Is $∅ ∈ ∅ $ true?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
If $ emptyset ∈ emptyset,emptyset $ is true, does it mean this $ emptyset in emptyset $ true ? If it is not, why it is false?
Also, does $ emptyset$ mean $emptyset,emptyset,emptyset$ ?
elementary-set-theory
edited Jan 21 at 9:19
Eevee Trainer
5,9471936
asked Jan 21 at 3:34
J.SJ.S
605
$endgroup$
add a comment |
$begingroup$
If $ emptyset ∈ emptyset,emptyset $ is true, does it mean this $ emptyset in emptyset $ true ? If it is not, why it is false?
Also, does $ emptyset$ mean $emptyset,emptyset,emptyset$ ?
elementary-set-theory
edited Jan 21 at 9:19
Eevee Trainer
5,9471936
asked Jan 21 at 3:34
J.SJ.S
605
$endgroup$
add a comment |
$begingroup$
If $ emptyset ∈ emptyset,emptyset $ is true, does it mean this $ emptyset in emptyset $ true ? If it is not, why it is false?
Also, does $ emptyset$ mean $emptyset,emptyset,emptyset$ ?
elementary-set-theory
edited Jan 21 at 9:19
Eevee Trainer
5,9471936
asked Jan 21 at 3:34
J.SJ.S
605
$endgroup$
If $ emptyset ∈ emptyset,emptyset $ is true, does it mean this $ emptyset in emptyset $ true ? If it is not, why it is false?
Also, does $ emptyset$ mean $emptyset,emptyset,emptyset$ ?
elementary-set-theory
elementary-set-theory
edited Jan 21 at 9:19
Eevee Trainer
5,9471936
asked Jan 21 at 3:34
J.SJ.S
605
edited Jan 21 at 9:19
Eevee Trainer
5,9471936
asked Jan 21 at 3:34
J.SJ.S
605
edited Jan 21 at 9:19
Eevee Trainer
5,9471936
edited Jan 21 at 9:19
Eevee Trainer
5,9471936
edited Jan 21 at 9:19
Eevee Trainer
5,9471936
5,9471936
asked Jan 21 at 3:34
J.SJ.S
605
asked Jan 21 at 3:34
J.SJ.S
605
asked Jan 21 at 3:34
J.SJ.S
605
605
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The notation $a in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$
The notation $emptyset, emptyset$ describes a set with exactly two elements.
The first element is $emptyset.$ The second element is $emptyset.$
Is one of those two elements exactly equal to $emptyset$?
The notation $ emptyset$ describes a set with one element.
That element is $emptyset.$
Which element of $ emptyset$ do you think is exactly equal to $emptyset$?
Hint: there's only one element you have to check.
The notation $emptyset, emptyset, emptyset$ again describes a set with two elements.
One element is $emptyset$ and the other is
$emptyset, emptyset.$
So this is definitely not the same thing as any set that has only one element.
answered Jan 21 at 3:51
David KDavid K
53.9k342116
$endgroup$
$begingroup$
This is not an answer but a counter-question.
$endgroup$
– rexkogitans
Jan 21 at 9:36
2
$begingroup$
@rexkogitans Also known as a "hint"
$endgroup$
– TreFox
Jan 21 at 19:36
add a comment |
$begingroup$
The set $emptyset$ is a set with one element: $emptyset$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).
Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.
This also explains why $emptyset notequiv emptyset,emptyset,emptyset$.
answered Jan 21 at 9:08
AdrianAdrian
562
$endgroup$
add a comment |
$begingroup$
Let us begin with naming the sets in the question: The set $B=a$ has a single element, which is $a$. Now, let $a=emptyset$, then we have $B=emptyset$, which is of course also a set with a single elementy, namely $emptyset$.
If we put $A$ into a set $C=B$, then we have $C=emptyset$. It can be easily seen that $a notin C$. Hence, $emptyset notin emptyset$.
The set $A=emptyset, b$ has two elements. Now, let $b=emptyset$. Then we have the set $A=emptyset, emptyset$. If we now take a look at $b$, then, since $bin A$, also $emptyset in A$, which is $emptyset in emptyset, emptyset$.
To answer if $emptyset$ means $emptyset,emptyset,emptyset$:
Look at $C$, it has only one element, which is $B$. It follows that $emptyset notin C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).
answered Jan 21 at 9:36
rexkogitansrexkogitans
1515
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add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
The notation $a in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$
The notation $emptyset, emptyset$ describes a set with exactly two elements.
The first element is $emptyset.$ The second element is $emptyset.$
Is one of those two elements exactly equal to $emptyset$?
The notation $ emptyset$ describes a set with one element.
