An uncountable set has uncountably many co-countable subsets containing an arbitrary point Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Uncountable minus countable set is uncountableCountable subset of a uncountable setCountable subsets of an uncountable setAre there countably or uncountably many infinite subsets of the positive even integers?A countable and an uncountable setIs there an infinite countable $sigma$-algebra on an uncountable setEvery infinite set has an infinite countable subset?Can subset of a countable set be uncountable?Uncountable Ordered Set Isomorphic to to Its Uncountable SubsetsIf $A$ is infinite, then $mathcalP(A)$ is uncountable
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An uncountable set has uncountably many co-countable subsets containing an arbitrary point
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Uncountable minus countable set is uncountableCountable subset of a uncountable setCountable subsets of an uncountable setAre there countably or uncountably many infinite subsets of the positive even integers?A countable and an uncountable setIs there an infinite countable $sigma$-algebra on an uncountable setEvery infinite set has an infinite countable subset?Can subset of a countable set be uncountable?Uncountable Ordered Set Isomorphic to to Its Uncountable SubsetsIf $A$ is infinite, then $mathcalP(A)$ is uncountable
$begingroup$
This theorem seems "obvious" to me, but I want to check my logic since I am un-familiar with un-countably infinite sets, and I know these can give rise to non-intuitive results. Any comments welcome.
Definitions
The cardinality of a set $A$ will be denoted $|A|$. If $|A|=|mathbbN|$, $A$ will be called countable or countably infinite and this will be denoted $|A|=aleph_0$. If $A$ is infinite and $|A|nealeph_0$, $A$ will be called uncountable or uncountably infinite.
Claim
If $A$ is un-countable then for any $ain A$, $A$ contains an uncountable number of disjoint co-countable sets that contain $a$. More formally;
For any $ain A$ where $|A|not=aleph_0$ there exist subsets
$B_betasubset A$, $betain I_beta$ for an index set
$I_beta$, $|I_beta|not=aleph_0$, such that for all $betain I_beta$
: $ain B_beta$ and $|A-B_beta|=aleph_0$.
Furthermore for any $betanot=beta'$, $B_betacap B_beta'=emptyset$
Proof
Without loss of generality let $I_beta=mathbbR$ and choose $betain I_beta$. Choose $a_beta,1in A$, $a_beta,1not=a$. Since $|A-a_beta,1|not=aleph_0$ there exists $a_beta,2inA-a_beta,1$, $a_beta,2not=a$. Since $|A-a_beta,1,a_beta,2|not=aleph_0$ there exists $a_beta,3inA-a_beta,1,a_beta,2$, $a_beta,3not=a$. Continuing in this way $n_beta$ times, $n_betainmathbbN$, we can construct a set $B_beta=A-a_beta,1,a_beta,2,...,a_beta,n_beta$ that satisfies $ain B_betasubset A$ and $|A-B_beta|=|a_beta,1,a_beta,2,...,a_beta,n_beta|=aleph_0$ so that $B_beta$ is a co-countable subset of $A$ containing $a$.
Choose another $beta'in I_beta$, $betanot=beta'$. Since $|B_beta|not=aleph_0$ there exists $a_beta',1in B_beta$, $a_beta',1not=a$. Since $|B_beta-a_beta',1|not=aleph_0$ there exists $a_beta',2in B_beta-a_beta',1$, $a_beta',2not=a$. Since $|B_beta-a_beta',1,a_beta',2|not=aleph_0$ there exists $a_beta',3in B_beta-a_beta',1,a_beta',2$, $a_beta',3not=a$. Continuing in this way $n_beta'$ times, $n_beta'inmathbbN$, we can construct a set $B_beta'=B_beta-a_beta',1,a_beta',2,...,a_beta',n_beta'$ that satisfies $ain B_beta'subset A$ and $|A-B_beta'|=|a_beta,1,a_beta,2,...,a_beta,n_betacupa_beta,1,a_beta,2,...,a_beta',n_beta'|=aleph_0+aleph_0=aleph_0$ so that $B_beta'$ is a co-countable subset of $A$ containing $a$.
