Cardinality: Injection between subsets of Uncountable set The 2019 Stack Overflow Developer Survey Results Are InProving a set is uncountableA countable and an uncountable setIs this set uncountable or countable?Real Analysis Cardinality ProofUncountable “relatively independent” subset of finite dimensional vector spaces over an uncountable fieldIf R is uncountable, then (0.1) is uncountable.Two well orders on an uncountable setCardinal arithmetic argument for countability of finite subsets of integers.Geometric solution to find a Set of totally ordered Subsets of $ mathbbN $ is uncountableHow many uncountable subsets of power set of integers are there?
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Cardinality: Injection between subsets of Uncountable set
The 2019 Stack Overflow Developer Survey Results Are InProving a set is uncountableA countable and an uncountable setIs this set uncountable or countable?Real Analysis Cardinality ProofUncountable “relatively independent” subset of finite dimensional vector spaces over an uncountable fieldIf R is uncountable, then (0.1) is uncountable.Two well orders on an uncountable setCardinal arithmetic argument for countability of finite subsets of integers.Geometric solution to find a Set of totally ordered Subsets of $ mathbbN $ is uncountableHow many uncountable subsets of power set of integers are there?
$begingroup$
assuming, S is infinite uncountable, I am trying to come up with injective f: (S union N) -> S. Where N is naturals.
So far I created S0 which consists of infinite sequence of elements of S, such that S0 = s1,s2,s3, .... that way I can have injective f1: N -> S0.
But I am having hard time trying to prove there exist injective f2: S -> S/S0.
functions elementary-set-theory cardinals
asked Mar 23 at 5:58
ActssassinActssassin
1
$endgroup$
add a comment |
$begingroup$
assuming, S is infinite uncountable, I am trying to come up with injective f: (S union N) -> S. Where N is naturals.
So far I created S0 which consists of infinite sequence of elements of S, such that S0 = s1,s2,s3, .... that way I can have injective f1: N -> S0.
But I am having hard time trying to prove there exist injective f2: S -> S/S0.
functions elementary-set-theory cardinals
asked Mar 23 at 5:58
ActssassinActssassin
1
$endgroup$
add a comment |
$begingroup$
assuming, S is infinite uncountable, I am trying to come up with injective f: (S union N) -> S. Where N is naturals.
So far I created S0 which consists of infinite sequence of elements of S, such that S0 = s1,s2,s3, .... that way I can have injective f1: N -> S0.
But I am having hard time trying to prove there exist injective f2: S -> S/S0.
functions elementary-set-theory cardinals
asked Mar 23 at 5:58
ActssassinActssassin
1
$endgroup$
assuming, S is infinite uncountable, I am trying to come up with injective f: (S union N) -> S. Where N is naturals.
So far I created S0 which consists of infinite sequence of elements of S, such that S0 = s1,s2,s3, .... that way I can have injective f1: N -> S0.
But I am having hard time trying to prove there exist injective f2: S -> S/S0.
functions elementary-set-theory cardinals
functions elementary-set-theory cardinals
asked Mar 23 at 5:58
ActssassinActssassin
1
asked Mar 23 at 5:58
ActssassinActssassin
1
asked Mar 23 at 5:58
ActssassinActssassin
1
asked Mar 23 at 5:58
ActssassinActssassin
1
asked Mar 23 at 5:58
ActssassinActssassin
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
Assume N and S are disjoint.
For each s in s1, s2, s3,.. pull from S a sequence
t$_s$1, t$_s$2, t$_s$3,.. with t$_s$1 = s.
Do this in a manner that none of the sequences have a common element.
This can be done because S is uncountable and all those sequences are requiring a mere countable number of elements.
To create a bijection from S - S0 to S, for each s in s1, s2, s3,..
map t$_s$2 to t$_s$1, t$_s$3 to t$_s$2, etc.
and for all of those elements not in any of the sequences, map them to themselves.
The case that S and N are not disjoint is left for the diligent reader.
answered Mar 23 at 11:17
William ElliotWilliam Elliot
9,1212820
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Assume N and S are disjoint.
For each s in s1, s2, s3,.. pull from S a sequence
t$_s$1, t$_s$2, t$_s$3,.. with t$_s$1 = s.
Do this in a manner that none of the sequences have a common element.
This can be done because S is uncountable and all those sequences are requiring a mere countable number of elements.
To create a bijection from S - S0 to S, for each s in s1, s2, s3,..
map t$_s$2 to t$_s$1, t$_s$3 to t$_s$2, etc.
and for all of those elements not in any of the sequences, map them to themselves.
The case that S and N are not disjoint is left for the diligent reader.
answered Mar 23 at 11:17
William ElliotWilliam Elliot
9,1212820
$endgroup$
add a comment |
$begingroup$
Assume N and S are disjoint.
For each s in s1, s2, s3,.. pull from S a sequence
t$_s$1, t$_s$2, t$_s$3,.. with t$_s$1 = s.
Do this in a manner that none of the sequences have a common element.
This can be done because S is uncountable and all those sequences are requiring a mere countable number of elements.
To create a bijection from S - S0 to S, for each s in s1, s2, s3,..
map t$_s$2 to t$_s$1, t$_s$3 to t$_s$2, etc.
and for all of those elements not in any of the sequences, map them to themselves.
The case that S and N are not disjoint is left for the diligent reader.
answered Mar 23 at 11:17
William ElliotWilliam Elliot
9,1212820
$endgroup$
add a comment |
$begingroup$
Assume N and S are disjoint.
For each s in s1, s2, s3,.. pull from S a sequence
t$_s$1, t$_s$2, t$_s$3,.. with t$_s$1 = s.
Do this in a manner that none of the sequences have a common element.
This can be done because S is uncountable and all those sequences are requiring a mere countable number of elements.
To create a bijection from S - S0 to S, for each s in s1, s2, s3,..
map t$_s$2 to t$_s$1, t$_s$3 to t$_s$2, etc.
and for all of those elements not in any of the sequences, map them to themselves.
The case that S and N are not disjoint is left for the diligent reader.
answered Mar 23 at 11:17
William ElliotWilliam Elliot
9,1212820
$endgroup$
Assume N and S are disjoint.
For each s in s1, s2, s3,.. pull from S a sequence
t$_s$1, t$_s$2, t$_s$3,.. with t$_s$1 = s.
Do this in a manner that none of the sequences have a common element.
This can be done because S is uncountable and all those sequences are requiring a mere countable number of elements.
To create a bijection from S - S0 to S, for each s in s1, s2, s3,..
map t$_s$2 to t$_s$1, t$_s$3 to t$_s$2, etc.
and for all of those elements not in any of the sequences, map them to themselves.
The case that S and N are not disjoint is left for the diligent reader.
answered Mar 23 at 11:17
William ElliotWilliam Elliot
9,1212820
answered Mar 23 at 11:17
William ElliotWilliam Elliot
9,1212820
answered Mar 23 at 11:17
William ElliotWilliam Elliot
9,1212820
answered Mar 23 at 11:17
William ElliotWilliam Elliot
9,1212820
9,1212820
add a comment |
add a comment |
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