I know this claim is false. So what is wrong with the proof? If $Asubseteq Bcup C$, then $A subseteq B$ or...
$begingroup$
If $Asubseteq Bcup C$, then $A subseteq B$ or $Asubseteq C$. A counterexample to this claim is: $A={ 2,3,4 }$, $B={1,2,3}$, $C={3,4,5}$. But I cannot find an error in this proof: Assume $Asubseteq Bcup C$. Let $xin A$. Then $xin Bcup C$. So $xin B$ or $xin C$. Case 1. $xin B$. We have $xin A rightarrow xin B$ Thus $Asubseteq B$. Case 2. $xin C$. We have $xin A rightarrow xin C$ Thus $Asubseteq C$. Hence, $Asubseteq B$ or $Asubseteq C$.
elementary-set-theory
asked Dec 8 '18 at 21:09
PlatonicTraditionPlatonicTradition
564
$endgroup$
|
show 1 more comment
$begingroup$
If $Asubseteq Bcup C$, then $A subseteq B$ or $Asubseteq C$. A counterexample to this claim is: $A={ 2,3,4 }$, $B={1,2,3}$, $C={3,4,5}$. But I cannot find an error in this proof: Assume $Asubseteq Bcup C$. Let $xin A$. Then $xin Bcup C$. So $xin B$ or $xin C$. Case 1. $xin B$. We have $xin A rightarrow xin B$ Thus $Asubseteq B$. Case 2. $xin C$. We have $xin A rightarrow xin C$ Thus $Asubseteq C$. Hence, $Asubseteq B$ or $Asubseteq C$.
elementary-set-theory
asked Dec 8 '18 at 21:09
PlatonicTraditionPlatonicTradition
564
$endgroup$
3
$begingroup$
$x$ is just one element. You have to prove one or the other for every element. You might have $xin A subset B; x not in C$ but then also have $y in A subset C; y not in C$.
$endgroup$
– fleablood
Dec 8 '18 at 21:15
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In the above proof, how is $xin A$ not an arbitrary element?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:42
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Because you said "suppose $x in B$". You can't say that for an arbitrary element. Once you suppose $x in B$ it is no longer an arbitrary element in $A$. It now has a specific property (that is is also in $B$) that not all elements have. SO it can not be arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:50
$begingroup$
That makes sense. But, since $xin B cup C$, is it unsafe to then say that $x$ has to be in either $B$ or $C$? And then prove by cases?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:54
$begingroup$
"arbitrary" a label some elements have. $2$ is not an arbitrary element, neither is $3$ or $4$. So ... none of the elements in $A$ are arbitrary? Well, that's word play. An "arbitrary element" is one that we don't KNOW any specific quality. One way say, let's suppose $x$ is even, we are assuming specifics and it can't claim it is arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:56
|
show 1 more comment
$begingroup$
If $Asubseteq Bcup C$, then $A subseteq B$ or $Asubseteq C$. A counterexample to this claim is: $A={ 2,3,4 }$, $B={1,2,3}$, $C={3,4,5}$. But I cannot find an error in this proof: Assume $Asubseteq Bcup C$. Let $xin A$. Then $xin Bcup C$. So $xin B$ or $xin C$. Case 1. $xin B$. We have $xin A rightarrow xin B$ Thus $Asubseteq B$. Case 2. $xin C$. We have $xin A rightarrow xin C$ Thus $Asubseteq C$. Hence, $Asubseteq B$ or $Asubseteq C$.
elementary-set-theory
asked Dec 8 '18 at 21:09
PlatonicTraditionPlatonicTradition
564
$endgroup$
If $Asubseteq Bcup C$, then $A subseteq B$ or $Asubseteq C$. A counterexample to this claim is: $A={ 2,3,4 }$, $B={1,2,3}$, $C={3,4,5}$. But I cannot find an error in this proof: Assume $Asubseteq Bcup C$. Let $xin A$. Then $xin Bcup C$. So $xin B$ or $xin C$. Case 1. $xin B$. We have $xin A rightarrow xin B$ Thus $Asubseteq B$. Case 2. $xin C$. We have $xin A rightarrow xin C$ Thus $Asubseteq C$. Hence, $Asubseteq B$ or $Asubseteq C$.
elementary-set-theory
elementary-set-theory
asked Dec 8 '18 at 21:09
PlatonicTraditionPlatonicTradition
564
asked Dec 8 '18 at 21:09
PlatonicTraditionPlatonicTradition
564
asked Dec 8 '18 at 21:09
PlatonicTraditionPlatonicTradition
564
asked Dec 8 '18 at 21:09
PlatonicTraditionPlatonicTradition
564
asked Dec 8 '18 at 21:09
PlatonicTraditionPlatonicTradition
564
564
3
$begingroup$
$x$ is just one element. You have to prove one or the other for every element. You might have $xin A subset B; x not in C$ but then also have $y in A subset C; y not in C$.