Since $B_betacap B_beta'=emptyset$ and since we can choose un-countably many $betain I_beta$ the result is proved $hspace5ptblacksquare$
elementary-set-theory cardinals
edited Mar 27 at 17:57
Andrés E. Caicedo
66.1k8160252
asked Mar 27 at 17:28
dandardandar
415616
$endgroup$
|
show 11 more comments
$begingroup$
This theorem seems "obvious" to me, but I want to check my logic since I am un-familiar with un-countably infinite sets, and I know these can give rise to non-intuitive results. Any comments welcome.
Definitions
The cardinality of a set $A$ will be denoted $|A|$. If $|A|=|mathbbN|$, $A$ will be called countable or countably infinite and this will be denoted $|A|=aleph_0$. If $A$ is infinite and $|A|nealeph_0$, $A$ will be called uncountable or uncountably infinite.
Claim
If $A$ is un-countable then for any $ain A$, $A$ contains an uncountable number of disjoint co-countable sets that contain $a$. More formally;
For any $ain A$ where $|A|not=aleph_0$ there exist subsets
$B_betasubset A$, $betain I_beta$ for an index set
$I_beta$, $|I_beta|not=aleph_0$, such that for all $betain I_beta$
: $ain B_beta$ and $|A-B_beta|=aleph_0$.
Furthermore for any $betanot=beta'$, $B_betacap B_beta'=emptyset$
Proof
Without loss of generality let $I_beta=mathbbR$ and choose $betain I_beta$. Choose $a_beta,1in A$, $a_beta,1not=a$. Since $|A-a_beta,1|not=aleph_0$ there exists $a_beta,2inA-a_beta,1$, $a_beta,2not=a$. Since $|A-a_beta,1,a_beta,2|not=aleph_0$ there exists $a_beta,3inA-a_beta,1,a_beta,2$, $a_beta,3not=a$. Continuing in this way $n_beta$ times, $n_betainmathbbN$, we can construct a set $B_beta=A-a_beta,1,a_beta,2,...,a_beta,n_beta$ that satisfies $ain B_betasubset A$ and $|A-B_beta|=|a_beta,1,a_beta,2,...,a_beta,n_beta|=aleph_0$ so that $B_beta$ is a co-countable subset of $A$ containing $a$.
Choose another $beta'in I_beta$, $betanot=beta'$. Since $|B_beta|not=aleph_0$ there exists $a_beta',1in B_beta$, $a_beta',1not=a$. Since $|B_beta-a_beta',1|not=aleph_0$ there exists $a_beta',2in B_beta-a_beta',1$, $a_beta',2not=a$. Since $|B_beta-a_beta',1,a_beta',2|not=aleph_0$ there exists $a_beta',3in B_beta-a_beta',1,a_beta',2$, $a_beta',3not=a$. Continuing in this way $n_beta'$ times, $n_beta'inmathbbN$, we can construct a set $B_beta'=B_beta-a_beta',1,a_beta',2,...,a_beta',n_beta'$ that satisfies $ain B_beta'subset A$ and $|A-B_beta'|=|a_beta,1,a_beta,2,...,a_beta,n_betacupa_beta,1,a_beta,2,...,a_beta',n_beta'|=aleph_0+aleph_0=aleph_0$ so that $B_beta'$ is a co-countable subset of $A$ containing $a$.
