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Cardinality of $A = varnothing, B = varnothing , C = varnothing$
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Given three sets $A = varnothing, B = varnothing , C = varnothing$ what are the cardinalities of those sets ?
Obviously cardinality of $A$ is $0$ and cardinality of $B$ is $1$, but I am not sure about set $C$, because some sources say that cardinality of such set is $2$. Can you please clarify this to me ?
discrete-mathematics elementary-set-theory
edited Jul 29 at 19:53
Cornman
2,30021027
asked Jul 29 at 19:41
Kevin
184
add a comment |Â
up vote
1
down vote
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Given three sets $A = varnothing, B = varnothing , C = varnothing$ what are the cardinalities of those sets ?
Obviously cardinality of $A$ is $0$ and cardinality of $B$ is $1$, but I am not sure about set $C$, because some sources say that cardinality of such set is $2$. Can you please clarify this to me ?
discrete-mathematics elementary-set-theory
edited Jul 29 at 19:53
Cornman
2,30021027
asked Jul 29 at 19:41
Kevin
184
Have you computed its power set?
– Javi
Jul 29 at 19:43
2
By power do you mean cardinality?
– user223391
Jul 29 at 19:45
@ZacharySelk yes, edited
– Kevin
Jul 29 at 19:47
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given three sets $A = varnothing, B = varnothing , C = varnothing$ what are the cardinalities of those sets ?
Obviously cardinality of $A$ is $0$ and cardinality of $B$ is $1$, but I am not sure about set $C$, because some sources say that cardinality of such set is $2$. Can you please clarify this to me ?
discrete-mathematics elementary-set-theory
edited Jul 29 at 19:53
Cornman
2,30021027
asked Jul 29 at 19:41
Kevin
184
Given three sets $A = varnothing, B = varnothing , C = varnothing$ what are the cardinalities of those sets ?
Obviously cardinality of $A$ is $0$ and cardinality of $B$ is $1$, but I am not sure about set $C$, because some sources say that cardinality of such set is $2$. Can you please clarify this to me ?
discrete-mathematics elementary-set-theory
edited Jul 29 at 19:53
Cornman
2,30021027
asked Jul 29 at 19:41
Kevin
184
edited Jul 29 at 19:53
Cornman
2,30021027
edited Jul 29 at 19:53
Cornman
2,30021027
edited Jul 29 at 19:53
Cornman
2,30021027
2,30021027
asked Jul 29 at 19:41
Kevin
184
asked Jul 29 at 19:41
Kevin
184
asked Jul 29 at 19:41
Kevin
184
184
Have you computed its power set?
– Javi
Jul 29 at 19:43
2
By power do you mean cardinality?
– user223391
Jul 29 at 19:45
@ZacharySelk yes, edited
– Kevin
Jul 29 at 19:47
add a comment |Â
Have you computed its power set?
– Javi
Jul 29 at 19:43
2
By power do you mean cardinality?
– user223391
Jul 29 at 19:45
@ZacharySelk yes, edited
– Kevin
Jul 29 at 19:47
Have you computed its power set?
– Javi
Jul 29 at 19:43
Have you computed its power set?
– Javi
Jul 29 at 19:43
2
2
By power do you mean cardinality?
– user223391
Jul 29 at 19:45
By power do you mean cardinality?
– user223391
Jul 29 at 19:45
@ZacharySelk yes, edited
– Kevin
Jul 29 at 19:47
@ZacharySelk yes, edited
– Kevin
Jul 29 at 19:47
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
$$A = varnothing, B = varnothing , C = varnothing$$
The way you have it $B$ and $C$ both have cardinality of $1$
My guess is that you wanted $$C = varnothing ,varnothing$$
Which has cardinality $2$.
answered Jul 29 at 19:54
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
1
down vote
The cardinality of $A$ is 0. The cardinality of $B$ and $C$ is both 1. For $B$ it is clear. $C$ just contains one element as well. This is the set that contains the emptyset $emptyset$. Namely $B$. One could write $C=B$, which makes it clear.
answered Jul 29 at 19:50
Cornman
2,30021027
add a comment |Â
up vote
1
down vote
The power set of $emptyset$ is indeed $lbrace emptyset rbrace$, as the only subset of $emptyset$ is $emptyset$ itself.
