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 1 hour ago
lthompsonlthompson
1169
$endgroup$
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 1 hour ago
lthompsonlthompson
1169
$endgroup$
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 1 hour ago
lthompsonlthompson
1169
$endgroup$
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 1 hour ago
lthompsonlthompson
1169
asked 1 hour ago
lthompsonlthompson
1169
asked 1 hour ago
lthompsonlthompson
1169
asked 1 hour ago
lthompsonlthompson
1169
asked 1 hour ago
lthompsonlthompson
1169
1169
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 56 mins ago
user458276user458276
743212
$endgroup$
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 56 mins ago
Ross MillikanRoss Millikan
298k24200373
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
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 56 mins ago
user458276user458276
743212
$endgroup$
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 56 mins ago
user458276user458276
743212
$endgroup$
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 56 mins ago
user458276user458276
743212
$endgroup$
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 56 mins ago
user458276user458276
743212
answered 56 mins ago
user458276user458276
743212
answered 56 mins ago
user458276user458276
743212
answered 56 mins ago
user458276user458276
743212
743212
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 56 mins 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 56 mins 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 56 mins 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 56 mins ago
Ross MillikanRoss Millikan
298k24200373
answered 56 mins ago
Ross MillikanRoss Millikan
298k24200373
answered 56 mins ago
Ross MillikanRoss Millikan
298k24200373
answered 56 mins ago
Ross MillikanRoss Millikan
298k24200373
298k24200373
add a comment |
add a comment |
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