Mixing
Partition of an infinite set into two infinite sets
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I wanted to try and prove this statement which looks seemingly true.
An infinite set $X$ can be partitioned in such a way that $X = X_1 cup X_2$ where $X_1$ and $X_2$ are infinite subsets of $X.$
Attempt :
case (i) If $X$ is countably infinite, then we can list the elements of $X$ as ${x_1,x_2,x_3,dots}.$ If we choose $X_1$ to be indexed by the odd natural numbers and $X_2$ to be indexed by the even natural numbers, we are done.
case (ii) If $X$ is uncountably infinite. Choose $X_1$ to be the countable infinite subset of $X.$ Then let $X_2 := X_1^mathsf{c}.$ This shall ensure that $X_2$ is uncountable. $spacespaceblacksquare$
But I think I'll have to be a bit clear for case (ii). How do I show that given an uncountable set $X,$ there always exists a countable infinite subset $X_1$ of $X.$
elementary-set-theory
asked Jan 20 at 10:41
Bijesh K.SBijesh K.S
996824
$endgroup$
add a comment |
$begingroup$
I wanted to try and prove this statement which looks seemingly true.
An infinite set $X$ can be partitioned in such a way that $X = X_1 cup X_2$ where $X_1$ and $X_2$ are infinite subsets of $X.$
Attempt :
case (i) If $X$ is countably infinite, then we can list the elements of $X$ as ${x_1,x_2,x_3,dots}.$ If we choose $X_1$ to be indexed by the odd natural numbers and $X_2$ to be indexed by the even natural numbers, we are done.
case (ii) If $X$ is uncountably infinite. Choose $X_1$ to be the countable infinite subset of $X.$ Then let $X_2 := X_1^mathsf{c}.$ This shall ensure that $X_2$ is uncountable. $spacespaceblacksquare$
But I think I'll have to be a bit clear for case (ii). How do I show that given an uncountable set $X,$ there always exists a countable infinite subset $X_1$ of $X.$
elementary-set-theory
asked Jan 20 at 10:41
Bijesh K.SBijesh K.S
996824
$endgroup$
1
$begingroup$
Every infinite set contains a countably infinite set (if we assume some axiom of choice). An uncountable set is in particular infinite. Also see this thread etc.
$endgroup$
– Henno Brandsma
Jan 20 at 10:47
add a comment |
$begingroup$
I wanted to try and prove this statement which looks seemingly true.
An infinite set $X$ can be partitioned in such a way that $X = X_1 cup X_2$ where $X_1$ and $X_2$ are infinite subsets of $X.$
Attempt :
case (i) If $X$ is countably infinite, then we can list the elements of $X$ as ${x_1,x_2,x_3,dots}.$ If we choose $X_1$ to be indexed by the odd natural numbers and $X_2$ to be indexed by the even natural numbers, we are done.
case (ii) If $X$ is uncountably infinite. Choose $X_1$ to be the countable infinite subset of $X.$ Then let $X_2 := X_1^mathsf{c}.$ This shall ensure that $X_2$ is uncountable. $spacespaceblacksquare$
But I think I'll have to be a bit clear for case (ii). How do I show that given an uncountable set $X,$ there always exists a countable infinite subset $X_1$ of $X.$
elementary-set-theory
asked Jan 20 at 10:41
Bijesh K.SBijesh K.S
996824
$endgroup$
I wanted to try and prove this statement which looks seemingly true.
