Proof that $sigma$-algebra is not countable (proof revision)
$begingroup$
This problem has already several answers here
Prove that an infinite sigma algebra contains an infinite sequence of disjoint sets and is uncountable
and it has been asked a lot times before.
But I'd like a revision of my proof. Given the complexity of the solutions I've seen I believe my proof is wrong, but I cannot determine what is my mistake.
This is the problem:
Let $mathcal{M}$ be an infinite $sigma$ algebra on a nonempty set $X$. Show that
(a) $mathcal{M}$ contains an infinite sequence of disjoint sets
(b) $mathcal{M}$ is not countable
So for $(a)$, let ${E_j}_{j=1}^infty subseteq mathcal{M}$ be a sequence of elements in $mathcal{M}$. Define $F_i$ as
$$F_k = E_kbackslash left( bigcup_{i=1}^{k-1} E_i right) = E_k
cap left( bigcup_{i=1}^{k-1} E_i right)^c$$
so clearly ${F_j}_{j=1}^infty subseteq mathcal{M}$ and they are disjoint.
For $(b)$, suppose there exists and injection $f$ from $mathcal{M}$ to $omega$. Then $bar{f}= frestriction_{ {F_i}_{i=1}^infty}$ is inyective and $Im(bar{f}) = omega$. But for any $k$ we have $F_k^c in mathcal{M}$ and $F_k^c notin {F_i}_{i=1}^infty$ (because $X = F_k cup F_k^c$ and the $F_i$ are disjoint). So $bar{f}(F_k^c) notin omega$ (since $bar{f}$ is an injection), but this contradicts $Im(bar{f}) = omega$.
Thanks in advance!
real-analysis measure-theory proof-verification
$endgroup$
add a comment |
$begingroup$
This problem has already several answers here
Prove that an infinite sigma algebra contains an infinite sequence of disjoint sets and is uncountable
and it has been asked a lot times before.
But I'd like a revision of my proof. Given the complexity of the solutions I've seen I believe my proof is wrong, but I cannot determine what is my mistake.
This is the problem:
Let $mathcal{M}$ be an infinite $sigma$ algebra on a nonempty set $X$. Show that
(a) $mathcal{M}$ contains an infinite sequence of disjoint sets
(b) $mathcal{M}$ is not countable
So for $(a)$, let ${E_j}_{j=1}^infty subseteq mathcal{M}$ be a sequence of elements in $mathcal{M}$. Define $F_i$ as
$$F_k = E_kbackslash left( bigcup_{i=1}^{k-1} E_i right) = E_k
cap left( bigcup_{i=1}^{k-1} E_i right)^c$$
so clearly ${F_j}_{j=1}^infty subseteq mathcal{M}$ and they are disjoint.
For $(b)$, suppose there exists and injection $f$ from $mathcal{M}$ to $omega$. Then $bar{f}= frestriction_{ {F_i}_{i=1}^infty}$ is inyective and $Im(bar{f}) = omega$. But for any $k$ we have $F_k^c in mathcal{M}$ and $F_k^c notin {F_i}_{i=1}^infty$ (because $X = F_k cup F_k^c$ and the $F_i$ are disjoint). So $bar{f}(F_k^c) notin omega$ (since $bar{f}$ is an injection), but this contradicts $Im(bar{f}) = omega$.
Thanks in advance!
real-analysis measure-theory proof-verification
$endgroup$
add a comment |
$begingroup$
This problem has already several answers here
Prove that an infinite sigma algebra contains an infinite sequence of disjoint sets and is uncountable
and it has been asked a lot times before.
But I'd like a revision of my proof. Given the complexity of the solutions I've seen I believe my proof is wrong, but I cannot determine what is my mistake.
This is the problem:
Let $mathcal{M}$ be an infinite $sigma$ algebra on a nonempty set $X$. Show that
(a) $mathcal{M}$ contains an infinite sequence of disjoint sets
(b) $mathcal{M}$ is not countable
So for $(a)$, let ${E_j}_{j=1}^infty subseteq mathcal{M}$ be a sequence of elements in $mathcal{M}$. Define $F_i$ as
$$F_k = E_kbackslash left( bigcup_{i=1}^{k-1} E_i right) = E_k
cap left( bigcup_{i=1}^{k-1} E_i right)^c$$
so clearly ${F_j}_{j=1}^infty subseteq mathcal{M}$ and they are disjoint.
For $(b)$, suppose there exists and injection $f$ from $mathcal{M}$ to $omega$. Then $bar{f}= frestriction_{ {F_i}_{i=1}^infty}$ is inyective and $Im(bar{f}) = omega$. But for any $k$ we have $F_k^c in mathcal{M}$ and $F_k^c notin {F_i}_{i=1}^infty$ (because $X = F_k cup F_k^c$ and the $F_i$ are disjoint). So $bar{f}(F_k^c) notin omega$ (since $bar{f}$ is an injection), but this contradicts $Im(bar{f}) = omega$.
Thanks in advance!
real-analysis measure-theory proof-verification
$endgroup$
This problem has already several answers here
Prove that an infinite sigma algebra contains an infinite sequence of disjoint sets and is uncountable
and it has been asked a lot times before.
But I'd like a revision of my proof. Given the complexity of the solutions I've seen I believe my proof is wrong, but I cannot determine what is my mistake.