$endgroup$
– fleablood
Dec 8 '18 at 21:15
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In the above proof, how is $xin A$ not an arbitrary element?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:42
$begingroup$
Because you said "suppose $x in B$". You can't say that for an arbitrary element. Once you suppose $x in B$ it is no longer an arbitrary element in $A$. It now has a specific property (that is is also in $B$) that not all elements have. SO it can not be arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:50
$begingroup$
That makes sense. But, since $xin B cup C$, is it unsafe to then say that $x$ has to be in either $B$ or $C$? And then prove by cases?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:54
$begingroup$
"arbitrary" a label some elements have. $2$ is not an arbitrary element, neither is $3$ or $4$. So ... none of the elements in $A$ are arbitrary? Well, that's word play. An "arbitrary element" is one that we don't KNOW any specific quality. One way say, let's suppose $x$ is even, we are assuming specifics and it can't claim it is arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:56
|
show 1 more comment
3
$begingroup$
$x$ is just one element. You have to prove one or the other for every element. You might have $xin A subset B; x not in C$ but then also have $y in A subset C; y not in C$.
$endgroup$
– fleablood
Dec 8 '18 at 21:15
$begingroup$
In the above proof, how is $xin A$ not an arbitrary element?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:42
$begingroup$
Because you said "suppose $x in B$". You can't say that for an arbitrary element. Once you suppose $x in B$ it is no longer an arbitrary element in $A$. It now has a specific property (that is is also in $B$) that not all elements have. SO it can not be arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:50
$begingroup$
That makes sense. But, since $xin B cup C$, is it unsafe to then say that $x$ has to be in either $B$ or $C$? And then prove by cases?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:54
$begingroup$
"arbitrary" a label some elements have. $2$ is not an arbitrary element, neither is $3$ or $4$. So ... none of the elements in $A$ are arbitrary? Well, that's word play. An "arbitrary element" is one that we don't KNOW any specific quality. One way say, let's suppose $x$ is even, we are assuming specifics and it can't claim it is arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:56
3
3
$begingroup$
$x$ is just one element. You have to prove one or the other for every element. You might have $xin A subset B; x not in C$ but then also have $y in A subset C; y not in C$.
$endgroup$
– fleablood
Dec 8 '18 at 21:15
$begingroup$
$x$ is just one element. You have to prove one or the other for every element. You might have $xin A subset B; x not in C$ but then also have $y in A subset C; y not in C$.
$endgroup$
– fleablood
Dec 8 '18 at 21:15
$begingroup$
In the above proof, how is $xin A$ not an arbitrary element?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:42
$begingroup$
In the above proof, how is $xin A$ not an arbitrary element?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:42
$begingroup$
Because you said "suppose $x in B$". You can't say that for an arbitrary element. Once you suppose $x in B$ it is no longer an arbitrary element in $A$. It now has a specific property (that is is also in $B$) that not all elements have. SO it can not be arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:50
$begingroup$
Because you said "suppose $x in B$". You can't say that for an arbitrary element. Once you suppose $x in B$ it is no longer an arbitrary element in $A$. It now has a specific property (that is is also in $B$) that not all elements have. SO it can not be arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:50
$begingroup$
That makes sense. But, since $xin B cup C$, is it unsafe to then say that $x$ has to be in either $B$ or $C$? And then prove by cases?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:54
$begingroup$
That makes sense. But, since $xin B cup C$, is it unsafe to then say that $x$ has to be in either $B$ or $C$? And then prove by cases?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:54
$begingroup$
"arbitrary" a label some elements have. $2$ is not an arbitrary element, neither is $3$ or $4$. So ... none of the elements in $A$ are arbitrary? Well, that's word play. An "arbitrary element" is one that we don't KNOW any specific quality. One way say, let's suppose $x$ is even, we are assuming specifics and it can't claim it is arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:56
$begingroup$
"arbitrary" a label some elements have. $2$ is not an arbitrary element, neither is $3$ or $4$. So ... none of the elements in $A$ are arbitrary? Well, that's word play. An "arbitrary element" is one that we don't KNOW any specific quality. One way say, let's suppose $x$ is even, we are assuming specifics and it can't claim it is arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:56
|
show 1 more comment
5 Answers
5
active
oldest
votes
$begingroup$
From the fact that one element $x$ of $A$ belongs to $B$, you can't deduce that $Asubset B$, since this means that every element of $A$ belongs to $B$..
answered Dec 8 '18 at 21:12
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
But in the proof, is $xin A$ not an arbitrary element of A? So that the proof applies for all $x in A$? Where does the proof go wrong?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:49
1
$begingroup$
Each individual element of $A$ belongs to $B$ or to $C$. It doesn't follow from that that all elements of $A$ belong to $B$ or that all elements of $A$ belongs to $C$. You provided an example yourself.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 21:52
$begingroup$
When you picked $x$ it was an arbitrary element. But once you picked it and started playing with it. It isn't arbitrary. It is now a specific element. Suppose you are told you can pick any cat from the pound. And you pick one at random and you picked Bert. And you said there's nothing special about picking Bert. Bert was an arbitrary cat. That's fine. Then you look at Bert and see he has tapeworms. You can't say "well, Bert was arbitrary so I can conclude all the cats had tapeworms."