Since $B_betacap B_beta'=emptyset$ and since we can choose un-countably many $betain I_beta$ the result is proved $hspace5ptblacksquare$
elementary-set-theory cardinals
edited Mar 27 at 17:57
Andrés E. Caicedo
66.1k8160252
asked Mar 27 at 17:28
dandardandar
415616
$endgroup$
3
$begingroup$
Your definition of "uncountably infinite" is not standard. Usually, it is an infinite set $S$ such that $|S|neq aleph_0.$ There are a lot of uncountably infinite cardinals other than $|mathbb R.|$
$endgroup$
– Thomas Andrews
Mar 27 at 17:31
1
$begingroup$
No, you are wrong, @AndrésE.Caicedo Given $ain A$ take a countable subset $Csubseteq A$ with $anotin C.$ Then take $Asetminus Tmid Tsubset C, text and T infinite$. There are uncountably many (countably) infinite subsets $T$ of $C$, and for each $ain A-T,$ and $A-T$ is co-countable.
$endgroup$
– Thomas Andrews
Mar 27 at 18:03
1
$begingroup$
@Thomas These sets are not disjoint.
$endgroup$
– Andrés E. Caicedo
Mar 27 at 18:05
1
$begingroup$
@dandar Without the requirement of disjointness, of course the claim holds, and it is straightforward. Thomas sketched in comments what to do. I do not see uncountably many sets produced using procedures along the lines your comment suggests. If besides disjointness we also ignore the requirement of having uncountably many such sets, then yes, your comment gives a possible approach.
$endgroup$
– Andrés E. Caicedo
Mar 28 at 16:27
1
$begingroup$
@dandar Yes, that's the argument.
$endgroup$
– Andrés E. Caicedo
Mar 28 at 18:33
|
show 11 more comments
$begingroup$
This theorem seems "obvious" to me, but I want to check my logic since I am un-familiar with un-countably infinite sets, and I know these can give rise to non-intuitive results. Any comments welcome.
Definitions
The cardinality of a set $A$ will be denoted $|A|$. If $|A|=|mathbbN|$, $A$ will be called countable or countably infinite and this will be denoted $|A|=aleph_0$. If $A$ is infinite and $|A|nealeph_0$, $A$ will be called uncountable or uncountably infinite.
Claim
If $A$ is un-countable then for any $ain A$, $A$ contains an uncountable number of disjoint co-countable sets that contain $a$. More formally;
For any $ain A$ where $|A|not=aleph_0$ there exist subsets
$B_betasubset A$, $betain I_beta$ for an index set
$I_beta$, $|I_beta|not=aleph_0$, such that for all $betain I_beta$
: $ain B_beta$ and $|A-B_beta|=aleph_0$.
Furthermore for any $betanot=beta'$, $B_betacap B_beta'=emptyset$
Proof
Without loss of generality let $I_beta=mathbbR$ and choose $betain I_beta$. Choose $a_beta,1in A$, $a_beta,1not=a$. Since $|A-a_beta,1|not=aleph_0$ there exists $a_beta,2inA-a_beta,1$, $a_beta,2not=a$. Since $|A-a_beta,1,a_beta,2|not=aleph_0$ there exists $a_beta,3inA-a_beta,1,a_beta,2$, $a_beta,3not=a$. Continuing in this way $n_beta$ times, $n_betainmathbbN$, we can construct a set $B_beta=A-a_beta,1,a_beta,2,...,a_beta,n_beta$ that satisfies $ain B_betasubset A$ and $|A-B_beta|=|a_beta,1,a_beta,2,...,a_beta,n_beta|=aleph_0$ so that $B_beta$ is a co-countable subset of $A$ containing $a$.
Choose another $beta'in I_beta$, $betanot=beta'$. Since $|B_beta|not=aleph_0$ there exists $a_beta',1in B_beta$, $a_beta',1not=a$. Since $|B_beta-a_beta',1|not=aleph_0$ there exists $a_beta',2in B_beta-a_beta',1$, $a_beta',2not=a$. Since $|B_beta-a_beta',1,a_beta',2|not=aleph_0$ there exists $a_beta',3in B_beta-a_beta',1,a_beta',2$, $a_beta',3not=a$. Continuing in this way $n_beta'$ times, $n_beta'inmathbbN$, we can construct a set $B_beta'=B_beta-a_beta',1,a_beta',2,...,a_beta',n_beta'$ that satisfies $ain B_beta'subset A$ and $|A-B_beta'|=|a_beta,1,a_beta,2,...,a_beta,n_betacupa_beta,1,a_beta,2,...,a_beta',n_beta'|=aleph_0+aleph_0=aleph_0$ so that $B_beta'$ is a co-countable subset of $A$ containing $a$.