This is the problem:
Let $mathcal{M}$ be an infinite $sigma$ algebra on a nonempty set $X$. Show that
(a) $mathcal{M}$ contains an infinite sequence of disjoint sets
(b) $mathcal{M}$ is not countable
So for $(a)$, let ${E_j}_{j=1}^infty subseteq mathcal{M}$ be a sequence of elements in $mathcal{M}$. Define $F_i$ as
$$F_k = E_kbackslash left( bigcup_{i=1}^{k-1} E_i right) = E_k
cap left( bigcup_{i=1}^{k-1} E_i right)^c$$
so clearly ${F_j}_{j=1}^infty subseteq mathcal{M}$ and they are disjoint.
For $(b)$, suppose there exists and injection $f$ from $mathcal{M}$ to $omega$. Then $bar{f}= frestriction_{ {F_i}_{i=1}^infty}$ is inyective and $Im(bar{f}) = omega$. But for any $k$ we have $F_k^c in mathcal{M}$ and $F_k^c notin {F_i}_{i=1}^infty$ (because $X = F_k cup F_k^c$ and the $F_i$ are disjoint). So $bar{f}(F_k^c) notin omega$ (since $bar{f}$ is an injection), but this contradicts $Im(bar{f}) = omega$.
Thanks in advance!
real-analysis measure-theory proof-verification
real-analysis measure-theory proof-verification
edited Dec 18 '18 at 7:37
Did
248k23224463
248k23224463
asked Dec 18 '18 at 6:37
mate89mate89
1819
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1 Answer
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$begingroup$
There is no guarantee that the $F_k$ will not be empty from some $k$ onwards.
You will need to do more work to ensure we get non-empty sets.
The proof of uncountability you gave makes no sense. Why is the image of $bar{f}$ equal to $omega$? You just claim that.
$endgroup$
$begingroup$
Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
$endgroup$
– mate89
Dec 19 '18 at 4:23
$begingroup$
@mate89 the image can be any infinite subset of $omega$
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:11
$begingroup$
Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
$endgroup$
– mate89
Dec 19 '18 at 5:22
$begingroup$
@mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:42
add a comment |
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1 Answer
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1 Answer
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$begingroup$
There is no guarantee that the $F_k$ will not be empty from some $k$ onwards.
You will need to do more work to ensure we get non-empty sets.
The proof of uncountability you gave makes no sense. Why is the image of $bar{f}$ equal to $omega$? You just claim that.
$endgroup$
$begingroup$
Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
$endgroup$
– mate89
Dec 19 '18 at 4:23
$begingroup$
@mate89 the image can be any infinite subset of $omega$
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:11
$begingroup$
Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
$endgroup$
– mate89
Dec 19 '18 at 5:22
$begingroup$
@mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:42
add a comment |
$begingroup$
There is no guarantee that the $F_k$ will not be empty from some $k$ onwards.
You will need to do more work to ensure we get non-empty sets.
The proof of uncountability you gave makes no sense. Why is the image of $bar{f}$ equal to $omega$? You just claim that.
$endgroup$
$begingroup$
Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
$endgroup$
– mate89
Dec 19 '18 at 4:23
$begingroup$
@mate89 the image can be any infinite subset of $omega$
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:11
$begingroup$
Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
$endgroup$
– mate89
Dec 19 '18 at 5:22
$begingroup$
@mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:42
add a comment |
$begingroup$
There is no guarantee that the $F_k$ will not be empty from some $k$ onwards.
You will need to do more work to ensure we get non-empty sets.
The proof of uncountability you gave makes no sense. Why is the image of $bar{f}$ equal to $omega$? You just claim that.
$endgroup$
There is no guarantee that the $F_k$ will not be empty from some $k$ onwards.
You will need to do more work to ensure we get non-empty sets.
The proof of uncountability you gave makes no sense. Why is the image of $bar{f}$ equal to $omega$? You just claim that.
answered Dec 18 '18 at 7:04
Henno BrandsmaHenno Brandsma
111k348118
111k348118
$begingroup$
Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
$endgroup$
– mate89
Dec 19 '18 at 4:23
$begingroup$
@mate89 the image can be any infinite subset of $omega$
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:11
$begingroup$
Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
$endgroup$
– mate89
Dec 19 '18 at 5:22
$begingroup$
@mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:42
add a comment |
$begingroup$
Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
$endgroup$
– mate89
Dec 19 '18 at 4:23
$begingroup$
@mate89 the image can be any infinite subset of $omega$
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:11
$begingroup$
Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
$endgroup$
– mate89
Dec 19 '18 at 5:22
$begingroup$
@mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:42
$begingroup$
Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
$endgroup$
– mate89
Dec 19 '18 at 4:23
$begingroup$
Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $omega$ (for any $k in omega$ there is a corresponding $F_k$), thus $Im(bar{f}) = omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks!
$endgroup$
– mate89
Dec 19 '18 at 4:23
$begingroup$
@mate89 the image can be any infinite subset of $omega$
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:11
$begingroup$
@mate89 the image can be any infinite subset of $omega$
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:11
$begingroup$
Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
$endgroup$
– mate89
Dec 19 '18 at 5:22
$begingroup$
Hmmm I see. One last question: if I have a sequence ${x_i}_{i=0}^infty$ where they are all disctinct ($x_i ne x_j$ for $j ne i$), and we set $A := {x_i}_{i=0}^infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = aleph_0$. Thanks again!
$endgroup$
– mate89
Dec 19 '18 at 5:22
$begingroup$
@mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:42
$begingroup$
@mate89 by definition $i to x_i$ is an injection so $A$ is trivially countably infinite.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 5:42
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