That element is $emptyset.$
Which element of $ emptyset$ do you think is exactly equal to $emptyset$?
Hint: there's only one element you have to check.
The notation $emptyset, emptyset, emptyset$ again describes a set with two elements.
One element is $emptyset$ and the other is
$emptyset, emptyset.$
So this is definitely not the same thing as any set that has only one element.
answered Jan 21 at 3:51
David KDavid K
53.9k342116
$endgroup$
$begingroup$
This is not an answer but a counter-question.
$endgroup$
– rexkogitans
Jan 21 at 9:36
2
$begingroup$
@rexkogitans Also known as a "hint"
$endgroup$
– TreFox
Jan 21 at 19:36
add a comment |
$begingroup$
The notation $a in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$
The notation $emptyset, emptyset$ describes a set with exactly two elements.
The first element is $emptyset.$ The second element is $emptyset.$
Is one of those two elements exactly equal to $emptyset$?
The notation $ emptyset$ describes a set with one element.
That element is $emptyset.$
Which element of $ emptyset$ do you think is exactly equal to $emptyset$?
Hint: there's only one element you have to check.
The notation $emptyset, emptyset, emptyset$ again describes a set with two elements.
One element is $emptyset$ and the other is
$emptyset, emptyset.$
So this is definitely not the same thing as any set that has only one element.
answered Jan 21 at 3:51
David KDavid K
53.9k342116
$endgroup$
$begingroup$
This is not an answer but a counter-question.
$endgroup$
– rexkogitans
Jan 21 at 9:36
2
$begingroup$
@rexkogitans Also known as a "hint"
$endgroup$
– TreFox
Jan 21 at 19:36
add a comment |
$begingroup$
The notation $a in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$
The notation $emptyset, emptyset$ describes a set with exactly two elements.
The first element is $emptyset.$ The second element is $emptyset.$
Is one of those two elements exactly equal to $emptyset$?
The notation $ emptyset$ describes a set with one element.
That element is $emptyset.$
Which element of $ emptyset$ do you think is exactly equal to $emptyset$?
Hint: there's only one element you have to check.
The notation $emptyset, emptyset, emptyset$ again describes a set with two elements.
One element is $emptyset$ and the other is
$emptyset, emptyset.$
So this is definitely not the same thing as any set that has only one element.
answered Jan 21 at 3:51
David KDavid K
53.9k342116
$endgroup$
The notation $a in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$
The notation $emptyset, emptyset$ describes a set with exactly two elements.
The first element is $emptyset.$ The second element is $emptyset.$
Is one of those two elements exactly equal to $emptyset$?
The notation $ emptyset$ describes a set with one element.
That element is $emptyset.$
Which element of $ emptyset$ do you think is exactly equal to $emptyset$?
Hint: there's only one element you have to check.
The notation $emptyset, emptyset, emptyset$ again describes a set with two elements.
One element is $emptyset$ and the other is
$emptyset, emptyset.$
So this is definitely not the same thing as any set that has only one element.
answered Jan 21 at 3:51
David KDavid K
53.9k342116
answered Jan 21 at 3:51
David KDavid K
53.9k342116
answered Jan 21 at 3:51
David KDavid K
53.9k342116
answered Jan 21 at 3:51
David KDavid K
53.9k342116
53.9k342116
$begingroup$
This is not an answer but a counter-question.
$endgroup$
– rexkogitans
Jan 21 at 9:36
2
$begingroup$
@rexkogitans Also known as a "hint"
$endgroup$
– TreFox
Jan 21 at 19:36
add a comment |
$begingroup$
This is not an answer but a counter-question.
$endgroup$
– rexkogitans
Jan 21 at 9:36
2
$begingroup$
@rexkogitans Also known as a "hint"
$endgroup$
– TreFox
Jan 21 at 19:36
$begingroup$
This is not an answer but a counter-question.
$endgroup$
– rexkogitans
Jan 21 at 9:36
$begingroup$
This is not an answer but a counter-question.
$endgroup$
– rexkogitans
Jan 21 at 9:36
2
2
$begingroup$
@rexkogitans Also known as a "hint"
$endgroup$
– TreFox
Jan 21 at 19:36
$begingroup$
@rexkogitans Also known as a "hint"
$endgroup$
– TreFox
Jan 21 at 19:36
add a comment |
$begingroup$
The set $emptyset$ is a set with one element: $emptyset$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).
Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.
This also explains why $emptyset notequiv emptyset,emptyset,emptyset$.
answered Jan 21 at 9:08
AdrianAdrian
562
$endgroup$
add a comment |
$begingroup$
The set $emptyset$ is a set with one element: $emptyset$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).
Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.