The power set of $lbrace emptyset rbrace$ indeed does have two elements: $emptyset$ and the set itself: $lbrace emptyset rbrace$, thus making it $lbrace lbrace emptyset rbrace, emptyset rbrace$.
answered Jul 29 at 19:53
Theo Bendit
11.8k1841
There was a typo in the question. It is asked for the cardinality and not the power set.
– Cornman
Jul 29 at 19:54
1
@Cornman I'm not convinced. I think the OP is confused between cardinality and power set. I think the heart of the confusion is that he thinks $lbrace emptyset rbrace$ might be a one-element set that breaks the rule $|mathcalP(S)| = 2^S$.
– Theo Bendit
Jul 29 at 19:56
Possible, thats why I upvoted your answer.
– Cornman
Jul 29 at 19:57
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$A = varnothing, B = varnothing , C = varnothing$$
The way you have it $B$ and $C$ both have cardinality of $1$
My guess is that you wanted $$C = varnothing ,varnothing$$
Which has cardinality $2$.
answered Jul 29 at 19:54
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
2
down vote
accepted
$$A = varnothing, B = varnothing , C = varnothing$$
The way you have it $B$ and $C$ both have cardinality of $1$
My guess is that you wanted $$C = varnothing ,varnothing$$
Which has cardinality $2$.
answered Jul 29 at 19:54
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$A = varnothing, B = varnothing , C = varnothing$$
The way you have it $B$ and $C$ both have cardinality of $1$
My guess is that you wanted $$C = varnothing ,varnothing$$
Which has cardinality $2$.
answered Jul 29 at 19:54
Mohammad Riazi-Kermani
27.3k41851
$$A = varnothing, B = varnothing , C = varnothing$$
The way you have it $B$ and $C$ both have cardinality of $1$
My guess is that you wanted $$C = varnothing ,varnothing$$
Which has cardinality $2$.
answered Jul 29 at 19:54
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 19:54
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 19:54
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 19:54
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
up vote
1
down vote
The cardinality of $A$ is 0. The cardinality of $B$ and $C$ is both 1. For $B$ it is clear. $C$ just contains one element as well. This is the set that contains the emptyset $emptyset$. Namely $B$. One could write $C=B$, which makes it clear.
answered Jul 29 at 19:50
Cornman
2,30021027
add a comment |Â
up vote
1
down vote
The cardinality of $A$ is 0. The cardinality of $B$ and $C$ is both 1. For $B$ it is clear. $C$ just contains one element as well. This is the set that contains the emptyset $emptyset$. Namely $B$. One could write $C=B$, which makes it clear.
answered Jul 29 at 19:50
Cornman
2,30021027
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The cardinality of $A$ is 0. The cardinality of $B$ and $C$ is both 1. For $B$ it is clear. $C$ just contains one element as well. This is the set that contains the emptyset $emptyset$. Namely $B$. One could write $C=B$, which makes it clear.
answered Jul 29 at 19:50
Cornman
2,30021027
The cardinality of $A$ is 0. The cardinality of $B$ and $C$ is both 1. For $B$ it is clear. $C$ just contains one element as well. This is the set that contains the emptyset $emptyset$. Namely $B$. One could write $C=B$, which makes it clear.
answered Jul 29 at 19:50
Cornman
2,30021027
answered Jul 29 at 19:50
Cornman
2,30021027
answered Jul 29 at 19:50
Cornman
2,30021027
answered Jul 29 at 19:50
Cornman
2,30021027
2,30021027
add a comment |Â
add a comment |Â
up vote
1
down vote
The power set of $emptyset$ is indeed $lbrace emptyset rbrace$, as the only subset of $emptyset$ is $emptyset$ itself.
The power set of $lbrace emptyset rbrace$ indeed does have two elements: $emptyset$ and the set itself: $lbrace emptyset rbrace$, thus making it $lbrace lbrace emptyset rbrace, emptyset rbrace$.
answered Jul 29 at 19:53
Theo Bendit
11.8k1841
There was a typo in the question. It is asked for the cardinality and not the power set.
– Cornman
Jul 29 at 19:54
1
@Cornman I'm not convinced. I think the OP is confused between cardinality and power set. I think the heart of the confusion is that he thinks $lbrace emptyset rbrace$ might be a one-element set that breaks the rule $|mathcalP(S)| = 2^S$.
– Theo Bendit
Jul 29 at 19:56
Possible, thats why I upvoted your answer.
