Is $∅ ∈ { {∅} }$ true?
$begingroup$
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
5,8011936
asked Jan 21 at 3:34
J.SJ.S
555
$endgroup$
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
5,8011936
asked Jan 21 at 3:34
J.SJ.S
555
$endgroup$
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
5,8011936
asked Jan 21 at 3:34
J.SJ.S
555
$endgroup$
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
5,8011936
asked Jan 21 at 3:34
J.SJ.S
555
edited Jan 21 at 9:19
Eevee Trainer
5,8011936
asked Jan 21 at 3:34
J.SJ.S
555
edited Jan 21 at 9:19
Eevee Trainer
5,8011936
edited Jan 21 at 9:19
Eevee Trainer
5,8011936
edited Jan 21 at 9:19
Eevee Trainer
5,8011936
5,8011936
asked Jan 21 at 3:34
J.SJ.S
555
asked Jan 21 at 3:34
J.SJ.S
555
asked Jan 21 at 3:34
J.SJ.S
555
555
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
462
$endgroup$
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 |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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
462
$endgroup$
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
462
$endgroup$
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
462
$endgroup$
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
462
answered Jan 21 at 9:08
AdrianAdrian
462
answered Jan 21 at 9:08
AdrianAdrian
462
answered Jan 21 at 9:08
AdrianAdrian
462
462
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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