$endgroup$
– fleablood
Dec 8 '18 at 22:03
add a comment |
$begingroup$
You have proved that
for every $x$, if $xin A$, then either $xin B$ or $xin C$
Your claim is
(for every $x$, if $xin A$, then $xin B$) or (for every $x$, if $xin A$, then $xin C$)
With logic symbols, the first formula is
$forall x,(xin Ato ( {xin B} lor {xin C}))$
and the second one is
$(forall x,(xin Ato xin B))lor(forall x,(xin Ato xin C))$
It should be clear that the two are quite different and, indeed, the simple example $A={1,2}$, $B={1}$, $C={2}$ makes the first true and the second false.
If you examine your proof with this example, you see that for $x=1$ we have $xin B$, but for $x=2$ we have $xin C$.
answered Dec 8 '18 at 21:23
egregegreg
184k1486205
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add a comment |
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Sometimes it is easier to prove something if you can see that it is true visually.
So try to draw two venn diagrams B and C and color their union. Now let A be another diagram inside the union of B and C, then what can you see from here?
answered Dec 8 '18 at 21:25
JensensJensens
245
$endgroup$
add a comment |
$begingroup$
If you want to get technical $x in A to xin B$ is ambiguous. It could mean either $exists x: xin A to x in B$ or it could mean $forall x: xin A to x in B$.
In your proof, you are saying "Case 1: $x in B$ therefore $exists x: xin A to xin B$". And that is true. But the definition of "$A subset B$" requires the statement "$forall x: xin A to x in B$". And that is false.
Consider this: $2in A$ . And $2in B$. So the sentence "$2in A to 2 in B$" is true (because $Mto N$ means ($M$ and $N$) or (not $M$)). So "$exists x: xin A to x in B$" is true. But that sure as heck does not mean "$forall x: x in A to x in B$"!
Obviously $x = 4$ would be a counter example.
But to say "$A subset B$" we must have "$forall x: xin A implies xin B$". And you just never had that.
Otherwise....
we could say:
$mathbb Z subset (0,3)$.
Pf: Let $x = 1$. $x in mathbb Z$ and $x in (0,3)$. So $x in mathbb Z to xin (0,3)$ so $mathbb Z subset (0,3)$.
answered Dec 8 '18 at 21:47
fleabloodfleablood
72.7k22788
$endgroup$
add a comment |
$begingroup$
Draw a Venn diagram where $B$ and $C$ are your normal two circles, and $A$ is a really thin ellipse passing through the middle of the two.
As far as the proof is concerned, you simply forgot that to prove a subset relation you need to show that all elements of one set belong to another. As soon you as you have chosen your $x$ you break into cases, and each time it may be different.
answered Dec 8 '18 at 21:12
RandomMathGuyRandomMathGuy
462
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"As you have chosen your $x$ you break into cases, and each time it may be different." What do you mean by this? It seems as though the $xin A$ is an arbitrary element of A, and hence the proof should apply to all $xin A$. I have already drawn that Venn diagram, and I understand the claim is false. I am still puzzled as to where the error is in the proof.
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– PlatonicTradition
Dec 8 '18 at 21:52
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@PlatonicTradition You did choose an arbitrary element of $A$, but what applies to all $xin A$ is the disjunction $xin B$ or $xin C$. When you move into your case by case analysis, you are no longer considering a truly arbitrary member of $A$.
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– RandomMathGuy
Dec 8 '18 at 21:58
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@PlatonicTradition Also, there was a missing word. I meant to say "As soon as you have chosen"
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– RandomMathGuy
Dec 8 '18 at 21:59
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@PlatonicTradition Additionally, your cases aren't technically correct either. Your first case can read $xin B$, but the second case should read $xnotin B$, thus $xin C$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 22:03
add a comment |
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5 Answers
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5 Answers
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$begingroup$
From the fact that one element $x$ of $A$ belongs to $B$, you can't deduce that $Asubset B$, since this means that every element of $A$ belongs to $B$..
answered Dec 8 '18 at 21:12
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
But in the proof, is $xin A$ not an arbitrary element of A? So that the proof applies for all $x in A$? Where does the proof go wrong?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:49
1
$begingroup$
Each individual element of $A$ belongs to $B$ or to $C$. It doesn't follow from that that all elements of $A$ belong to $B$ or that all elements of $A$ belongs to $C$. You provided an example yourself.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 21:52
$begingroup$
When you picked $x$ it was an arbitrary element. But once you picked it and started playing with it. It isn't arbitrary. It is now a specific element. Suppose you are told you can pick any cat from the pound. And you pick one at random and you picked Bert. And you said there's nothing special about picking Bert. Bert was an arbitrary cat. That's fine. Then you look at Bert and see he has tapeworms. You can't say "well, Bert was arbitrary so I can conclude all the cats had tapeworms."
