap C^c = Acap B^c cap C^c$
whereas
$(A-C) - (B-C) = (Acap C^c)cap (Bcap C^c)^c=$
$(Acap C^c) cap (B^c cup C)= [(Acap C^c) cap B^c] cup [(Acap C^c) cap C] = $
$(Acap B^c cap C^c) cup [Acap (C^c cap C)]=$
$(Acap B^c cap C^c) cup [Acap emptyset]=$
$(Acap B^c cap C^c) cup emptyset=$
$Acap B^c cap C^c$ also.
answered Jan 23 at 22:55
fleabloodfleablood
72.3k22687
$endgroup$
add a comment |
$begingroup$
$(A-C)-(B-C)=(Acapbar C)capoverline{(Bcapbar C)}=(Acapbar C)cap(bar Bcup C)=underbrace{(Acapbar Ccapbar B)}_{(A-B)-C}cupunderbrace{(Acapunderbrace{bar Ccap C}_{varnothing})}_{varnothing}$
answered Jan 23 at 22:56
zwimzwim
12.5k831
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Turn $A-B$ into $A capbar{B}$ and use De Morgan's rules.
answered Jan 23 at 22:26
lightxbulblightxbulb
1,125311
$endgroup$
add a comment |
$begingroup$
Turn $A-B$ into $A capbar{B}$ and use De Morgan's rules.
answered Jan 23 at 22:26
lightxbulblightxbulb
1,125311
$endgroup$
add a comment |
$begingroup$
Turn $A-B$ into $A capbar{B}$ and use De Morgan's rules.
answered Jan 23 at 22:26
lightxbulblightxbulb
1,125311
$endgroup$
Turn $A-B$ into $A capbar{B}$ and use De Morgan's rules.
answered Jan 23 at 22:26
lightxbulblightxbulb
1,125311
answered Jan 23 at 22:26
lightxbulblightxbulb
1,125311
answered Jan 23 at 22:26
lightxbulblightxbulb
1,125311
answered Jan 23 at 22:26
lightxbulblightxbulb
1,125311
1,125311
add a comment |
add a comment |
$begingroup$
Element chase:
If $x in (A-B)-C$ then $x in A$ and $x not in C$ so $xin (A-C$.
And if $x in (A-B) -C$ then $x not in B$ so $x not in B-C subset B$. So $x in (A-C) -(B-C)$.
So $(A-B)-C subset (A-C) - (B-C)$.
Likewise if $y in (A-C)-(B-C)$ then $x in A$. $x not in B-C$ so if $x in B$ then $x in C$. But $x in A-C$ so $x not in C$. So $x not in B$. So $x in A-B$. And $x not in C$ so $x in (A-B) -C$.
So $(A-C) -(B-C) subset (A-B) -C$.
So $(A-B) - C = (A-C) -(B-C)$.
... Another way is to consider which elements are in which sets based on whether they are or are not in $A,B$ or $C$:
Take an element $x in (A-B) -C$.
Is $x in A$? Yes. Is $xin B$? No. Is $x in C$? No.
Take an element $y in (A-C) - (B-C)$.
Is $x in A$? Yes. Is $x in B$? Hmmm. Not if it's not in $C$ But it could be that it is in $B$ and $C$. Is $x in C$. No.... Oh, wait, if it's not $C$ then it can't be in both $B$ and $C$ and since it can't be in $B-C$, it can't be in $B$ after all.
So in both cases $(A-B)-C$ and $(A-C)-(B-C)$ are both precisely the elements in $A$ that aren't in $B$ or $C$.
...
And if we want to do it formally:
$(A-B) - C = (Acap B^c) cap C^c = Acap B^c cap C^c$
whereas
$(A-C) - (B-C) = (Acap C^c)cap (Bcap C^c)^c=$
$(Acap C^c) cap (B^c cup C)= [(Acap C^c) cap B^c] cup [(Acap C^c) cap C] = $
$(Acap B^c cap C^c) cup [Acap (C^c cap C)]=$
$(Acap B^c cap C^c) cup [Acap emptyset]=$
$(Acap B^c cap C^c) cup emptyset=$
$Acap B^c cap C^c$ also.
answered Jan 23 at 22:55
fleabloodfleablood
72.3k22687
$endgroup$
add a comment |
$begingroup$
Element chase:
If $x in (A-B)-C$ then $x in A$ and $x not in C$ so $xin (A-C$.