This also explains why $emptyset notequiv emptyset,emptyset,emptyset$.
answered Jan 21 at 9:08
AdrianAdrian
562
$endgroup$
add a comment |
$begingroup$
The set $emptyset$ is a set with one element: $emptyset$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).
Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.
This also explains why $emptyset notequiv emptyset,emptyset,emptyset$.
answered Jan 21 at 9:08
AdrianAdrian
562
$endgroup$
The set $emptyset$ is a set with one element: $emptyset$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).
Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.
This also explains why $emptyset notequiv emptyset,emptyset,emptyset$.
answered Jan 21 at 9:08
AdrianAdrian
562
answered Jan 21 at 9:08
AdrianAdrian
562
answered Jan 21 at 9:08
AdrianAdrian
562
answered Jan 21 at 9:08
AdrianAdrian
562
562
add a comment |
add a comment |
$begingroup$
Let us begin with naming the sets in the question: The set $B=a$ has a single element, which is $a$. Now, let $a=emptyset$, then we have $B=emptyset$, which is of course also a set with a single elementy, namely $emptyset$.
If we put $A$ into a set $C=B$, then we have $C=emptyset$. It can be easily seen that $a notin C$. Hence, $emptyset notin emptyset$.
The set $A=emptyset, b$ has two elements. Now, let $b=emptyset$. Then we have the set $A=emptyset, emptyset$. If we now take a look at $b$, then, since $bin A$, also $emptyset in A$, which is $emptyset in emptyset, emptyset$.
To answer if $emptyset$ means $emptyset,emptyset,emptyset$:
Look at $C$, it has only one element, which is $B$. It follows that $emptyset notin C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).
answered Jan 21 at 9:36
rexkogitansrexkogitans
1515
$endgroup$
add a comment |
$begingroup$
Let us begin with naming the sets in the question: The set $B=a$ has a single element, which is $a$. Now, let $a=emptyset$, then we have $B=emptyset$, which is of course also a set with a single elementy, namely $emptyset$.
If we put $A$ into a set $C=B$, then we have $C=emptyset$. It can be easily seen that $a notin C$. Hence, $emptyset notin emptyset$.
The set $A=emptyset, b$ has two elements. Now, let $b=emptyset$. Then we have the set $A=emptyset, emptyset$. If we now take a look at $b$, then, since $bin A$, also $emptyset in A$, which is $emptyset in emptyset, emptyset$.
To answer if $emptyset$ means $emptyset,emptyset,emptyset$:
Look at $C$, it has only one element, which is $B$. It follows that $emptyset notin C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).
answered Jan 21 at 9:36
rexkogitansrexkogitans
1515
$endgroup$
add a comment |
$begingroup$
Let us begin with naming the sets in the question: The set $B=a$ has a single element, which is $a$. Now, let $a=emptyset$, then we have $B=emptyset$, which is of course also a set with a single elementy, namely $emptyset$.
If we put $A$ into a set $C=B$, then we have $C=emptyset$. It can be easily seen that $a notin C$. Hence, $emptyset notin emptyset$.
The set $A=emptyset, b$ has two elements. Now, let $b=emptyset$. Then we have the set $A=emptyset, emptyset$. If we now take a look at $b$, then, since $bin A$, also $emptyset in A$, which is $emptyset in emptyset, emptyset$.
To answer if $emptyset$ means $emptyset,emptyset,emptyset$:
Look at $C$, it has only one element, which is $B$. It follows that $emptyset notin C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).
answered Jan 21 at 9:36
rexkogitansrexkogitans
1515
$endgroup$
Let us begin with naming the sets in the question: The set $B=a$ has a single element, which is $a$. Now, let $a=emptyset$, then we have $B=emptyset$, which is of course also a set with a single elementy, namely $emptyset$.
If we put $A$ into a set $C=B$, then we have $C=emptyset$. It can be easily seen that $a notin C$. Hence, $emptyset notin emptyset$.
The set $A=emptyset, b$ has two elements. Now, let $b=emptyset$. Then we have the set $A=emptyset, emptyset$. If we now take a look at $b$, then, since $bin A$, also $emptyset in A$, which is $emptyset in emptyset, emptyset$.
To answer if $emptyset$ means $emptyset,emptyset,emptyset$:
Look at $C$, it has only one element, which is $B$. It follows that $emptyset notin C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).
answered Jan 21 at 9:36
rexkogitansrexkogitans
1515
answered Jan 21 at 9:36
rexkogitansrexkogitans
1515
answered Jan 21 at 9:36
rexkogitansrexkogitans
1515
answered Jan 21 at 9:36
rexkogitansrexkogitans
1515
1515
add a comment |
add a comment |
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