$endgroup$
– fleablood
Dec 8 '18 at 22:03
add a comment |
$begingroup$
From the fact that one element $x$ of $A$ belongs to $B$, you can't deduce that $Asubset B$, since this means that every element of $A$ belongs to $B$..
answered Dec 8 '18 at 21:12
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
But in the proof, is $xin A$ not an arbitrary element of A? So that the proof applies for all $x in A$? Where does the proof go wrong?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:49
1
$begingroup$
Each individual element of $A$ belongs to $B$ or to $C$. It doesn't follow from that that all elements of $A$ belong to $B$ or that all elements of $A$ belongs to $C$. You provided an example yourself.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 21:52
$begingroup$
When you picked $x$ it was an arbitrary element. But once you picked it and started playing with it. It isn't arbitrary. It is now a specific element. Suppose you are told you can pick any cat from the pound. And you pick one at random and you picked Bert. And you said there's nothing special about picking Bert. Bert was an arbitrary cat. That's fine. Then you look at Bert and see he has tapeworms. You can't say "well, Bert was arbitrary so I can conclude all the cats had tapeworms."
$endgroup$
– fleablood
Dec 8 '18 at 22:03
add a comment |
$begingroup$
From the fact that one element $x$ of $A$ belongs to $B$, you can't deduce that $Asubset B$, since this means that every element of $A$ belongs to $B$..
answered Dec 8 '18 at 21:12
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
From the fact that one element $x$ of $A$ belongs to $B$, you can't deduce that $Asubset B$, since this means that every element of $A$ belongs to $B$..
answered Dec 8 '18 at 21:12
José Carlos SantosJosé Carlos Santos
168k23132236
answered Dec 8 '18 at 21:12
José Carlos SantosJosé Carlos Santos
168k23132236
answered Dec 8 '18 at 21:12
José Carlos SantosJosé Carlos Santos
168k23132236
answered Dec 8 '18 at 21:12
José Carlos SantosJosé Carlos Santos
168k23132236
168k23132236
$begingroup$
But in the proof, is $xin A$ not an arbitrary element of A? So that the proof applies for all $x in A$? Where does the proof go wrong?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:49
1
$begingroup$
Each individual element of $A$ belongs to $B$ or to $C$. It doesn't follow from that that all elements of $A$ belong to $B$ or that all elements of $A$ belongs to $C$. You provided an example yourself.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 21:52
$begingroup$
When you picked $x$ it was an arbitrary element. But once you picked it and started playing with it. It isn't arbitrary. It is now a specific element. Suppose you are told you can pick any cat from the pound. And you pick one at random and you picked Bert. And you said there's nothing special about picking Bert. Bert was an arbitrary cat. That's fine. Then you look at Bert and see he has tapeworms. You can't say "well, Bert was arbitrary so I can conclude all the cats had tapeworms."
$endgroup$
– fleablood
Dec 8 '18 at 22:03
add a comment |
$begingroup$
But in the proof, is $xin A$ not an arbitrary element of A? So that the proof applies for all $x in A$? Where does the proof go wrong?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:49
1
$begingroup$
Each individual element of $A$ belongs to $B$ or to $C$. It doesn't follow from that that all elements of $A$ belong to $B$ or that all elements of $A$ belongs to $C$. You provided an example yourself.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 21:52
$begingroup$
When you picked $x$ it was an arbitrary element. But once you picked it and started playing with it. It isn't arbitrary. It is now a specific element. Suppose you are told you can pick any cat from the pound. And you pick one at random and you picked Bert. And you said there's nothing special about picking Bert. Bert was an arbitrary cat. That's fine. Then you look at Bert and see he has tapeworms. You can't say "well, Bert was arbitrary so I can conclude all the cats had tapeworms."
$endgroup$
– fleablood
Dec 8 '18 at 22:03
$begingroup$
But in the proof, is $xin A$ not an arbitrary element of A? So that the proof applies for all $x in A$? Where does the proof go wrong?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:49
$begingroup$
But in the proof, is $xin A$ not an arbitrary element of A? So that the proof applies for all $x in A$? Where does the proof go wrong?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:49
1
1
$begingroup$
Each individual element of $A$ belongs to $B$ or to $C$. It doesn't follow from that that all elements of $A$ belong to $B$ or that all elements of $A$ belongs to $C$. You provided an example yourself.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 21:52
$begingroup$
Each individual element of $A$ belongs to $B$ or to $C$. It doesn't follow from that that all elements of $A$ belong to $B$ or that all elements of $A$ belongs to $C$. You provided an example yourself.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 21:52
$begingroup$
When you picked $x$ it was an arbitrary element. But once you picked it and started playing with it. It isn't arbitrary. It is now a specific element. Suppose you are told you can pick any cat from the pound. And you pick one at random and you picked Bert. And you said there's nothing special about picking Bert. Bert was an arbitrary cat. That's fine. Then you look at Bert and see he has tapeworms. You can't say "well, Bert was arbitrary so I can conclude all the cats had tapeworms."