And if $x in (A-B) -C$ then $x not in B$ so $x not in B-C subset B$. So $x in (A-C) -(B-C)$.
So $(A-B)-C subset (A-C) - (B-C)$.
Likewise if $y in (A-C)-(B-C)$ then $x in A$. $x not in B-C$ so if $x in B$ then $x in C$. But $x in A-C$ so $x not in C$. So $x not in B$. So $x in A-B$. And $x not in C$ so $x in (A-B) -C$.
So $(A-C) -(B-C) subset (A-B) -C$.
So $(A-B) - C = (A-C) -(B-C)$.
... Another way is to consider which elements are in which sets based on whether they are or are not in $A,B$ or $C$:
Take an element $x in (A-B) -C$.
Is $x in A$? Yes. Is $xin B$? No. Is $x in C$? No.
Take an element $y in (A-C) - (B-C)$.
Is $x in A$? Yes. Is $x in B$? Hmmm. Not if it's not in $C$ But it could be that it is in $B$ and $C$. Is $x in C$. No.... Oh, wait, if it's not $C$ then it can't be in both $B$ and $C$ and since it can't be in $B-C$, it can't be in $B$ after all.
So in both cases $(A-B)-C$ and $(A-C)-(B-C)$ are both precisely the elements in $A$ that aren't in $B$ or $C$.
...
And if we want to do it formally:
$(A-B) - C = (Acap B^c) cap C^c = Acap B^c cap C^c$
whereas
$(A-C) - (B-C) = (Acap C^c)cap (Bcap C^c)^c=$
$(Acap C^c) cap (B^c cup C)= [(Acap C^c) cap B^c] cup [(Acap C^c) cap C] = $
$(Acap B^c cap C^c) cup [Acap (C^c cap C)]=$
$(Acap B^c cap C^c) cup [Acap emptyset]=$
$(Acap B^c cap C^c) cup emptyset=$
$Acap B^c cap C^c$ also.
answered Jan 23 at 22:55
fleabloodfleablood
72.3k22687
$endgroup$
add a comment |
$begingroup$
Element chase:
If $x in (A-B)-C$ then $x in A$ and $x not in C$ so $xin (A-C$.
And if $x in (A-B) -C$ then $x not in B$ so $x not in B-C subset B$. So $x in (A-C) -(B-C)$.
So $(A-B)-C subset (A-C) - (B-C)$.
Likewise if $y in (A-C)-(B-C)$ then $x in A$. $x not in B-C$ so if $x in B$ then $x in C$. But $x in A-C$ so $x not in C$. So $x not in B$. So $x in A-B$. And $x not in C$ so $x in (A-B) -C$.
So $(A-C) -(B-C) subset (A-B) -C$.
So $(A-B) - C = (A-C) -(B-C)$.
... Another way is to consider which elements are in which sets based on whether they are or are not in $A,B$ or $C$:
Take an element $x in (A-B) -C$.
Is $x in A$? Yes. Is $xin B$? No. Is $x in C$? No.
Take an element $y in (A-C) - (B-C)$.
Is $x in A$? Yes. Is $x in B$? Hmmm. Not if it's not in $C$ But it could be that it is in $B$ and $C$. Is $x in C$. No.... Oh, wait, if it's not $C$ then it can't be in both $B$ and $C$ and since it can't be in $B-C$, it can't be in $B$ after all.
So in both cases $(A-B)-C$ and $(A-C)-(B-C)$ are both precisely the elements in $A$ that aren't in $B$ or $C$.
...
And if we want to do it formally:
$(A-B) - C = (Acap B^c) cap C^c = Acap B^c cap C^c$
whereas
$(A-C) - (B-C) = (Acap C^c)cap (Bcap C^c)^c=$
$(Acap C^c) cap (B^c cup C)= [(Acap C^c) cap B^c] cup [(Acap C^c) cap C] = $
$(Acap B^c cap C^c) cup [Acap (C^c cap C)]=$
$(Acap B^c cap C^c) cup [Acap emptyset]=$
$(Acap B^c cap C^c) cup emptyset=$
$Acap B^c cap C^c$ also.
answered Jan 23 at 22:55
fleabloodfleablood
72.3k22687
$endgroup$
Element chase:
If $x in (A-B)-C$ then $x in A$ and $x not in C$ so $xin (A-C$.