– Cornman
Jul 29 at 19:57
add a comment |Â
up vote
1
down vote
The power set of $emptyset$ is indeed $lbrace emptyset rbrace$, as the only subset of $emptyset$ is $emptyset$ itself.
The power set of $lbrace emptyset rbrace$ indeed does have two elements: $emptyset$ and the set itself: $lbrace emptyset rbrace$, thus making it $lbrace lbrace emptyset rbrace, emptyset rbrace$.
answered Jul 29 at 19:53
Theo Bendit
11.8k1841
There was a typo in the question. It is asked for the cardinality and not the power set.
– Cornman
Jul 29 at 19:54
1
@Cornman I'm not convinced. I think the OP is confused between cardinality and power set. I think the heart of the confusion is that he thinks $lbrace emptyset rbrace$ might be a one-element set that breaks the rule $|mathcalP(S)| = 2^S$.
– Theo Bendit
Jul 29 at 19:56
Possible, thats why I upvoted your answer.
– Cornman
Jul 29 at 19:57
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The power set of $emptyset$ is indeed $lbrace emptyset rbrace$, as the only subset of $emptyset$ is $emptyset$ itself.
The power set of $lbrace emptyset rbrace$ indeed does have two elements: $emptyset$ and the set itself: $lbrace emptyset rbrace$, thus making it $lbrace lbrace emptyset rbrace, emptyset rbrace$.
answered Jul 29 at 19:53
Theo Bendit
11.8k1841
The power set of $emptyset$ is indeed $lbrace emptyset rbrace$, as the only subset of $emptyset$ is $emptyset$ itself.
The power set of $lbrace emptyset rbrace$ indeed does have two elements: $emptyset$ and the set itself: $lbrace emptyset rbrace$, thus making it $lbrace lbrace emptyset rbrace, emptyset rbrace$.
answered Jul 29 at 19:53
Theo Bendit
11.8k1841
answered Jul 29 at 19:53
Theo Bendit
11.8k1841
answered Jul 29 at 19:53
Theo Bendit
11.8k1841
answered Jul 29 at 19:53
Theo Bendit
11.8k1841
11.8k1841
There was a typo in the question. It is asked for the cardinality and not the power set.
– Cornman
Jul 29 at 19:54
1
@Cornman I'm not convinced. I think the OP is confused between cardinality and power set. I think the heart of the confusion is that he thinks $lbrace emptyset rbrace$ might be a one-element set that breaks the rule $|mathcalP(S)| = 2^S$.
– Theo Bendit
Jul 29 at 19:56
Possible, thats why I upvoted your answer.
– Cornman
Jul 29 at 19:57
add a comment |Â
There was a typo in the question. It is asked for the cardinality and not the power set.
– Cornman
Jul 29 at 19:54
1
@Cornman I'm not convinced. I think the OP is confused between cardinality and power set. I think the heart of the confusion is that he thinks $lbrace emptyset rbrace$ might be a one-element set that breaks the rule $|mathcalP(S)| = 2^S$.
– Theo Bendit
Jul 29 at 19:56
Possible, thats why I upvoted your answer.
– Cornman
Jul 29 at 19:57
There was a typo in the question. It is asked for the cardinality and not the power set.
– Cornman
Jul 29 at 19:54
There was a typo in the question. It is asked for the cardinality and not the power set.
– Cornman
Jul 29 at 19:54
1
1
@Cornman I'm not convinced. I think the OP is confused between cardinality and power set. I think the heart of the confusion is that he thinks $lbrace emptyset rbrace$ might be a one-element set that breaks the rule $|mathcalP(S)| = 2^S$.
– Theo Bendit
Jul 29 at 19:56
@Cornman I'm not convinced. I think the OP is confused between cardinality and power set. I think the heart of the confusion is that he thinks $lbrace emptyset rbrace$ might be a one-element set that breaks the rule $|mathcalP(S)| = 2^S$.
– Theo Bendit
Jul 29 at 19:56
Possible, thats why I upvoted your answer.
– Cornman
Jul 29 at 19:57
Possible, thats why I upvoted your answer.
– Cornman
Jul 29 at 19:57
add a comment |Â
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Have you computed its power set?
– Javi
Jul 29 at 19:43
2
By power do you mean cardinality?
– user223391
Jul 29 at 19:45
@ZacharySelk yes, edited
– Kevin
Jul 29 at 19:47