Does the empty set count as an element? [duplicate]
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What is the cardinality of the set of the empty set?
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We're asked the number of elements in a power set and I for {} is the number of elements 1 or 0?
discrete-mathematics
asked Nov 30 '18 at 19:04
happysainthappysaint
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Nov 30 '18 at 22:24
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This question already has an answer here:
What is the cardinality of the set of the empty set?
3 answers
We're asked the number of elements in a power set and I for {} is the number of elements 1 or 0?
discrete-mathematics
asked Nov 30 '18 at 19:04
happysainthappysaint
62
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marked as duplicate by caverac, GNUSupporter 8964民主女神 地下教會, Dietrich Burde, Martin Sleziak, amWhy discrete-mathematics
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Nov 30 '18 at 22:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Yes, it counts. The empty set is a subset of any set, so it will be in the power set of any set.
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– saulspatz
Nov 30 '18 at 19:07
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This question already has an answer here:
What is the cardinality of the set of the empty set?
3 answers
We're asked the number of elements in a power set and I for {} is the number of elements 1 or 0?
discrete-mathematics
asked Nov 30 '18 at 19:04
happysainthappysaint
62
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This question already has an answer here:
What is the cardinality of the set of the empty set?
3 answers
We're asked the number of elements in a power set and I for {} is the number of elements 1 or 0?
This question already has an answer here:
What is the cardinality of the set of the empty set?
3 answers
discrete-mathematics
discrete-mathematics
asked Nov 30 '18 at 19:04
happysainthappysaint
62
asked Nov 30 '18 at 19:04
happysainthappysaint
62
asked Nov 30 '18 at 19:04
happysainthappysaint
62
asked Nov 30 '18 at 19:04
happysainthappysaint
62
asked Nov 30 '18 at 19:04
happysainthappysaint
62
62
marked as duplicate by caverac, GNUSupporter 8964民主女神 地下教會, Dietrich Burde, Martin Sleziak, amWhy discrete-mathematics
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Nov 30 '18 at 22:24
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Nov 30 '18 at 22:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Yes, it counts. The empty set is a subset of any set, so it will be in the power set of any set.
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– saulspatz
Nov 30 '18 at 19:07
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Yes, it counts. The empty set is a subset of any set, so it will be in the power set of any set.
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– saulspatz
Nov 30 '18 at 19:07
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Yes, it counts. The empty set is a subset of any set, so it will be in the power set of any set.
$endgroup$
– saulspatz
Nov 30 '18 at 19:07
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Yes, it counts. The empty set is a subset of any set, so it will be in the power set of any set.
$endgroup$
– saulspatz
Nov 30 '18 at 19:07
add a comment |
1 Answer
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No, an empty set is a subset of every set, not necessarily an element. However, the empty set is an element of the power set of any set.
Notice: $Xsubset Yiff forall x(xin X implies x in Y)$ which is vacously true for any arbitrary set and $emptyset$ since $lnot exists x(xin emptyset)$. However, to say $forall A(emptyset in A)$ implies for a Set $B={a,b,c}$ its $|B|=4$, which is obviously false.
Hope that answers your question.
answered Nov 30 '18 at 19:12
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
387114
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1 Answer
1
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1 Answer
1
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active
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active
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$begingroup$
No, an empty set is a subset of every set, not necessarily an element. However, the empty set is an element of the power set of any set.
Notice: $Xsubset Yiff forall x(xin X implies x in Y)$ which is vacously true for any arbitrary set and $emptyset$ since $lnot exists x(xin emptyset)$. However, to say $forall A(emptyset in A)$ implies for a Set $B={a,b,c}$ its $|B|=4$, which is obviously false.
Hope that answers your question.
answered Nov 30 '18 at 19:12
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
387114
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add a comment |
$begingroup$
No, an empty set is a subset of every set, not necessarily an element. However, the empty set is an element of the power set of any set.
Notice: $Xsubset Yiff forall x(xin X implies x in Y)$ which is vacously true for any arbitrary set and $emptyset$ since $lnot exists x(xin emptyset)$. However, to say $forall A(emptyset in A)$ implies for a Set $B={a,b,c}$ its $|B|=4$, which is obviously false.
Hope that answers your question.
answered Nov 30 '18 at 19:12
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
387114
$endgroup$
add a comment |
$begingroup$
No, an empty set is a subset of every set, not necessarily an element. However, the empty set is an element of the power set of any set.
Notice: $Xsubset Yiff forall x(xin X implies x in Y)$ which is vacously true for any arbitrary set and $emptyset$ since $lnot exists x(xin emptyset)$. However, to say $forall A(emptyset in A)$ implies for a Set $B={a,b,c}$ its $|B|=4$, which is obviously false.
Hope that answers your question.
answered Nov 30 '18 at 19:12
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
387114
$endgroup$
No, an empty set is a subset of every set, not necessarily an element. However, the empty set is an element of the power set of any set.
Notice: $Xsubset Yiff forall x(xin X implies x in Y)$ which is vacously true for any arbitrary set and $emptyset$ since $lnot exists x(xin emptyset)$. However, to say $forall A(emptyset in A)$ implies for a Set $B={a,b,c}$ its $|B|=4$, which is obviously false.
Hope that answers your question.
answered Nov 30 '18 at 19:12
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
387114
edited Nov 30 '18 at 19:22
answered Nov 30 '18 at 19:12
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
387114
answered Nov 30 '18 at 19:12
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
387114
answered Nov 30 '18 at 19:12
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
387114
387114
add a comment |
add a comment |
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Yes, it counts. The empty set is a subset of any set, so it will be in the power set of any set.
$endgroup$
– saulspatz
Nov 30 '18 at 19:07