Mixing
Proving that $(omega_n)^omega=omega_n$ providing CH but not GCH
$begingroup$
This is an exercise from a book from Kunen - SET THEORY, An Introduction to Independence Proofs
Assume CH but don't assume GCH. Show that $(omega_n)^omega=omega_n$ for $1 le n < omega$.
I don't have a clue for a good starting point...
set-theory cardinals
edited Aug 3 '15 at 14:29
Asaf Karagila♦
308k33441774
asked Aug 3 '15 at 12:54
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
This is an exercise from a book from Kunen - SET THEORY, An Introduction to Independence Proofs
Assume CH but don't assume GCH. Show that $(omega_n)^omega=omega_n$ for $1 le n < omega$.
I don't have a clue for a good starting point...
set-theory cardinals
edited Aug 3 '15 at 14:29
Asaf Karagila♦
308k33441774
asked Aug 3 '15 at 12:54
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
If you have heard of the Hausdorff formula this follows immediately. If not, a good way to proceed is by induction. The case $n=1$ follows by CH. Now consider $omega_{n+1}^omega$ as the set of functions from $omega$ into $omega_{n+1}$. But $omega_{n+1}$ is regular, so...
$endgroup$
– Miha Habič
Aug 3 '15 at 13:25
add a comment |
$begingroup$
This is an exercise from a book from Kunen - SET THEORY, An Introduction to Independence Proofs
Assume CH but don't assume GCH. Show that $(omega_n)^omega=omega_n$ for $1 le n < omega$.
I don't have a clue for a good starting point...
set-theory cardinals
edited Aug 3 '15 at 14:29
Asaf Karagila♦
308k33441774
asked Aug 3 '15 at 12:54
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
This is an exercise from a book from Kunen - SET THEORY, An Introduction to Independence Proofs
Assume CH but don't assume GCH. Show that $(omega_n)^omega=omega_n$ for $1 le n < omega$.
I don't have a clue for a good starting point...
set-theory cardinals
set-theory cardinals
edited Aug 3 '15 at 14:29
Asaf Karagila♦
308k33441774
asked Aug 3 '15 at 12:54
mathcounterexamples.netmathcounterexamples.net
26.9k22158
edited Aug 3 '15 at 14:29
Asaf Karagila♦
308k33441774
asked Aug 3 '15 at 12:54
mathcounterexamples.netmathcounterexamples.net
26.9k22158
edited Aug 3 '15 at 14:29
Asaf Karagila♦
308k33441774
edited Aug 3 '15 at 14:29
Asaf Karagila♦
308k33441774
edited Aug 3 '15 at 14:29
Asaf Karagila♦
308k33441774
308k33441774
asked Aug 3 '15 at 12:54
mathcounterexamples.netmathcounterexamples.net
26.9k22158
asked Aug 3 '15 at 12:54
mathcounterexamples.netmathcounterexamples.net
26.9k22158
asked Aug 3 '15 at 12:54
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
$begingroup$
If you have heard of the Hausdorff formula this follows immediately. If not, a good way to proceed is by induction. The case $n=1$ follows by CH. Now consider $omega_{n+1}^omega$ as the set of functions from $omega$ into $omega_{n+1}$. But $omega_{n+1}$ is regular, so...
$endgroup$
– Miha Habič
Aug 3 '15 at 13:25
add a comment |
$begingroup$
If you have heard of the Hausdorff formula this follows immediately. If not, a good way to proceed is by induction. The case $n=1$ follows by CH. Now consider $omega_{n+1}^omega$ as the set of functions from $omega$ into $omega_{n+1}$. But $omega_{n+1}$ is regular, so...
$endgroup$
– Miha Habič
Aug 3 '15 at 13:25
$begingroup$
If you have heard of the Hausdorff formula this follows immediately. If not, a good way to proceed is by induction. The case $n=1$ follows by CH. Now consider $omega_{n+1}^omega$ as the set of functions from $omega$ into $omega_{n+1}$. But $omega_{n+1}$ is regular, so...
$endgroup$
– Miha Habič
Aug 3 '15 at 13:25
$begingroup$
If you have heard of the Hausdorff formula this follows immediately. If not, a good way to proceed is by induction. The case $n=1$ follows by CH. Now consider $omega_{n+1}^omega$ as the set of functions from $omega$ into $omega_{n+1}$. But $omega_{n+1}$ is regular, so...
