Solving system of set equations
Given
$$
begin{cases}
A setminus X = B\
X setminus A = C
end{cases}
B subseteq A, A cap C neq emptyset
$$
I'm asked to find $X$.
My attempt:
$A setminus X = A setminus ((X setminus A) cup (X cap A)) = B$
$A setminus (C cup (X cap A)) = B$
$A = B cup C cup (X cap A)$
$A setminus (B cup C) = X setminus overline{A}$
$(A setminus (B cup C)) cup overline{A} = X$
Is everything correct?
If not, what is the correct solution?
discrete-mathematics proof-verification elementary-set-theory
edited Nov 28 '18 at 2:14
Andrés E. Caicedo
64.8k8158246
asked Nov 28 '18 at 1:25
False Promise
6581715
add a comment |
Given
$$
begin{cases}
A setminus X = B\
X setminus A = C
end{cases}
B subseteq A, A cap C neq emptyset
$$
I'm asked to find $X$.
My attempt:
$A setminus X = A setminus ((X setminus A) cup (X cap A)) = B$
$A setminus (C cup (X cap A)) = B$
$A = B cup C cup (X cap A)$
$A setminus (B cup C) = X setminus overline{A}$
$(A setminus (B cup C)) cup overline{A} = X$
Is everything correct?
If not, what is the correct solution?
discrete-mathematics proof-verification elementary-set-theory
edited Nov 28 '18 at 2:14
Andrés E. Caicedo
64.8k8158246
asked Nov 28 '18 at 1:25
False Promise
6581715
Have you tried drawing a Venn diagram?
– Y. Forman
Nov 28 '18 at 1:27
@Y.Forman Ahh...I didn't because I don't want to draw for all possible set configurations.
– False Promise
Nov 28 '18 at 1:28
Try drawing just $A$ and $X$ in the general configuration, then identifying $B$ and $C$ on the diagram.
– Y. Forman
Nov 28 '18 at 1:29
add a comment |
Given
$$
begin{cases}
A setminus X = B\
X setminus A = C
end{cases}
B subseteq A, A cap C neq emptyset
$$
I'm asked to find $X$.
My attempt:
$A setminus X = A setminus ((X setminus A) cup (X cap A)) = B$
$A setminus (C cup (X cap A)) = B$
$A = B cup C cup (X cap A)$
$A setminus (B cup C) = X setminus overline{A}$
$(A setminus (B cup C)) cup overline{A} = X$
Is everything correct?
If not, what is the correct solution?
discrete-mathematics proof-verification elementary-set-theory
edited Nov 28 '18 at 2:14
Andrés E. Caicedo
64.8k8158246
asked Nov 28 '18 at 1:25
False Promise
6581715
Given
$$
begin{cases}
A setminus X = B\
X setminus A = C
end{cases}
B subseteq A, A cap C neq emptyset
$$
I'm asked to find $X$.
My attempt:
$A setminus X = A setminus ((X setminus A) cup (X cap A)) = B$
$A setminus (C cup (X cap A)) = B$
$A = B cup C cup (X cap A)$
$A setminus (B cup C) = X setminus overline{A}$
$(A setminus (B cup C)) cup overline{A} = X$
Is everything correct?
If not, what is the correct solution?
discrete-mathematics proof-verification elementary-set-theory
discrete-mathematics proof-verification elementary-set-theory
edited Nov 28 '18 at 2:14
Andrés E. Caicedo
64.8k8158246
asked Nov 28 '18 at 1:25
False Promise
6581715
edited Nov 28 '18 at 2:14
Andrés E. Caicedo
64.8k8158246
asked Nov 28 '18 at 1:25
False Promise
6581715
edited Nov 28 '18 at 2:14
Andrés E. Caicedo
64.8k8158246
edited Nov 28 '18 at 2:14
Andrés E. Caicedo
64.8k8158246
edited Nov 28 '18 at 2:14
Andrés E. Caicedo
64.8k8158246
64.8k8158246
asked Nov 28 '18 at 1:25
False Promise
6581715
asked Nov 28 '18 at 1:25
False Promise
6581715
asked Nov 28 '18 at 1:25
False Promise
6581715
6581715
Have you tried drawing a Venn diagram?
– Y. Forman
Nov 28 '18 at 1:27
@Y.Forman Ahh...I didn't because I don't want to draw for all possible set configurations.
– False Promise
Nov 28 '18 at 1:28
Try drawing just $A$ and $X$ in the general configuration, then identifying $B$ and $C$ on the diagram.
– Y. Forman
Nov 28 '18 at 1:29
add a comment |
Have you tried drawing a Venn diagram?
