My proof that the empty set is unique
$begingroup$
I'm trying to prove that the empty set is unique.
Proof:
Let $U = { a }$ be the universal set.
Assume $a notin emptyset '$ and $a notin emptyset$.
Without loss of generality, since $a notin emptyset'$, $emptyset '$ does not contain any elements. Since $emptyset '$ does not contain any elements, it must by default be a subset of $emptyset$, since the conditional statement
$$a in emptyset ' Rightarrow emptyset ' subseteq emptyset$$
is vacuously true.
Therefore, since $emptyset' subseteq emptyset$ and $emptyset subseteq emptyset '$, we have that $emptyset ' = emptyset$. $tag*{$blacksquare$}$
I would appreciate it if people could please provide feedback as to the correctness of my proof.
EDIT: Please be specific about what is incorrect and why. That way, I can learn what I did wrong and improve much more effectively.
proof-verification elementary-set-theory
asked Jan 6 at 17:11
The PointerThe Pointer
2,48731641
$endgroup$
add a comment |
$begingroup$
I'm trying to prove that the empty set is unique.
Proof:
Let $U = { a }$ be the universal set.
Assume $a notin emptyset '$ and $a notin emptyset$.
Without loss of generality, since $a notin emptyset'$, $emptyset '$ does not contain any elements. Since $emptyset '$ does not contain any elements, it must by default be a subset of $emptyset$, since the conditional statement
$$a in emptyset ' Rightarrow emptyset ' subseteq emptyset$$
is vacuously true.
Therefore, since $emptyset' subseteq emptyset$ and $emptyset subseteq emptyset '$, we have that $emptyset ' = emptyset$. $tag*{$blacksquare$}$
I would appreciate it if people could please provide feedback as to the correctness of my proof.
EDIT: Please be specific about what is incorrect and why. That way, I can learn what I did wrong and improve much more effectively.
proof-verification elementary-set-theory
asked Jan 6 at 17:11
The PointerThe Pointer
2,48731641
$endgroup$
4
$begingroup$
You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
$endgroup$
– Hagen von Eitzen
Jan 6 at 17:22
$begingroup$
@HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
$endgroup$
– The Pointer
Jan 6 at 17:25
add a comment |
$begingroup$
I'm trying to prove that the empty set is unique.
Proof:
Let $U = { a }$ be the universal set.
Assume $a notin emptyset '$ and $a notin emptyset$.
Without loss of generality, since $a notin emptyset'$, $emptyset '$ does not contain any elements. Since $emptyset '$ does not contain any elements, it must by default be a subset of $emptyset$, since the conditional statement
$$a in emptyset ' Rightarrow emptyset ' subseteq emptyset$$
is vacuously true.
Therefore, since $emptyset' subseteq emptyset$ and $emptyset subseteq emptyset '$, we have that $emptyset ' = emptyset$. $tag*{$blacksquare$}$
I would appreciate it if people could please provide feedback as to the correctness of my proof.
EDIT: Please be specific about what is incorrect and why. That way, I can learn what I did wrong and improve much more effectively.
proof-verification elementary-set-theory
asked Jan 6 at 17:11
The PointerThe Pointer
2,48731641
$endgroup$
I'm trying to prove that the empty set is unique.
Proof:
Let $U = { a }$ be the universal set.
Assume $a notin emptyset '$ and $a notin emptyset$.
Without loss of generality, since $a notin emptyset'$, $emptyset '$ does not contain any elements. Since $emptyset '$ does not contain any elements, it must by default be a subset of $emptyset$, since the conditional statement
$$a in emptyset ' Rightarrow emptyset ' subseteq emptyset$$
is vacuously true.
Therefore, since $emptyset' subseteq emptyset$ and $emptyset subseteq emptyset '$, we have that $emptyset ' = emptyset$. $tag*{$blacksquare$}$
I would appreciate it if people could please provide feedback as to the correctness of my proof.
EDIT: Please be specific about what is incorrect and why. That way, I can learn what I did wrong and improve much more effectively.
proof-verification elementary-set-theory
proof-verification elementary-set-theory
asked Jan 6 at 17:11
The PointerThe Pointer
2,48731641
asked Jan 6 at 17:11
The PointerThe Pointer
2,48731641
edited Jan 6 at 17:37
The Pointer
asked Jan 6 at 17:11
The PointerThe Pointer
2,48731641
asked Jan 6 at 17:11
The PointerThe Pointer
2,48731641
asked Jan 6 at 17:11
The PointerThe Pointer
2,48731641
2,48731641
4
$begingroup$
You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
$endgroup$
– Hagen von Eitzen
Jan 6 at 17:22
$begingroup$
@HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
$endgroup$
– The Pointer
Jan 6 at 17:25
add a comment |
4
$begingroup$
You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
$endgroup$
– Hagen von Eitzen
Jan 6 at 17:22
$begingroup$
@HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
$endgroup$
– The Pointer
Jan 6 at 17:25
4
4
$begingroup$
You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
$endgroup$
– Hagen von Eitzen
Jan 6 at 17:22
$begingroup$
You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
$endgroup$
– Hagen von Eitzen
Jan 6 at 17:22
$begingroup$
@HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
$endgroup$
– The Pointer
Jan 6 at 17:25
$begingroup$
@HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
$endgroup$
– The Pointer
Jan 6 at 17:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A subseteq B$. Similarly, $B subseteq A$. Hence $A=B$.
EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $anotin emptyset'$ and $emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.
answered Jan 6 at 17:32
EuxhenHEuxhenH
484211
$endgroup$
add a comment |
$begingroup$
Let $A$ and $B$ be two empty sets.
Then the assertions $xin A$ and $xin B$ are logically equivalent.
By the definition of equality of sets, $A=B$ iff $forall x(xin ALongleftrightarrow xin B)$, it follows that $A=B$.
answered Jan 6 at 17:23
WuestenfuxWuestenfux
5,4841513
$endgroup$
1
$begingroup$
This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
$endgroup$
– Holo
Jan 6 at 17:25
$begingroup$
@SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
$endgroup$
– Holo
Jan 6 at 17:45
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A subseteq B$. Similarly, $B subseteq A$. Hence $A=B$.
EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $anotin emptyset'$ and $emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.
answered Jan 6 at 17:32
EuxhenHEuxhenH
484211
$endgroup$
add a comment |
$begingroup$
The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A subseteq B$. Similarly, $B subseteq A$. Hence $A=B$.
EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $anotin emptyset'$ and $emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.
answered Jan 6 at 17:32
EuxhenHEuxhenH
484211
$endgroup$
add a comment |
$begingroup$
The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A subseteq B$. Similarly, $B subseteq A$. Hence $A=B$.
EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $anotin emptyset'$ and $emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.
answered Jan 6 at 17:32
EuxhenHEuxhenH
484211
$endgroup$
The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A subseteq B$. Similarly, $B subseteq A$. Hence $A=B$.
EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $anotin emptyset'$ and $emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.
answered Jan 6 at 17:32
EuxhenHEuxhenH
484211
edited Jan 7 at 16:37
answered Jan 6 at 17:32
EuxhenHEuxhenH
484211
answered Jan 6 at 17:32
EuxhenHEuxhenH
484211
answered Jan 6 at 17:32
EuxhenHEuxhenH
484211
484211
add a comment |
add a comment |
$begingroup$
Let $A$ and $B$ be two empty sets.
Then the assertions $xin A$ and $xin B$ are logically equivalent.
By the definition of equality of sets, $A=B$ iff $forall x(xin ALongleftrightarrow xin B)$, it follows that $A=B$.
answered Jan 6 at 17:23
WuestenfuxWuestenfux
5,4841513
$endgroup$
1
$begingroup$
This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
$endgroup$
– Holo
Jan 6 at 17:25
$begingroup$
@SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
$endgroup$
– Holo
Jan 6 at 17:45
add a comment |
$begingroup$
Let $A$ and $B$ be two empty sets.
Then the assertions $xin A$ and $xin B$ are logically equivalent.
By the definition of equality of sets, $A=B$ iff $forall x(xin ALongleftrightarrow xin B)$, it follows that $A=B$.
answered Jan 6 at 17:23
WuestenfuxWuestenfux
5,4841513
$endgroup$
1
$begingroup$
This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
$endgroup$
– Holo
Jan 6 at 17:25
$begingroup$
@SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
$endgroup$
– Holo
Jan 6 at 17:45
add a comment |
$begingroup$
Let $A$ and $B$ be two empty sets.
Then the assertions $xin A$ and $xin B$ are logically equivalent.
By the definition of equality of sets, $A=B$ iff $forall x(xin ALongleftrightarrow xin B)$, it follows that $A=B$.
answered Jan 6 at 17:23
WuestenfuxWuestenfux
5,4841513
$endgroup$
Let $A$ and $B$ be two empty sets.
Then the assertions $xin A$ and $xin B$ are logically equivalent.
By the definition of equality of sets, $A=B$ iff $forall x(xin ALongleftrightarrow xin B)$, it follows that $A=B$.
answered Jan 6 at 17:23
WuestenfuxWuestenfux
5,4841513
answered Jan 6 at 17:23
WuestenfuxWuestenfux
5,4841513
answered Jan 6 at 17:23
WuestenfuxWuestenfux
5,4841513
answered Jan 6 at 17:23
WuestenfuxWuestenfux
5,4841513
5,4841513
1
$begingroup$
This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
$endgroup$
– Holo
Jan 6 at 17:25
$begingroup$
@SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
$endgroup$
– Holo
Jan 6 at 17:45
add a comment |
1
$begingroup$
This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
$endgroup$
– Holo
Jan 6 at 17:25
$begingroup$
@SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
$endgroup$
– Holo
Jan 6 at 17:45
1
1
$begingroup$
This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
$endgroup$
– Holo
Jan 6 at 17:25
$begingroup$
This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
$endgroup$
– Holo
Jan 6 at 17:25
$begingroup$
@SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
$endgroup$
– Holo
Jan 6 at 17:45
$begingroup$
@SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
$endgroup$
– Holo
Jan 6 at 17:45
add a comment |
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$begingroup$
You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
$endgroup$
– Hagen von Eitzen
Jan 6 at 17:22
$begingroup$
@HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
$endgroup$
– The Pointer
Jan 6 at 17:25