$endgroup$
– Miha Habič
Aug 3 '15 at 13:25
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
HINT: Use Hausdorff's formula which states that $aleph_{alpha+1}^{aleph_beta}=aleph_alpha^{aleph_beta}cdotaleph_{alpha+1}$. And induction.
answered Aug 3 '15 at 14:28
Asaf Karagila♦Asaf Karagila
308k33441774
$endgroup$
add a comment |
$begingroup$
Recall that $omega_1$ is a regular cardinal. Now, consider $(omega_1)^omega$ and let $A in (omega_1)^omega$. Define the $sup(A) = alpha$ such that $alpha$ is the smallest ordinal where every $ beta in A$, $ beta < alpha$. Since $cof(omega_1) = omega_1$, we note that $alpha$ must always be a countable ordinal. Partition $ (omega_1)^omega$ into equivalence classes where $A$~ $B$ if $sup(A) = sup(B)$. Denote the equivalence class for some ordinal $alpha$ and $O_alpha$. Notice that $|O_alpha| = 2^{aleph_0} = aleph_1 $ for $omega<alpha <omega_1$ (and $|O_alpha| = emptyset$ for any finite $alpha$).
Thus, $(|omega_1)^omega| = |bigcup_{omega<alpha<omega_1} O_alpha|$. But the RHS is just the $omega_1$ union of sets of size $omega_1$ (which has cardinality $omega_1$).
Now you can finish the problem by induction on $n$.
answered Aug 3 '15 at 13:33
KyleKyle
5,29221334
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add a comment |
$begingroup$
By CH, we know that $2^omega = omega_1$. Thus, ${omega_1}^omega = (2^omega)^omega = 2^{omegacdotomega} = 2^omega = omega_1$.
Now assume that ${omega_n}^omega = omega_n$. Because $omega_{n+1}$ is regular, any map from $omega$ to $omega_{n+1}$ cannot be cofinal. Thus the set of maps from $omega$ to $omega_{n+1}$ is the union of the set of maps from $omega$ to $alpha$ where the cardinality of $alpha$ is $omega_n$. By our inductive hypothesis, each element of this union has size $omega_n$ and there are $omega_{n+1}$ members of this union, so the set of all maps from $omega$ to $omega_{n+1}$ has size no greater than $omega_{n+1}$. Q.E.D.
answered Jan 31 at 2:08
Robert ShoreRobert Shore
3,611324
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
HINT: Use Hausdorff's formula which states that $aleph_{alpha+1}^{aleph_beta}=aleph_alpha^{aleph_beta}cdotaleph_{alpha+1}$. And induction.
answered Aug 3 '15 at 14:28
Asaf Karagila♦Asaf Karagila
308k33441774
$endgroup$
add a comment |
$begingroup$
HINT: Use Hausdorff's formula which states that $aleph_{alpha+1}^{aleph_beta}=aleph_alpha^{aleph_beta}cdotaleph_{alpha+1}$. And induction.
answered Aug 3 '15 at 14:28
Asaf Karagila♦Asaf Karagila
308k33441774
$endgroup$
add a comment |
$begingroup$
HINT: Use Hausdorff's formula which states that $aleph_{alpha+1}^{aleph_beta}=aleph_alpha^{aleph_beta}cdotaleph_{alpha+1}$. And induction.
answered Aug 3 '15 at 14:28
Asaf Karagila♦Asaf Karagila
308k33441774
$endgroup$
HINT: Use Hausdorff's formula which states that $aleph_{alpha+1}^{aleph_beta}=aleph_alpha^{aleph_beta}cdotaleph_{alpha+1}$. And induction.
answered Aug 3 '15 at 14:28
Asaf Karagila♦Asaf Karagila
308k33441774
answered Aug 3 '15 at 14:28
Asaf Karagila♦Asaf Karagila
308k33441774
answered Aug 3 '15 at 14:28
Asaf Karagila♦Asaf Karagila
308k33441774
answered Aug 3 '15 at 14:28
Asaf Karagila♦Asaf Karagila
308k33441774
308k33441774
add a comment |
add a comment |
$begingroup$
Recall that $omega_1$ is a regular cardinal. Now, consider $(omega_1)^omega$ and let $A in (omega_1)^omega$. Define the $sup(A) = alpha$ such that $alpha$ is the smallest ordinal where every $ beta in A$, $ beta < alpha$. Since $cof(omega_1) = omega_1$, we note that $alpha$ must always be a countable ordinal. Partition $ (omega_1)^omega$ into equivalence classes where $A$~ $B$ if $sup(A) = sup(B)$. Denote the equivalence class for some ordinal $alpha$ and $O_alpha$. Notice that $|O_alpha| = 2^{aleph_0} = aleph_1 $ for $omega<alpha <omega_1$ (and $|O_alpha| = emptyset$ for any finite $alpha$).
