Every subset equal to original set?
$begingroup$
Is there any set whose every subset is equal to the set itself? It seems like this isn't possible, but maybe something similar is possible.
elementary-set-theory
asked 2 hours ago
lthompsonlthompson
1269
$endgroup$
add a comment |
$begingroup$
Is there any set whose every subset is equal to the set itself? It seems like this isn't possible, but maybe something similar is possible.
elementary-set-theory
asked 2 hours ago
lthompsonlthompson
1269
$endgroup$
add a comment |
$begingroup$
Is there any set whose every subset is equal to the set itself? It seems like this isn't possible, but maybe something similar is possible.
elementary-set-theory
asked 2 hours ago
lthompsonlthompson
1269
$endgroup$
Is there any set whose every subset is equal to the set itself? It seems like this isn't possible, but maybe something similar is possible.
elementary-set-theory
elementary-set-theory
asked 2 hours ago
lthompsonlthompson
1269
asked 2 hours ago
lthompsonlthompson
1269
asked 2 hours ago
lthompsonlthompson
1269
asked 2 hours ago
lthompsonlthompson
1269
asked 2 hours ago
lthompsonlthompson
1269
1269
add a comment |
add a comment |
2 Answers
2
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$begingroup$
In standard foundations (by which I mean ZF, or ZFC) the empty set works:
If $Ssubset emptyset$, then $S = emptyset$.
If you wish to do so otherwise, you’d violate the Axiom of Extensionality.
answered 1 hour ago
user458276user458276
763212
$endgroup$
add a comment |
$begingroup$
The empty set has only itself as a subset. This is the only example because every set has the empty set as a subset.
answered 1 hour ago
Ross MillikanRoss Millikan
298k24200373
$endgroup$
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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$begingroup$
In standard foundations (by which I mean ZF, or ZFC) the empty set works:
If $Ssubset emptyset$, then $S = emptyset$.
If you wish to do so otherwise, you’d violate the Axiom of Extensionality.
answered 1 hour ago
user458276user458276
763212
$endgroup$
add a comment |
$begingroup$
In standard foundations (by which I mean ZF, or ZFC) the empty set works:
If $Ssubset emptyset$, then $S = emptyset$.
If you wish to do so otherwise, you’d violate the Axiom of Extensionality.
answered 1 hour ago
user458276user458276
763212
$endgroup$
add a comment |
$begingroup$
In standard foundations (by which I mean ZF, or ZFC) the empty set works:
If $Ssubset emptyset$, then $S = emptyset$.
If you wish to do so otherwise, you’d violate the Axiom of Extensionality.
answered 1 hour ago
user458276user458276
763212
$endgroup$
In standard foundations (by which I mean ZF, or ZFC) the empty set works:
If $Ssubset emptyset$, then $S = emptyset$.
If you wish to do so otherwise, you’d violate the Axiom of Extensionality.
answered 1 hour ago
user458276user458276
763212
answered 1 hour ago
user458276user458276
763212
answered 1 hour ago
user458276user458276
763212
answered 1 hour ago
user458276user458276
763212
763212
add a comment |
add a comment |
$begingroup$
The empty set has only itself as a subset. This is the only example because every set has the empty set as a subset.
answered 1 hour ago
Ross MillikanRoss Millikan
298k24200373
$endgroup$
add a comment |
$begingroup$
The empty set has only itself as a subset. This is the only example because every set has the empty set as a subset.
answered 1 hour ago
Ross MillikanRoss Millikan
298k24200373
$endgroup$
add a comment |
$begingroup$
The empty set has only itself as a subset. This is the only example because every set has the empty set as a subset.
answered 1 hour ago
Ross MillikanRoss Millikan
298k24200373
$endgroup$
The empty set has only itself as a subset. This is the only example because every set has the empty set as a subset.
answered 1 hour ago
Ross MillikanRoss Millikan
298k24200373
answered 1 hour ago
Ross MillikanRoss Millikan
298k24200373
answered 1 hour ago
Ross MillikanRoss Millikan
298k24200373
answered 1 hour ago
Ross MillikanRoss Millikan
298k24200373
298k24200373
add a comment |
add a comment |
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