$endgroup$
– fleablood
Dec 8 '18 at 22:03
$begingroup$
When you picked $x$ it was an arbitrary element. But once you picked it and started playing with it. It isn't arbitrary. It is now a specific element. Suppose you are told you can pick any cat from the pound. And you pick one at random and you picked Bert. And you said there's nothing special about picking Bert. Bert was an arbitrary cat. That's fine. Then you look at Bert and see he has tapeworms. You can't say "well, Bert was arbitrary so I can conclude all the cats had tapeworms."
$endgroup$
– fleablood
Dec 8 '18 at 22:03
add a comment |
$begingroup$
You have proved that
for every $x$, if $xin A$, then either $xin B$ or $xin C$
Your claim is
(for every $x$, if $xin A$, then $xin B$) or (for every $x$, if $xin A$, then $xin C$)
With logic symbols, the first formula is
$forall x,(xin Ato ( {xin B} lor {xin C}))$
and the second one is
$(forall x,(xin Ato xin B))lor(forall x,(xin Ato xin C))$
It should be clear that the two are quite different and, indeed, the simple example $A={1,2}$, $B={1}$, $C={2}$ makes the first true and the second false.
If you examine your proof with this example, you see that for $x=1$ we have $xin B$, but for $x=2$ we have $xin C$.
answered Dec 8 '18 at 21:23
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
You have proved that
for every $x$, if $xin A$, then either $xin B$ or $xin C$
Your claim is
(for every $x$, if $xin A$, then $xin B$) or (for every $x$, if $xin A$, then $xin C$)
With logic symbols, the first formula is
$forall x,(xin Ato ( {xin B} lor {xin C}))$
and the second one is
$(forall x,(xin Ato xin B))lor(forall x,(xin Ato xin C))$
It should be clear that the two are quite different and, indeed, the simple example $A={1,2}$, $B={1}$, $C={2}$ makes the first true and the second false.
If you examine your proof with this example, you see that for $x=1$ we have $xin B$, but for $x=2$ we have $xin C$.
answered Dec 8 '18 at 21:23
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
You have proved that
for every $x$, if $xin A$, then either $xin B$ or $xin C$
Your claim is
(for every $x$, if $xin A$, then $xin B$) or (for every $x$, if $xin A$, then $xin C$)
With logic symbols, the first formula is
$forall x,(xin Ato ( {xin B} lor {xin C}))$
and the second one is
$(forall x,(xin Ato xin B))lor(forall x,(xin Ato xin C))$
It should be clear that the two are quite different and, indeed, the simple example $A={1,2}$, $B={1}$, $C={2}$ makes the first true and the second false.
If you examine your proof with this example, you see that for $x=1$ we have $xin B$, but for $x=2$ we have $xin C$.
answered Dec 8 '18 at 21:23
egregegreg
184k1486205
$endgroup$
You have proved that
for every $x$, if $xin A$, then either $xin B$ or $xin C$
Your claim is
(for every $x$, if $xin A$, then $xin B$) or (for every $x$, if $xin A$, then $xin C$)
With logic symbols, the first formula is
$forall x,(xin Ato ( {xin B} lor {xin C}))$
and the second one is
$(forall x,(xin Ato xin B))lor(forall x,(xin Ato xin C))$
It should be clear that the two are quite different and, indeed, the simple example $A={1,2}$, $B={1}$, $C={2}$ makes the first true and the second false.
If you examine your proof with this example, you see that for $x=1$ we have $xin B$, but for $x=2$ we have $xin C$.
answered Dec 8 '18 at 21:23
egregegreg
184k1486205
answered Dec 8 '18 at 21:23
egregegreg
184k1486205
answered Dec 8 '18 at 21:23
egregegreg
184k1486205
answered Dec 8 '18 at 21:23
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
$begingroup$
Sometimes it is easier to prove something if you can see that it is true visually.
So try to draw two venn diagrams B and C and color their union. Now let A be another diagram inside the union of B and C, then what can you see from here?
answered Dec 8 '18 at 21:25
JensensJensens
245
$endgroup$
add a comment |
$begingroup$
Sometimes it is easier to prove something if you can see that it is true visually.