Since $B_betacap B_beta'=emptyset$ and since we can choose un-countably many $betain I_beta$ the result is proved $hspace5ptblacksquare$
elementary-set-theory cardinals
edited Mar 27 at 17:57
Andrés E. Caicedo
66.1k8160252
asked Mar 27 at 17:28
dandardandar
415616
$endgroup$
This theorem seems "obvious" to me, but I want to check my logic since I am un-familiar with un-countably infinite sets, and I know these can give rise to non-intuitive results. Any comments welcome.
Definitions
The cardinality of a set $A$ will be denoted $|A|$. If $|A|=|mathbbN|$, $A$ will be called countable or countably infinite and this will be denoted $|A|=aleph_0$. If $A$ is infinite and $|A|nealeph_0$, $A$ will be called uncountable or uncountably infinite.
Claim
If $A$ is un-countable then for any $ain A$, $A$ contains an uncountable number of disjoint co-countable sets that contain $a$. More formally;
For any $ain A$ where $|A|not=aleph_0$ there exist subsets
$B_betasubset A$, $betain I_beta$ for an index set
$I_beta$, $|I_beta|not=aleph_0$, such that for all $betain I_beta$
: $ain B_beta$ and $|A-B_beta|=aleph_0$.
Furthermore for any $betanot=beta'$, $B_betacap B_beta'=emptyset$
Proof
Without loss of generality let $I_beta=mathbbR$ and choose $betain I_beta$. Choose $a_beta,1in A$, $a_beta,1not=a$. Since $|A-a_beta,1|not=aleph_0$ there exists $a_beta,2inA-a_beta,1$, $a_beta,2not=a$. Since $|A-a_beta,1,a_beta,2|not=aleph_0$ there exists $a_beta,3inA-a_beta,1,a_beta,2$, $a_beta,3not=a$. Continuing in this way $n_beta$ times, $n_betainmathbbN$, we can construct a set $B_beta=A-a_beta,1,a_beta,2,...,a_beta,n_beta$ that satisfies $ain B_betasubset A$ and $|A-B_beta|=|a_beta,1,a_beta,2,...,a_beta,n_beta|=aleph_0$ so that $B_beta$ is a co-countable subset of $A$ containing $a$.
Choose another $beta'in I_beta$, $betanot=beta'$. Since $|B_beta|not=aleph_0$ there exists $a_beta',1in B_beta$, $a_beta',1not=a$. Since $|B_beta-a_beta',1|not=aleph_0$ there exists $a_beta',2in B_beta-a_beta',1$, $a_beta',2not=a$. Since $|B_beta-a_beta',1,a_beta',2|not=aleph_0$ there exists $a_beta',3in B_beta-a_beta',1,a_beta',2$, $a_beta',3not=a$. Continuing in this way $n_beta'$ times, $n_beta'inmathbbN$, we can construct a set $B_beta'=B_beta-a_beta',1,a_beta',2,...,a_beta',n_beta'$ that satisfies $ain B_beta'subset A$ and $|A-B_beta'|=|a_beta,1,a_beta,2,...,a_beta,n_betacupa_beta,1,a_beta,2,...,a_beta',n_beta'|=aleph_0+aleph_0=aleph_0$ so that $B_beta'$ is a co-countable subset of $A$ containing $a$.