And if $x in (A-B) -C$ then $x not in B$ so $x not in B-C subset B$. So $x in (A-C) -(B-C)$.
So $(A-B)-C subset (A-C) - (B-C)$.
Likewise if $y in (A-C)-(B-C)$ then $x in A$. $x not in B-C$ so if $x in B$ then $x in C$. But $x in A-C$ so $x not in C$. So $x not in B$. So $x in A-B$. And $x not in C$ so $x in (A-B) -C$.
So $(A-C) -(B-C) subset (A-B) -C$.
So $(A-B) - C = (A-C) -(B-C)$.
... Another way is to consider which elements are in which sets based on whether they are or are not in $A,B$ or $C$:
Take an element $x in (A-B) -C$.
Is $x in A$? Yes. Is $xin B$? No. Is $x in C$? No.
Take an element $y in (A-C) - (B-C)$.
Is $x in A$? Yes. Is $x in B$? Hmmm. Not if it's not in $C$ But it could be that it is in $B$ and $C$. Is $x in C$. No.... Oh, wait, if it's not $C$ then it can't be in both $B$ and $C$ and since it can't be in $B-C$, it can't be in $B$ after all.
So in both cases $(A-B)-C$ and $(A-C)-(B-C)$ are both precisely the elements in $A$ that aren't in $B$ or $C$.
...
And if we want to do it formally:
$(A-B) - C = (Acap B^c) cap C^c = Acap B^c cap C^c$
whereas
$(A-C) - (B-C) = (Acap C^c)cap (Bcap C^c)^c=$
$(Acap C^c) cap (B^c cup C)= [(Acap C^c) cap B^c] cup [(Acap C^c) cap C] = $
$(Acap B^c cap C^c) cup [Acap (C^c cap C)]=$
$(Acap B^c cap C^c) cup [Acap emptyset]=$
$(Acap B^c cap C^c) cup emptyset=$
$Acap B^c cap C^c$ also.
answered Jan 23 at 22:55
fleabloodfleablood
72.3k22687
answered Jan 23 at 22:55
fleabloodfleablood
72.3k22687
answered Jan 23 at 22:55
fleabloodfleablood
72.3k22687
answered Jan 23 at 22:55
fleabloodfleablood
72.3k22687
72.3k22687
add a comment |
add a comment |
$begingroup$
$(A-C)-(B-C)=(Acapbar C)capoverline{(Bcapbar C)}=(Acapbar C)cap(bar Bcup C)=underbrace{(Acapbar Ccapbar B)}_{(A-B)-C}cupunderbrace{(Acapunderbrace{bar Ccap C}_{varnothing})}_{varnothing}$
answered Jan 23 at 22:56
zwimzwim
12.5k831
$endgroup$
add a comment |
$begingroup$
$(A-C)-(B-C)=(Acapbar C)capoverline{(Bcapbar C)}=(Acapbar C)cap(bar Bcup C)=underbrace{(Acapbar Ccapbar B)}_{(A-B)-C}cupunderbrace{(Acapunderbrace{bar Ccap C}_{varnothing})}_{varnothing}$
answered Jan 23 at 22:56
zwimzwim
12.5k831
$endgroup$
add a comment |
$begingroup$
$(A-C)-(B-C)=(Acapbar C)capoverline{(Bcapbar C)}=(Acapbar C)cap(bar Bcup C)=underbrace{(Acapbar Ccapbar B)}_{(A-B)-C}cupunderbrace{(Acapunderbrace{bar Ccap C}_{varnothing})}_{varnothing}$
answered Jan 23 at 22:56
zwimzwim
12.5k831
$endgroup$
$(A-C)-(B-C)=(Acapbar C)capoverline{(Bcapbar C)}=(Acapbar C)cap(bar Bcup C)=underbrace{(Acapbar Ccapbar B)}_{(A-B)-C}cupunderbrace{(Acapunderbrace{bar Ccap C}_{varnothing})}_{varnothing}$
answered Jan 23 at 22:56
zwimzwim
12.5k831
answered Jan 23 at 22:56
zwimzwim
12.5k831
answered Jan 23 at 22:56
zwimzwim
12.5k831
answered Jan 23 at 22:56
zwimzwim
12.5k831
12.5k831
add a comment |
add a comment |
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$begingroup$
You can use a venn diagram with 3 circles.
$endgroup$
– Math Lover
Jan 23 at 23:46