– Y. Forman
Nov 28 '18 at 1:27
@Y.Forman Ahh...I didn't because I don't want to draw for all possible set configurations.
– False Promise
Nov 28 '18 at 1:28
Try drawing just $A$ and $X$ in the general configuration, then identifying $B$ and $C$ on the diagram.
– Y. Forman
Nov 28 '18 at 1:29
Have you tried drawing a Venn diagram?
– Y. Forman
Nov 28 '18 at 1:27
Have you tried drawing a Venn diagram?
– Y. Forman
Nov 28 '18 at 1:27
@Y.Forman Ahh...I didn't because I don't want to draw for all possible set configurations.
– False Promise
Nov 28 '18 at 1:28
@Y.Forman Ahh...I didn't because I don't want to draw for all possible set configurations.
– False Promise
Nov 28 '18 at 1:28
Try drawing just $A$ and $X$ in the general configuration, then identifying $B$ and $C$ on the diagram.
– Y. Forman
Nov 28 '18 at 1:29
Try drawing just $A$ and $X$ in the general configuration, then identifying $B$ and $C$ on the diagram.
– Y. Forman
Nov 28 '18 at 1:29
add a comment |
1 Answer
1
active
oldest
votes
Your jump from the second to the third line is unjustified - and the third line clearly cannot be true if $C cap A = varnothing$
Instead, start from $X = (Xsetminus A) cup(Xcap A)$. The first term is $C$; the second is related to $B$, see if you can figure out how.
As I mentioned in the comments, I recommend following along a Venn diagram as you derive these equations.
answered Nov 28 '18 at 1:33
Y. Forman
11.4k523
@FalsePromise But $A setminus X = B$ implies $B$ and $X$ are disjoint, so we cannot have $X cap A=B$
– Y. Forman
Nov 28 '18 at 1:54
Yea, that was a mistake. I actually ended up with a non convex circle for $X$, i.e. $X$ lies on the outermost level having a gap in the middle which is $B$, this is why I got such result.
– False Promise
Nov 28 '18 at 1:59
@FalsePromise I'm not sure what you mean. In any case, whatever you see in the diagram, try to follow the reasoning through to the equation manipulation.
– Y. Forman
Nov 28 '18 at 2:01
Thank you! Finally came up with answer by drawing diagrams right: $X = C cup (A setminus B)$.
– False Promise
Nov 28 '18 at 2:36
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your jump from the second to the third line is unjustified - and the third line clearly cannot be true if $C cap A = varnothing$
Instead, start from $X = (Xsetminus A) cup(Xcap A)$. The first term is $C$; the second is related to $B$, see if you can figure out how.
As I mentioned in the comments, I recommend following along a Venn diagram as you derive these equations.
answered Nov 28 '18 at 1:33
Y. Forman
11.4k523
@FalsePromise But $A setminus X = B$ implies $B$ and $X$ are disjoint, so we cannot have $X cap A=B$
– Y. Forman
Nov 28 '18 at 1:54
Yea, that was a mistake. I actually ended up with a non convex circle for $X$, i.e. $X$ lies on the outermost level having a gap in the middle which is $B$, this is why I got such result.
– False Promise
Nov 28 '18 at 1:59
@FalsePromise I'm not sure what you mean. In any case, whatever you see in the diagram, try to follow the reasoning through to the equation manipulation.
– Y. Forman
Nov 28 '18 at 2:01
Thank you! Finally came up with answer by drawing diagrams right: $X = C cup (A setminus B)$.
– False Promise
Nov 28 '18 at 2:36
add a comment |
Your jump from the second to the third line is unjustified - and the third line clearly cannot be true if $C cap A = varnothing$
Instead, start from $X = (Xsetminus A) cup(Xcap A)$. The first term is $C$; the second is related to $B$, see if you can figure out how.
As I mentioned in the comments, I recommend following along a Venn diagram as you derive these equations.
answered Nov 28 '18 at 1:33
Y. Forman
11.4k523
@FalsePromise But $A setminus X = B$ implies $B$ and $X$ are disjoint, so we cannot have $X cap A=B$
– Y. Forman
Nov 28 '18 at 1:54
Yea, that was a mistake. I actually ended up with a non convex circle for $X$, i.e. $X$ lies on the outermost level having a gap in the middle which is $B$, this is why I got such result.
– False Promise
Nov 28 '18 at 1:59
@FalsePromise I'm not sure what you mean. In any case, whatever you see in the diagram, try to follow the reasoning through to the equation manipulation.
– Y. Forman
Nov 28 '18 at 2:01
Thank you! Finally came up with answer by drawing diagrams right: $X = C cup (A setminus B)$.