Thus, $(|omega_1)^omega| = |bigcup_{omega<alpha<omega_1} O_alpha|$. But the RHS is just the $omega_1$ union of sets of size $omega_1$ (which has cardinality $omega_1$).
Now you can finish the problem by induction on $n$.
answered Aug 3 '15 at 13:33
KyleKyle
5,29221334
$endgroup$
add a comment |
$begingroup$
Recall that $omega_1$ is a regular cardinal. Now, consider $(omega_1)^omega$ and let $A in (omega_1)^omega$. Define the $sup(A) = alpha$ such that $alpha$ is the smallest ordinal where every $ beta in A$, $ beta < alpha$. Since $cof(omega_1) = omega_1$, we note that $alpha$ must always be a countable ordinal. Partition $ (omega_1)^omega$ into equivalence classes where $A$~ $B$ if $sup(A) = sup(B)$. Denote the equivalence class for some ordinal $alpha$ and $O_alpha$. Notice that $|O_alpha| = 2^{aleph_0} = aleph_1 $ for $omega<alpha <omega_1$ (and $|O_alpha| = emptyset$ for any finite $alpha$).
Thus, $(|omega_1)^omega| = |bigcup_{omega<alpha<omega_1} O_alpha|$. But the RHS is just the $omega_1$ union of sets of size $omega_1$ (which has cardinality $omega_1$).
Now you can finish the problem by induction on $n$.
answered Aug 3 '15 at 13:33
KyleKyle
5,29221334
$endgroup$
add a comment |
$begingroup$
Recall that $omega_1$ is a regular cardinal. Now, consider $(omega_1)^omega$ and let $A in (omega_1)^omega$. Define the $sup(A) = alpha$ such that $alpha$ is the smallest ordinal where every $ beta in A$, $ beta < alpha$. Since $cof(omega_1) = omega_1$, we note that $alpha$ must always be a countable ordinal. Partition $ (omega_1)^omega$ into equivalence classes where $A$~ $B$ if $sup(A) = sup(B)$. Denote the equivalence class for some ordinal $alpha$ and $O_alpha$. Notice that $|O_alpha| = 2^{aleph_0} = aleph_1 $ for $omega<alpha <omega_1$ (and $|O_alpha| = emptyset$ for any finite $alpha$).
Thus, $(|omega_1)^omega| = |bigcup_{omega<alpha<omega_1} O_alpha|$. But the RHS is just the $omega_1$ union of sets of size $omega_1$ (which has cardinality $omega_1$).
Now you can finish the problem by induction on $n$.
answered Aug 3 '15 at 13:33
KyleKyle
5,29221334
$endgroup$
Recall that $omega_1$ is a regular cardinal. Now, consider $(omega_1)^omega$ and let $A in (omega_1)^omega$. Define the $sup(A) = alpha$ such that $alpha$ is the smallest ordinal where every $ beta in A$, $ beta < alpha$. Since $cof(omega_1) = omega_1$, we note that $alpha$ must always be a countable ordinal. Partition $ (omega_1)^omega$ into equivalence classes where $A$~ $B$ if $sup(A) = sup(B)$. Denote the equivalence class for some ordinal $alpha$ and $O_alpha$. Notice that $|O_alpha| = 2^{aleph_0} = aleph_1 $ for $omega<alpha <omega_1$ (and $|O_alpha| = emptyset$ for any finite $alpha$).
Thus, $(|omega_1)^omega| = |bigcup_{omega<alpha<omega_1} O_alpha|$. But the RHS is just the $omega_1$ union of sets of size $omega_1$ (which has cardinality $omega_1$).