An infinite set $X$ can be partitioned in such a way that $X = X_1 cup X_2$ where $X_1$ and $X_2$ are infinite subsets of $X.$
Attempt :
case (i) If $X$ is countably infinite, then we can list the elements of $X$ as ${x_1,x_2,x_3,dots}.$ If we choose $X_1$ to be indexed by the odd natural numbers and $X_2$ to be indexed by the even natural numbers, we are done.
case (ii) If $X$ is uncountably infinite. Choose $X_1$ to be the countable infinite subset of $X.$ Then let $X_2 := X_1^mathsf{c}.$ This shall ensure that $X_2$ is uncountable. $spacespaceblacksquare$
But I think I'll have to be a bit clear for case (ii). How do I show that given an uncountable set $X,$ there always exists a countable infinite subset $X_1$ of $X.$
elementary-set-theory
elementary-set-theory
asked Jan 20 at 10:41
Bijesh K.SBijesh K.S
996824
asked Jan 20 at 10:41
Bijesh K.SBijesh K.S
996824
asked Jan 20 at 10:41
Bijesh K.SBijesh K.S
996824
asked Jan 20 at 10:41
Bijesh K.SBijesh K.S
996824
asked Jan 20 at 10:41
Bijesh K.SBijesh K.S
996824
996824
1
$begingroup$
Every infinite set contains a countably infinite set (if we assume some axiom of choice). An uncountable set is in particular infinite. Also see this thread etc.
$endgroup$
– Henno Brandsma
Jan 20 at 10:47
add a comment |
1
$begingroup$
Every infinite set contains a countably infinite set (if we assume some axiom of choice). An uncountable set is in particular infinite. Also see this thread etc.
$endgroup$
– Henno Brandsma
Jan 20 at 10:47
1
1
$begingroup$
Every infinite set contains a countably infinite set (if we assume some axiom of choice). An uncountable set is in particular infinite. Also see this thread etc.
$endgroup$
– Henno Brandsma
Jan 20 at 10:47
$begingroup$
Every infinite set contains a countably infinite set (if we assume some axiom of choice). An uncountable set is in particular infinite. Also see this thread etc.
$endgroup$
– Henno Brandsma
Jan 20 at 10:47
add a comment |
1 Answer
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Case 2. Well order X and take the first $omega_0$ elements.
Or recursively construct a denumerable set. Pick any element a$_1$ from X. Having picked S = { $a_1,.. a_n$ }, pick any
a$_{n+1}$ from uncountable X - S.
answered Jan 20 at 11:08
William ElliotWilliam Elliot
8,3822720
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add a comment |
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$begingroup$
Case 2. Well order X and take the first $omega_0$ elements.
Or recursively construct a denumerable set. Pick any element a$_1$ from X. Having picked S = { $a_1,.. a_n$ }, pick any
a$_{n+1}$ from uncountable X - S.
answered Jan 20 at 11:08
William ElliotWilliam Elliot
8,3822720
$endgroup$
add a comment |
$begingroup$
Case 2. Well order X and take the first $omega_0$ elements.
Or recursively construct a denumerable set. Pick any element a$_1$ from X. Having picked S = { $a_1,.. a_n$ }, pick any
a$_{n+1}$ from uncountable X - S.
answered Jan 20 at 11:08
William ElliotWilliam Elliot
8,3822720
$endgroup$
add a comment |
$begingroup$
Case 2. Well order X and take the first $omega_0$ elements.
Or recursively construct a denumerable set. Pick any element a$_1$ from X. Having picked S = { $a_1,.. a_n$ }, pick any
a$_{n+1}$ from uncountable X - S.
answered Jan 20 at 11:08
William ElliotWilliam Elliot
8,3822720
$endgroup$
Case 2. Well order X and take the first $omega_0$ elements.
Or recursively construct a denumerable set. Pick any element a$_1$ from X. Having picked S = { $a_1,.. a_n$ }, pick any
a$_{n+1}$ from uncountable X - S.
answered Jan 20 at 11:08
William ElliotWilliam Elliot
8,3822720
answered Jan 20 at 11:08
William ElliotWilliam Elliot
8,3822720
answered Jan 20 at 11:08
William ElliotWilliam Elliot
8,3822720
answered Jan 20 at 11:08
William ElliotWilliam Elliot
8,3822720
8,3822720
add a comment |
add a comment |
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$begingroup$
Every infinite set contains a countably infinite set (if we assume some axiom of choice). An uncountable set is in particular infinite. Also see this thread etc.
$endgroup$
– Henno Brandsma
Jan 20 at 10:47