– False Promise
Nov 28 '18 at 2:36
add a comment |
Your jump from the second to the third line is unjustified - and the third line clearly cannot be true if $C cap A = varnothing$
Instead, start from $X = (Xsetminus A) cup(Xcap A)$. The first term is $C$; the second is related to $B$, see if you can figure out how.
As I mentioned in the comments, I recommend following along a Venn diagram as you derive these equations.
answered Nov 28 '18 at 1:33
Y. Forman
11.4k523
Your jump from the second to the third line is unjustified - and the third line clearly cannot be true if $C cap A = varnothing$
Instead, start from $X = (Xsetminus A) cup(Xcap A)$. The first term is $C$; the second is related to $B$, see if you can figure out how.
As I mentioned in the comments, I recommend following along a Venn diagram as you derive these equations.
answered Nov 28 '18 at 1:33
Y. Forman
11.4k523
answered Nov 28 '18 at 1:33
Y. Forman
11.4k523
answered Nov 28 '18 at 1:33
Y. Forman
11.4k523
answered Nov 28 '18 at 1:33
Y. Forman
11.4k523
11.4k523
@FalsePromise But $A setminus X = B$ implies $B$ and $X$ are disjoint, so we cannot have $X cap A=B$
– Y. Forman
Nov 28 '18 at 1:54
Yea, that was a mistake. I actually ended up with a non convex circle for $X$, i.e. $X$ lies on the outermost level having a gap in the middle which is $B$, this is why I got such result.
– False Promise
Nov 28 '18 at 1:59
@FalsePromise I'm not sure what you mean. In any case, whatever you see in the diagram, try to follow the reasoning through to the equation manipulation.
– Y. Forman
Nov 28 '18 at 2:01
Thank you! Finally came up with answer by drawing diagrams right: $X = C cup (A setminus B)$.
– False Promise
Nov 28 '18 at 2:36
add a comment |
@FalsePromise But $A setminus X = B$ implies $B$ and $X$ are disjoint, so we cannot have $X cap A=B$
– Y. Forman
Nov 28 '18 at 1:54
Yea, that was a mistake. I actually ended up with a non convex circle for $X$, i.e. $X$ lies on the outermost level having a gap in the middle which is $B$, this is why I got such result.
– False Promise
Nov 28 '18 at 1:59
@FalsePromise I'm not sure what you mean. In any case, whatever you see in the diagram, try to follow the reasoning through to the equation manipulation.
– Y. Forman
Nov 28 '18 at 2:01
Thank you! Finally came up with answer by drawing diagrams right: $X = C cup (A setminus B)$.
– False Promise
Nov 28 '18 at 2:36
@FalsePromise But $A setminus X = B$ implies $B$ and $X$ are disjoint, so we cannot have $X cap A=B$
– Y. Forman
Nov 28 '18 at 1:54
@FalsePromise But $A setminus X = B$ implies $B$ and $X$ are disjoint, so we cannot have $X cap A=B$
– Y. Forman
Nov 28 '18 at 1:54
Yea, that was a mistake. I actually ended up with a non convex circle for $X$, i.e. $X$ lies on the outermost level having a gap in the middle which is $B$, this is why I got such result.
– False Promise
Nov 28 '18 at 1:59
Yea, that was a mistake. I actually ended up with a non convex circle for $X$, i.e. $X$ lies on the outermost level having a gap in the middle which is $B$, this is why I got such result.
– False Promise
Nov 28 '18 at 1:59
@FalsePromise I'm not sure what you mean. In any case, whatever you see in the diagram, try to follow the reasoning through to the equation manipulation.
– Y. Forman
Nov 28 '18 at 2:01
@FalsePromise I'm not sure what you mean. In any case, whatever you see in the diagram, try to follow the reasoning through to the equation manipulation.
– Y. Forman
Nov 28 '18 at 2:01
Thank you! Finally came up with answer by drawing diagrams right: $X = C cup (A setminus B)$.
– False Promise
Nov 28 '18 at 2:36
Thank you! Finally came up with answer by drawing diagrams right: $X = C cup (A setminus B)$.
– False Promise
Nov 28 '18 at 2:36
add a comment |
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Have you tried drawing a Venn diagram?
– Y. Forman
Nov 28 '18 at 1:27
@Y.Forman Ahh...I didn't because I don't want to draw for all possible set configurations.
– False Promise
Nov 28 '18 at 1:28
Try drawing just $A$ and $X$ in the general configuration, then identifying $B$ and $C$ on the diagram.
– Y. Forman
Nov 28 '18 at 1:29