Since $B_betacap B_beta'=emptyset$ and since we can choose un-countably many $betain I_beta$ the result is proved $hspace5ptblacksquare$
elementary-set-theory cardinals
elementary-set-theory cardinals
edited Mar 27 at 17:57
Andrés E. Caicedo
66.1k8160252
asked Mar 27 at 17:28
dandardandar
415616
edited Mar 27 at 17:57
Andrés E. Caicedo
66.1k8160252
asked Mar 27 at 17:28
dandardandar
415616
edited Mar 27 at 17:57
Andrés E. Caicedo
66.1k8160252
edited Mar 27 at 17:57
Andrés E. Caicedo
66.1k8160252
edited Mar 27 at 17:57
Andrés E. Caicedo
66.1k8160252
66.1k8160252
asked Mar 27 at 17:28
dandardandar
415616
asked Mar 27 at 17:28
dandardandar
415616
asked Mar 27 at 17:28
dandardandar
415616
415616
3
$begingroup$
Your definition of "uncountably infinite" is not standard. Usually, it is an infinite set $S$ such that $|S|neq aleph_0.$ There are a lot of uncountably infinite cardinals other than $|mathbb R.|$
$endgroup$
– Thomas Andrews
Mar 27 at 17:31
1
$begingroup$
No, you are wrong, @AndrésE.Caicedo Given $ain A$ take a countable subset $Csubseteq A$ with $anotin C.$ Then take $Asetminus Tmid Tsubset C, text and T infinite$. There are uncountably many (countably) infinite subsets $T$ of $C$, and for each $ain A-T,$ and $A-T$ is co-countable.
$endgroup$
– Thomas Andrews
Mar 27 at 18:03
1
$begingroup$
@Thomas These sets are not disjoint.
$endgroup$
– Andrés E. Caicedo
Mar 27 at 18:05
1
$begingroup$
@dandar Without the requirement of disjointness, of course the claim holds, and it is straightforward. Thomas sketched in comments what to do. I do not see uncountably many sets produced using procedures along the lines your comment suggests. If besides disjointness we also ignore the requirement of having uncountably many such sets, then yes, your comment gives a possible approach.
$endgroup$
– Andrés E. Caicedo
Mar 28 at 16:27
1
$begingroup$
@dandar Yes, that's the argument.
$endgroup$
– Andrés E. Caicedo
Mar 28 at 18:33
|
show 11 more comments
3
$begingroup$
Your definition of "uncountably infinite" is not standard. Usually, it is an infinite set $S$ such that $|S|neq aleph_0.$ There are a lot of uncountably infinite cardinals other than $|mathbb R.|$
$endgroup$
– Thomas Andrews
Mar 27 at 17:31
1
$begingroup$
No, you are wrong, @AndrésE.Caicedo Given $ain A$ take a countable subset $Csubseteq A$ with $anotin C.$ Then take $Asetminus Tmid Tsubset C, text and T infinite$. There are uncountably many (countably) infinite subsets $T$ of $C$, and for each $ain A-T,$ and $A-T$ is co-countable.
$endgroup$
– Thomas Andrews
Mar 27 at 18:03
1
$begingroup$
@Thomas These sets are not disjoint.
$endgroup$
– Andrés E. Caicedo
Mar 27 at 18:05
1
$begingroup$
@dandar Without the requirement of disjointness, of course the claim holds, and it is straightforward. Thomas sketched in comments what to do. I do not see uncountably many sets produced using procedures along the lines your comment suggests. If besides disjointness we also ignore the requirement of having uncountably many such sets, then yes, your comment gives a possible approach.
$endgroup$
– Andrés E. Caicedo
Mar 28 at 16:27
1
$begingroup$
@dandar Yes, that's the argument.