So try to draw two venn diagrams B and C and color their union. Now let A be another diagram inside the union of B and C, then what can you see from here?
answered Dec 8 '18 at 21:25
JensensJensens
245
$endgroup$
add a comment |
$begingroup$
Sometimes it is easier to prove something if you can see that it is true visually.
So try to draw two venn diagrams B and C and color their union. Now let A be another diagram inside the union of B and C, then what can you see from here?
answered Dec 8 '18 at 21:25
JensensJensens
245
$endgroup$
Sometimes it is easier to prove something if you can see that it is true visually.
So try to draw two venn diagrams B and C and color their union. Now let A be another diagram inside the union of B and C, then what can you see from here?
answered Dec 8 '18 at 21:25
JensensJensens
245
answered Dec 8 '18 at 21:25
JensensJensens
245
answered Dec 8 '18 at 21:25
JensensJensens
245
answered Dec 8 '18 at 21:25
JensensJensens
245
245
add a comment |
add a comment |
$begingroup$
If you want to get technical $x in A to xin B$ is ambiguous. It could mean either $exists x: xin A to x in B$ or it could mean $forall x: xin A to x in B$.
In your proof, you are saying "Case 1: $x in B$ therefore $exists x: xin A to xin B$". And that is true. But the definition of "$A subset B$" requires the statement "$forall x: xin A to x in B$". And that is false.
Consider this: $2in A$ . And $2in B$. So the sentence "$2in A to 2 in B$" is true (because $Mto N$ means ($M$ and $N$) or (not $M$)). So "$exists x: xin A to x in B$" is true. But that sure as heck does not mean "$forall x: x in A to x in B$"!
Obviously $x = 4$ would be a counter example.
But to say "$A subset B$" we must have "$forall x: xin A implies xin B$". And you just never had that.
Otherwise....
we could say:
$mathbb Z subset (0,3)$.
Pf: Let $x = 1$. $x in mathbb Z$ and $x in (0,3)$. So $x in mathbb Z to xin (0,3)$ so $mathbb Z subset (0,3)$.
answered Dec 8 '18 at 21:47
fleabloodfleablood
72.7k22788
$endgroup$
add a comment |
$begingroup$
If you want to get technical $x in A to xin B$ is ambiguous. It could mean either $exists x: xin A to x in B$ or it could mean $forall x: xin A to x in B$.
In your proof, you are saying "Case 1: $x in B$ therefore $exists x: xin A to xin B$". And that is true. But the definition of "$A subset B$" requires the statement "$forall x: xin A to x in B$". And that is false.
Consider this: $2in A$ . And $2in B$. So the sentence "$2in A to 2 in B$" is true (because $Mto N$ means ($M$ and $N$) or (not $M$)). So "$exists x: xin A to x in B$" is true. But that sure as heck does not mean "$forall x: x in A to x in B$"!
Obviously $x = 4$ would be a counter example.
But to say "$A subset B$" we must have "$forall x: xin A implies xin B$". And you just never had that.
Otherwise....
we could say:
$mathbb Z subset (0,3)$.
Pf: Let $x = 1$. $x in mathbb Z$ and $x in (0,3)$. So $x in mathbb Z to xin (0,3)$ so $mathbb Z subset (0,3)$.
answered Dec 8 '18 at 21:47
fleabloodfleablood
72.7k22788
$endgroup$
add a comment |
$begingroup$
If you want to get technical $x in A to xin B$ is ambiguous. It could mean either $exists x: xin A to x in B$ or it could mean $forall x: xin A to x in B$.
In your proof, you are saying "Case 1: $x in B$ therefore $exists x: xin A to xin B$". And that is true. But the definition of "$A subset B$" requires the statement "$forall x: xin A to x in B$". And that is false.
Consider this: $2in A$ . And $2in B$. So the sentence "$2in A to 2 in B$" is true (because $Mto N$ means ($M$ and $N$) or (not $M$)). So "$exists x: xin A to x in B$" is true. But that sure as heck does not mean "$forall x: x in A to x in B$"!
Obviously $x = 4$ would be a counter example.
But to say "$A subset B$" we must have "$forall x: xin A implies xin B$". And you just never had that.
Otherwise....
we could say:
$mathbb Z subset (0,3)$.
Pf: Let $x = 1$. $x in mathbb Z$ and $x in (0,3)$. So $x in mathbb Z to xin (0,3)$ so $mathbb Z subset (0,3)$.
answered Dec 8 '18 at 21:47
fleabloodfleablood
72.7k22788
$endgroup$
If you want to get technical $x in A to xin B$ is ambiguous. It could mean either $exists x: xin A to x in B$ or it could mean $forall x: xin A to x in B$.
In your proof, you are saying "Case 1: $x in B$ therefore $exists x: xin A to xin B$". And that is true. But the definition of "$A subset B$" requires the statement "$forall x: xin A to x in B$". And that is false.