Now you can finish the problem by induction on $n$.
answered Aug 3 '15 at 13:33
KyleKyle
5,29221334
edited Aug 3 '15 at 13:40
answered Aug 3 '15 at 13:33
KyleKyle
5,29221334
answered Aug 3 '15 at 13:33
KyleKyle
5,29221334
answered Aug 3 '15 at 13:33
KyleKyle
5,29221334
5,29221334
add a comment |
add a comment |
$begingroup$
By CH, we know that $2^omega = omega_1$. Thus, ${omega_1}^omega = (2^omega)^omega = 2^{omegacdotomega} = 2^omega = omega_1$.
Now assume that ${omega_n}^omega = omega_n$. Because $omega_{n+1}$ is regular, any map from $omega$ to $omega_{n+1}$ cannot be cofinal. Thus the set of maps from $omega$ to $omega_{n+1}$ is the union of the set of maps from $omega$ to $alpha$ where the cardinality of $alpha$ is $omega_n$. By our inductive hypothesis, each element of this union has size $omega_n$ and there are $omega_{n+1}$ members of this union, so the set of all maps from $omega$ to $omega_{n+1}$ has size no greater than $omega_{n+1}$. Q.E.D.
answered Jan 31 at 2:08
Robert ShoreRobert Shore
3,611324
$endgroup$
add a comment |
$begingroup$
By CH, we know that $2^omega = omega_1$. Thus, ${omega_1}^omega = (2^omega)^omega = 2^{omegacdotomega} = 2^omega = omega_1$.
Now assume that ${omega_n}^omega = omega_n$. Because $omega_{n+1}$ is regular, any map from $omega$ to $omega_{n+1}$ cannot be cofinal. Thus the set of maps from $omega$ to $omega_{n+1}$ is the union of the set of maps from $omega$ to $alpha$ where the cardinality of $alpha$ is $omega_n$. By our inductive hypothesis, each element of this union has size $omega_n$ and there are $omega_{n+1}$ members of this union, so the set of all maps from $omega$ to $omega_{n+1}$ has size no greater than $omega_{n+1}$. Q.E.D.
answered Jan 31 at 2:08
Robert ShoreRobert Shore
3,611324
$endgroup$
add a comment |
$begingroup$
By CH, we know that $2^omega = omega_1$. Thus, ${omega_1}^omega = (2^omega)^omega = 2^{omegacdotomega} = 2^omega = omega_1$.
Now assume that ${omega_n}^omega = omega_n$. Because $omega_{n+1}$ is regular, any map from $omega$ to $omega_{n+1}$ cannot be cofinal. Thus the set of maps from $omega$ to $omega_{n+1}$ is the union of the set of maps from $omega$ to $alpha$ where the cardinality of $alpha$ is $omega_n$. By our inductive hypothesis, each element of this union has size $omega_n$ and there are $omega_{n+1}$ members of this union, so the set of all maps from $omega$ to $omega_{n+1}$ has size no greater than $omega_{n+1}$. Q.E.D.
answered Jan 31 at 2:08
Robert ShoreRobert Shore
3,611324
$endgroup$
By CH, we know that $2^omega = omega_1$. Thus, ${omega_1}^omega = (2^omega)^omega = 2^{omegacdotomega} = 2^omega = omega_1$.
Now assume that ${omega_n}^omega = omega_n$. Because $omega_{n+1}$ is regular, any map from $omega$ to $omega_{n+1}$ cannot be cofinal. Thus the set of maps from $omega$ to $omega_{n+1}$ is the union of the set of maps from $omega$ to $alpha$ where the cardinality of $alpha$ is $omega_n$. By our inductive hypothesis, each element of this union has size $omega_n$ and there are $omega_{n+1}$ members of this union, so the set of all maps from $omega$ to $omega_{n+1}$ has size no greater than $omega_{n+1}$. Q.E.D.
answered Jan 31 at 2:08
Robert ShoreRobert Shore
3,611324
answered Jan 31 at 2:08
Robert ShoreRobert Shore
3,611324
answered Jan 31 at 2:08
Robert ShoreRobert Shore
3,611324
answered Jan 31 at 2:08
Robert ShoreRobert Shore
3,611324
3,611324
add a comment |
add a comment |
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$begingroup$
If you have heard of the Hausdorff formula this follows immediately. If not, a good way to proceed is by induction. The case $n=1$ follows by CH. Now consider $omega_{n+1}^omega$ as the set of functions from $omega$ into $omega_{n+1}$. But $omega_{n+1}$ is regular, so...
$endgroup$
– Miha Habič
Aug 3 '15 at 13:25