$endgroup$
– Andrés E. Caicedo
Mar 28 at 18:33
3
3
$begingroup$
Your definition of "uncountably infinite" is not standard. Usually, it is an infinite set $S$ such that $|S|neq aleph_0.$ There are a lot of uncountably infinite cardinals other than $|mathbb R.|$
$endgroup$
– Thomas Andrews
Mar 27 at 17:31
$begingroup$
Your definition of "uncountably infinite" is not standard. Usually, it is an infinite set $S$ such that $|S|neq aleph_0.$ There are a lot of uncountably infinite cardinals other than $|mathbb R.|$
$endgroup$
– Thomas Andrews
Mar 27 at 17:31
1
1
$begingroup$
No, you are wrong, @AndrésE.Caicedo Given $ain A$ take a countable subset $Csubseteq A$ with $anotin C.$ Then take $Asetminus Tmid Tsubset C, text and T infinite$. There are uncountably many (countably) infinite subsets $T$ of $C$, and for each $ain A-T,$ and $A-T$ is co-countable.
$endgroup$
– Thomas Andrews
Mar 27 at 18:03
$begingroup$
No, you are wrong, @AndrésE.Caicedo Given $ain A$ take a countable subset $Csubseteq A$ with $anotin C.$ Then take $Asetminus Tmid Tsubset C, text and T infinite$. There are uncountably many (countably) infinite subsets $T$ of $C$, and for each $ain A-T,$ and $A-T$ is co-countable.
$endgroup$
– Thomas Andrews
Mar 27 at 18:03
1
1
$begingroup$
@Thomas These sets are not disjoint.
$endgroup$
– Andrés E. Caicedo
Mar 27 at 18:05
$begingroup$
@Thomas These sets are not disjoint.
$endgroup$
– Andrés E. Caicedo
Mar 27 at 18:05
1
1
$begingroup$
@dandar Without the requirement of disjointness, of course the claim holds, and it is straightforward. Thomas sketched in comments what to do. I do not see uncountably many sets produced using procedures along the lines your comment suggests. If besides disjointness we also ignore the requirement of having uncountably many such sets, then yes, your comment gives a possible approach.
$endgroup$
– Andrés E. Caicedo
Mar 28 at 16:27
$begingroup$
@dandar Without the requirement of disjointness, of course the claim holds, and it is straightforward. Thomas sketched in comments what to do. I do not see uncountably many sets produced using procedures along the lines your comment suggests. If besides disjointness we also ignore the requirement of having uncountably many such sets, then yes, your comment gives a possible approach.
$endgroup$
– Andrés E. Caicedo
Mar 28 at 16:27
1
1
$begingroup$
@dandar Yes, that's the argument.
$endgroup$
– Andrés E. Caicedo
Mar 28 at 18:33
$begingroup$
@dandar Yes, that's the argument.
$endgroup$
– Andrés E. Caicedo
Mar 28 at 18:33
|
show 11 more comments
0
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oldest
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EgtCmnn,ZZAQqthrct5i,zpxPPIV6oEPQ,WbEhEsPqG0FS2Kpqj4EPxHSovB5T,x
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Your definition of "uncountably infinite" is not standard. Usually, it is an infinite set $S$ such that $|S|neq aleph_0.$ There are a lot of uncountably infinite cardinals other than $|mathbb R.|$
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– Thomas Andrews
Mar 27 at 17:31
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No, you are wrong, @AndrésE.Caicedo Given $ain A$ take a countable subset $Csubseteq A$ with $anotin C.$ Then take $Asetminus Tmid Tsubset C, text and T infinite$. There are uncountably many (countably) infinite subsets $T$ of $C$, and for each $ain A-T,$ and $A-T$ is co-countable.
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– Thomas Andrews
Mar 27 at 18:03
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@Thomas These sets are not disjoint.
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– Andrés E. Caicedo
Mar 27 at 18:05
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@dandar Without the requirement of disjointness, of course the claim holds, and it is straightforward. Thomas sketched in comments what to do. I do not see uncountably many sets produced using procedures along the lines your comment suggests. If besides disjointness we also ignore the requirement of having uncountably many such sets, then yes, your comment gives a possible approach.
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– Andrés E. Caicedo
Mar 28 at 16:27
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@dandar Yes, that's the argument.
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– Andrés E. Caicedo
Mar 28 at 18:33