Consider this: $2in A$ . And $2in B$. So the sentence "$2in A to 2 in B$" is true (because $Mto N$ means ($M$ and $N$) or (not $M$)). So "$exists x: xin A to x in B$" is true. But that sure as heck does not mean "$forall x: x in A to x in B$"!
Obviously $x = 4$ would be a counter example.
But to say "$A subset B$" we must have "$forall x: xin A implies xin B$". And you just never had that.
Otherwise....
we could say:
$mathbb Z subset (0,3)$.
Pf: Let $x = 1$. $x in mathbb Z$ and $x in (0,3)$. So $x in mathbb Z to xin (0,3)$ so $mathbb Z subset (0,3)$.
answered Dec 8 '18 at 21:47
fleabloodfleablood
72.7k22788
answered Dec 8 '18 at 21:47
fleabloodfleablood
72.7k22788
answered Dec 8 '18 at 21:47
fleabloodfleablood
72.7k22788
answered Dec 8 '18 at 21:47
fleabloodfleablood
72.7k22788
72.7k22788
add a comment |
add a comment |
$begingroup$
Draw a Venn diagram where $B$ and $C$ are your normal two circles, and $A$ is a really thin ellipse passing through the middle of the two.
As far as the proof is concerned, you simply forgot that to prove a subset relation you need to show that all elements of one set belong to another. As soon you as you have chosen your $x$ you break into cases, and each time it may be different.
answered Dec 8 '18 at 21:12
RandomMathGuyRandomMathGuy
462
$endgroup$
$begingroup$
"As you have chosen your $x$ you break into cases, and each time it may be different." What do you mean by this? It seems as though the $xin A$ is an arbitrary element of A, and hence the proof should apply to all $xin A$. I have already drawn that Venn diagram, and I understand the claim is false. I am still puzzled as to where the error is in the proof.
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:52
$begingroup$
@PlatonicTradition You did choose an arbitrary element of $A$, but what applies to all $xin A$ is the disjunction $xin B$ or $xin C$. When you move into your case by case analysis, you are no longer considering a truly arbitrary member of $A$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:58
$begingroup$
@PlatonicTradition Also, there was a missing word. I meant to say "As soon as you have chosen"
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:59
$begingroup$
@PlatonicTradition Additionally, your cases aren't technically correct either. Your first case can read $xin B$, but the second case should read $xnotin B$, thus $xin C$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 22:03
add a comment |
$begingroup$
Draw a Venn diagram where $B$ and $C$ are your normal two circles, and $A$ is a really thin ellipse passing through the middle of the two.
As far as the proof is concerned, you simply forgot that to prove a subset relation you need to show that all elements of one set belong to another. As soon you as you have chosen your $x$ you break into cases, and each time it may be different.
answered Dec 8 '18 at 21:12
RandomMathGuyRandomMathGuy
462
$endgroup$
$begingroup$
"As you have chosen your $x$ you break into cases, and each time it may be different." What do you mean by this? It seems as though the $xin A$ is an arbitrary element of A, and hence the proof should apply to all $xin A$. I have already drawn that Venn diagram, and I understand the claim is false. I am still puzzled as to where the error is in the proof.
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:52
$begingroup$
@PlatonicTradition You did choose an arbitrary element of $A$, but what applies to all $xin A$ is the disjunction $xin B$ or $xin C$. When you move into your case by case analysis, you are no longer considering a truly arbitrary member of $A$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:58
$begingroup$
@PlatonicTradition Also, there was a missing word. I meant to say "As soon as you have chosen"
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:59
$begingroup$
@PlatonicTradition Additionally, your cases aren't technically correct either. Your first case can read $xin B$, but the second case should read $xnotin B$, thus $xin C$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 22:03
add a comment |
$begingroup$
Draw a Venn diagram where $B$ and $C$ are your normal two circles, and $A$ is a really thin ellipse passing through the middle of the two.
As far as the proof is concerned, you simply forgot that to prove a subset relation you need to show that all elements of one set belong to another. As soon you as you have chosen your $x$ you break into cases, and each time it may be different.
answered Dec 8 '18 at 21:12
RandomMathGuyRandomMathGuy
462
$endgroup$
Draw a Venn diagram where $B$ and $C$ are your normal two circles, and $A$ is a really thin ellipse passing through the middle of the two.
As far as the proof is concerned, you simply forgot that to prove a subset relation you need to show that all elements of one set belong to another. As soon you as you have chosen your $x$ you break into cases, and each time it may be different.
answered Dec 8 '18 at 21:12
RandomMathGuyRandomMathGuy
462
edited Dec 8 '18 at 21:58
answered Dec 8 '18 at 21:12
RandomMathGuyRandomMathGuy
462
answered Dec 8 '18 at 21:12
RandomMathGuyRandomMathGuy
462
answered Dec 8 '18 at 21:12
RandomMathGuyRandomMathGuy
462
462
$begingroup$
"As you have chosen your $x$ you break into cases, and each time it may be different." What do you mean by this? It seems as though the $xin A$ is an arbitrary element of A, and hence the proof should apply to all $xin A$. I have already drawn that Venn diagram, and I understand the claim is false. I am still puzzled as to where the error is in the proof.
