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Every subset equal to original set?
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Every subset equal to original set?
Is the void set (∅) a proper subset of every set?Set theory: difference between belong/contained and includes/subset?Proving every infinite set is a subset of some denumerable set and vice versaIf the empty set is a subset of every set, why isn't ${emptyset,{a}}={{a}}$?What is the reason behind calling $emptyset$ improper subset of any non-empty set.?Can subset of a countable set be uncountable?Set Theory Subset QuestionIs every empty set equal?Why a set that is subset/equal to infinite set isn't infinite? (by definition)Why the empty set is a subset of every 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 32 mins 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 32 mins 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 32 mins 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 32 mins ago
lthompsonlthompson
1169
asked 32 mins ago
lthompsonlthompson
1169
asked 32 mins ago
lthompsonlthompson
1169
asked 32 mins ago
lthompsonlthompson
1169
asked 32 mins ago
lthompsonlthompson
1169
1169
add a comment |
add a comment |
2 Answers
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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 26 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 26 mins ago
Ross MillikanRoss Millikan
298k24200373
$endgroup$
add a comment |
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2 Answers
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2 Answers
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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 26 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 26 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 26 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 26 mins ago
user458276user458276
743212
answered 26 mins ago
user458276user458276
743212
answered 26 mins ago
user458276user458276
743212
answered 26 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 26 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 26 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 26 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 26 mins ago
Ross MillikanRoss Millikan
298k24200373
answered 26 mins ago
Ross MillikanRoss Millikan
298k24200373
answered 26 mins ago
Ross MillikanRoss Millikan
298k24200373
answered 26 mins ago
Ross MillikanRoss Millikan
298k24200373
298k24200373
add a comment |
add a comment |
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