Proving set identities: empty set case.
$begingroup$
I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$
I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.
We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.
But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.
In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?
If so, some suggestions as to how to incorporate them into proofs would be helpful :-).
elementary-set-theory proof-writing
asked Dec 22 '18 at 18:24
E-muE-mu
767417
$endgroup$
add a comment |
$begingroup$
I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$
I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.
We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.
But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.
In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?
If so, some suggestions as to how to incorporate them into proofs would be helpful :-).
elementary-set-theory proof-writing
asked Dec 22 '18 at 18:24
E-muE-mu
767417
$endgroup$
5
$begingroup$
Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
$endgroup$
– DonAntonio
Dec 22 '18 at 18:35
$begingroup$
Words to live by: "vacuous truth". I am almost tempted to say this is a duplicate of many other questions that were asked before. For example, math.stackexchange.com/questions/734418/… math.stackexchange.com/questions/2723860/… math.stackexchange.com/questions/1953218/… and there are many others which are similar.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 10:24
add a comment |
$begingroup$
I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$
I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.
We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.
But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.
In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?
If so, some suggestions as to how to incorporate them into proofs would be helpful :-).
elementary-set-theory proof-writing
asked Dec 22 '18 at 18:24
E-muE-mu
767417
$endgroup$
I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$
I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.
We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.
But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.
In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?
If so, some suggestions as to how to incorporate them into proofs would be helpful :-).
elementary-set-theory proof-writing
elementary-set-theory proof-writing
asked Dec 22 '18 at 18:24
E-muE-mu
767417
asked Dec 22 '18 at 18:24
E-muE-mu
767417
asked Dec 22 '18 at 18:24
E-muE-mu
767417
asked Dec 22 '18 at 18:24
E-muE-mu
767417
asked Dec 22 '18 at 18:24
E-muE-mu
767417
767417
5
$begingroup$
Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
$endgroup$
– DonAntonio
Dec 22 '18 at 18:35
$begingroup$
Words to live by: "vacuous truth". I am almost tempted to say this is a duplicate of many other questions that were asked before. For example, math.stackexchange.com/questions/734418/… math.stackexchange.com/questions/2723860/… math.stackexchange.com/questions/1953218/… and there are many others which are similar.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 10:24
add a comment |
5
$begingroup$
Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
$endgroup$
– DonAntonio
Dec 22 '18 at 18:35
$begingroup$
Words to live by: "vacuous truth". I am almost tempted to say this is a duplicate of many other questions that were asked before. For example, math.stackexchange.com/questions/734418/… math.stackexchange.com/questions/2723860/… math.stackexchange.com/questions/1953218/… and there are many others which are similar.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 10:24
5
5
$begingroup$
Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
$endgroup$
– DonAntonio
Dec 22 '18 at 18:35
$begingroup$
Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
$endgroup$
– DonAntonio
Dec 22 '18 at 18:35
$begingroup$
Words to live by: "vacuous truth". I am almost tempted to say this is a duplicate of many other questions that were asked before. For example, math.stackexchange.com/questions/734418/… math.stackexchange.com/questions/2723860/… math.stackexchange.com/questions/1953218/… and there are many others which are similar.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 10:24
$begingroup$
Words to live by: "vacuous truth". I am almost tempted to say this is a duplicate of many other questions that were asked before. For example, math.stackexchange.com/questions/734418/… math.stackexchange.com/questions/2723860/… math.stackexchange.com/questions/1953218/… and there are many others which are similar.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 10:24
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that
for every $x$, if $xin X$, then $xin Y$.
Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when
either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true
If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.
The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.
answered Dec 22 '18 at 18:43
egregegreg
180k1485202
$endgroup$
1
$begingroup$
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
$endgroup$
– E-mu
Dec 22 '18 at 19:39
$begingroup$
@E-mu Basically so
$endgroup$
– egreg
Dec 22 '18 at 20:27
add a comment |
$begingroup$
If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.
It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
answered Dec 22 '18 at 18:45
drhabdrhab
99.1k544130
$endgroup$
add a comment |
$begingroup$
Since the empty set is a subset of any set, there is no need of including that in a formal proof.
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .
answered Dec 22 '18 at 18:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
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
$begingroup$
In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that
for every $x$, if $xin X$, then $xin Y$.
Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when
either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true
If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.
The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.
answered Dec 22 '18 at 18:43
egregegreg
180k1485202
$endgroup$
1
$begingroup$
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
$endgroup$
– E-mu
Dec 22 '18 at 19:39
$begingroup$
@E-mu Basically so
$endgroup$
– egreg
Dec 22 '18 at 20:27
add a comment |
$begingroup$
In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that
for every $x$, if $xin X$, then $xin Y$.
Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when
either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true
If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.
The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.
answered Dec 22 '18 at 18:43
egregegreg
180k1485202
$endgroup$
1
$begingroup$
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
$endgroup$
– E-mu
Dec 22 '18 at 19:39
$begingroup$
@E-mu Basically so
$endgroup$
– egreg
Dec 22 '18 at 20:27
add a comment |
$begingroup$
In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that
for every $x$, if $xin X$, then $xin Y$.
Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when
either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true
If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.
The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.
answered Dec 22 '18 at 18:43
egregegreg
180k1485202
$endgroup$
In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that
for every $x$, if $xin X$, then $xin Y$.
Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when
either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true
If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.
The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.
answered Dec 22 '18 at 18:43
egregegreg
180k1485202
answered Dec 22 '18 at 18:43
egregegreg
180k1485202
answered Dec 22 '18 at 18:43
egregegreg
180k1485202
answered Dec 22 '18 at 18:43
egregegreg
180k1485202
180k1485202
1
$begingroup$
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
$endgroup$
– E-mu
Dec 22 '18 at 19:39
$begingroup$
@E-mu Basically so
$endgroup$
– egreg
Dec 22 '18 at 20:27
add a comment |
1
$begingroup$
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
$endgroup$
– E-mu
Dec 22 '18 at 19:39
$begingroup$
@E-mu Basically so
$endgroup$
– egreg
Dec 22 '18 at 20:27
1
1
$begingroup$
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
$endgroup$
– E-mu
Dec 22 '18 at 19:39
$begingroup$
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
$endgroup$
– E-mu
Dec 22 '18 at 19:39
$begingroup$
@E-mu Basically so
$endgroup$
– egreg
Dec 22 '18 at 20:27
$begingroup$
@E-mu Basically so
$endgroup$
– egreg
Dec 22 '18 at 20:27
add a comment |
$begingroup$
If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.
It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
answered Dec 22 '18 at 18:45
drhabdrhab
99.1k544130
$endgroup$
add a comment |
$begingroup$
If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.
It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
answered Dec 22 '18 at 18:45
drhabdrhab
99.1k544130
$endgroup$
add a comment |
$begingroup$
If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.
It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
answered Dec 22 '18 at 18:45
drhabdrhab
99.1k544130
$endgroup$
If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.
It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
answered Dec 22 '18 at 18:45
drhabdrhab
99.1k544130
edited Dec 22 '18 at 18:46
answered Dec 22 '18 at 18:45
drhabdrhab
99.1k544130
answered Dec 22 '18 at 18:45
drhabdrhab
99.1k544130
answered Dec 22 '18 at 18:45
drhabdrhab
99.1k544130
99.1k544130
add a comment |
add a comment |
$begingroup$
Since the empty set is a subset of any set, there is no need of including that in a formal proof.
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .
answered Dec 22 '18 at 18:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
Since the empty set is a subset of any set, there is no need of including that in a formal proof.
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .
answered Dec 22 '18 at 18:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
Since the empty set is a subset of any set, there is no need of including that in a formal proof.
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .
answered Dec 22 '18 at 18:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
Since the empty set is a subset of any set, there is no need of including that in a formal proof.
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .
answered Dec 22 '18 at 18:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Dec 22 '18 at 18:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Dec 22 '18 at 18:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Dec 22 '18 at 18:39
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
41.5k42061
add a comment |
add a comment |
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Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
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– DonAntonio
Dec 22 '18 at 18:35
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Words to live by: "vacuous truth". I am almost tempted to say this is a duplicate of many other questions that were asked before. For example, math.stackexchange.com/questions/734418/… math.stackexchange.com/questions/2723860/… math.stackexchange.com/questions/1953218/… and there are many others which are similar.
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– Asaf Karagila♦
Dec 23 '18 at 10:24