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:52
$begingroup$
@PlatonicTradition You did choose an arbitrary element of $A$, but what applies to all $xin A$ is the disjunction $xin B$ or $xin C$. When you move into your case by case analysis, you are no longer considering a truly arbitrary member of $A$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:58
$begingroup$
@PlatonicTradition Also, there was a missing word. I meant to say "As soon as you have chosen"
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:59
$begingroup$
@PlatonicTradition Additionally, your cases aren't technically correct either. Your first case can read $xin B$, but the second case should read $xnotin B$, thus $xin C$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 22:03
add a comment |
$begingroup$
"As you have chosen your $x$ you break into cases, and each time it may be different." What do you mean by this? It seems as though the $xin A$ is an arbitrary element of A, and hence the proof should apply to all $xin A$. I have already drawn that Venn diagram, and I understand the claim is false. I am still puzzled as to where the error is in the proof.
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:52
$begingroup$
@PlatonicTradition You did choose an arbitrary element of $A$, but what applies to all $xin A$ is the disjunction $xin B$ or $xin C$. When you move into your case by case analysis, you are no longer considering a truly arbitrary member of $A$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:58
$begingroup$
@PlatonicTradition Also, there was a missing word. I meant to say "As soon as you have chosen"
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:59
$begingroup$
@PlatonicTradition Additionally, your cases aren't technically correct either. Your first case can read $xin B$, but the second case should read $xnotin B$, thus $xin C$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 22:03
$begingroup$
"As you have chosen your $x$ you break into cases, and each time it may be different." What do you mean by this? It seems as though the $xin A$ is an arbitrary element of A, and hence the proof should apply to all $xin A$. I have already drawn that Venn diagram, and I understand the claim is false. I am still puzzled as to where the error is in the proof.
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:52
$begingroup$
"As you have chosen your $x$ you break into cases, and each time it may be different." What do you mean by this? It seems as though the $xin A$ is an arbitrary element of A, and hence the proof should apply to all $xin A$. I have already drawn that Venn diagram, and I understand the claim is false. I am still puzzled as to where the error is in the proof.
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:52
$begingroup$
@PlatonicTradition You did choose an arbitrary element of $A$, but what applies to all $xin A$ is the disjunction $xin B$ or $xin C$. When you move into your case by case analysis, you are no longer considering a truly arbitrary member of $A$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:58
$begingroup$
@PlatonicTradition You did choose an arbitrary element of $A$, but what applies to all $xin A$ is the disjunction $xin B$ or $xin C$. When you move into your case by case analysis, you are no longer considering a truly arbitrary member of $A$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:58
$begingroup$
@PlatonicTradition Also, there was a missing word. I meant to say "As soon as you have chosen"
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:59
$begingroup$
@PlatonicTradition Also, there was a missing word. I meant to say "As soon as you have chosen"
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:59
$begingroup$
@PlatonicTradition Additionally, your cases aren't technically correct either. Your first case can read $xin B$, but the second case should read $xnotin B$, thus $xin C$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 22:03
$begingroup$
@PlatonicTradition Additionally, your cases aren't technically correct either. Your first case can read $xin B$, but the second case should read $xnotin B$, thus $xin C$.
$endgroup$
– RandomMathGuy
Dec 8 '18 at 22:03
add a comment |
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$begingroup$
$x$ is just one element. You have to prove one or the other for every element. You might have $xin A subset B; x not in C$ but then also have $y in A subset C; y not in C$.
$endgroup$
– fleablood
Dec 8 '18 at 21:15
$begingroup$
In the above proof, how is $xin A$ not an arbitrary element?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:42
$begingroup$
Because you said "suppose $x in B$". You can't say that for an arbitrary element. Once you suppose $x in B$ it is no longer an arbitrary element in $A$. It now has a specific property (that is is also in $B$) that not all elements have. SO it can not be arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:50
$begingroup$
That makes sense. But, since $xin B cup C$, is it unsafe to then say that $x$ has to be in either $B$ or $C$? And then prove by cases?
$endgroup$
– PlatonicTradition
Dec 8 '18 at 21:54
$begingroup$
"arbitrary" a label some elements have. $2$ is not an arbitrary element, neither is $3$ or $4$. So ... none of the elements in $A$ are arbitrary? Well, that's word play. An "arbitrary element" is one that we don't KNOW any specific quality. One way say, let's suppose $x$ is even, we are assuming specifics and it can't claim it is arbitrary.
$endgroup$
– fleablood
Dec 8 '18 at 21:56