Is it true that for any cardinal $kappa$ that in injects into $aleph_{omega}$, $(kappa)^{+}$ injects into...
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Here, $(kappa)^+$ is the Hartog's Number, the least cardinal $lambda$ so that there is no injection from $lambda$ to $kappa$.
My feeling is that this is true, since for any ordinal $alpha < omega $,
$ alpha+1<omega$, so for each infinite cardinal $aleph_alpha<aleph_omega$, $ aleph_{alpha+1}=(aleph_{alpha})^+<aleph_omega$, but I feel I am making some leaps in logic here or have missed something and this doesn't constitute any kind of a proof.
set-theory cardinals ordinals
asked Nov 24 at 15:07
MMR
267
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up vote
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Here, $(kappa)^+$ is the Hartog's Number, the least cardinal $lambda$ so that there is no injection from $lambda$ to $kappa$.
My feeling is that this is true, since for any ordinal $alpha < omega $,
$ alpha+1<omega$, so for each infinite cardinal $aleph_alpha<aleph_omega$, $ aleph_{alpha+1}=(aleph_{alpha})^+<aleph_omega$, but I feel I am making some leaps in logic here or have missed something and this doesn't constitute any kind of a proof.
set-theory cardinals ordinals
asked Nov 24 at 15:07
MMR
267
Doesn't $aleph_omega$ inject into $aleph_omega$?
– Andrés E. Caicedo
Nov 24 at 15:14
Yes, you are right. Thank you
– MMR
Nov 24 at 15:30
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Here, $(kappa)^+$ is the Hartog's Number, the least cardinal $lambda$ so that there is no injection from $lambda$ to $kappa$.
My feeling is that this is true, since for any ordinal $alpha < omega $,
$ alpha+1<omega$, so for each infinite cardinal $aleph_alpha<aleph_omega$, $ aleph_{alpha+1}=(aleph_{alpha})^+<aleph_omega$, but I feel I am making some leaps in logic here or have missed something and this doesn't constitute any kind of a proof.
set-theory cardinals ordinals
asked Nov 24 at 15:07
MMR
267
Here, $(kappa)^+$ is the Hartog's Number, the least cardinal $lambda$ so that there is no injection from $lambda$ to $kappa$.
My feeling is that this is true, since for any ordinal $alpha < omega $,
$ alpha+1<omega$, so for each infinite cardinal $aleph_alpha<aleph_omega$, $ aleph_{alpha+1}=(aleph_{alpha})^+<aleph_omega$, but I feel I am making some leaps in logic here or have missed something and this doesn't constitute any kind of a proof.
set-theory cardinals ordinals
set-theory cardinals ordinals
asked Nov 24 at 15:07
MMR
267
asked Nov 24 at 15:07
MMR
267
asked Nov 24 at 15:07
MMR
267
asked Nov 24 at 15:07
MMR
267
asked Nov 24 at 15:07
MMR
267
267
Doesn't $aleph_omega$ inject into $aleph_omega$?
– Andrés E. Caicedo
Nov 24 at 15:14
Yes, you are right. Thank you
– MMR
Nov 24 at 15:30
add a comment |
Doesn't $aleph_omega$ inject into $aleph_omega$?
– Andrés E. Caicedo
Nov 24 at 15:14
Yes, you are right. Thank you
– MMR
Nov 24 at 15:30
Doesn't $aleph_omega$ inject into $aleph_omega$?
– Andrés E. Caicedo
Nov 24 at 15:14
Doesn't $aleph_omega$ inject into $aleph_omega$?
– Andrés E. Caicedo
Nov 24 at 15:14
Yes, you are right. Thank you
– MMR
Nov 24 at 15:30
Yes, you are right. Thank you
– MMR
Nov 24 at 15:30
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
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accepted
It is true that $aleph_alpha<aleph_omegaimpliesaleph_{alpha+1}<aleph_omega$ - this is true simply because $omega$ is a limit ordinal - but "$aleph_alpha$ injects into $aleph_omega$" does not imply $aleph_alpha<aleph_omega$ (e.g. $aleph_omega$ injects into $aleph_omega$, obviously).
answered Nov 24 at 15:16
Noah Schweber
118k10146278
Ah yes, I see. How would you find an example of a cardinal with this property?
– MMR
Nov 24 at 15:32
@Matthew No cardinal has that property: $kappa$ always injects into $kappa$.
– Noah Schweber
Nov 24 at 15:35
Would it be true instead for the well ordering on ordinals i.e. If $kappa<aleph_omega$, then $(kappa)^+<aleph_omega$? Since in this case $aleph_omega$ is not strictly less than itself?
– MMR
Nov 24 at 15:49
1
@Matthew Yes. You can also phrase it in terms of cardinalities the same way.
– Andrés E. Caicedo
Nov 24 at 15:53
@Matthew And note that you can turn that into an embedding-phrased condition: if $kappa$ injects as a proper initial segment of $aleph_omega$, then $kappa^+<aleph_omega$.
– Noah Schweber
Nov 24 at 16:14
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It is true that $aleph_alpha<aleph_omegaimpliesaleph_{alpha+1}<aleph_omega$ - this is true simply because $omega$ is a limit ordinal - but "$aleph_alpha$ injects into $aleph_omega$" does not imply $aleph_alpha<aleph_omega$ (e.g. $aleph_omega$ injects into $aleph_omega$, obviously).
answered Nov 24 at 15:16
Noah Schweber
118k10146278
Ah yes, I see. How would you find an example of a cardinal with this property?
– MMR
Nov 24 at 15:32
@Matthew No cardinal has that property: $kappa$ always injects into $kappa$.
– Noah Schweber
Nov 24 at 15:35
Would it be true instead for the well ordering on ordinals i.e. If $kappa<aleph_omega$, then $(kappa)^+<aleph_omega$? Since in this case $aleph_omega$ is not strictly less than itself?
– MMR
Nov 24 at 15:49
1
@Matthew Yes. You can also phrase it in terms of cardinalities the same way.
– Andrés E. Caicedo
Nov 24 at 15:53
@Matthew And note that you can turn that into an embedding-phrased condition: if $kappa$ injects as a proper initial segment of $aleph_omega$, then $kappa^+<aleph_omega$.
– Noah Schweber
Nov 24 at 16:14
|
show 1 more comment
up vote
0
down vote
accepted
It is true that $aleph_alpha<aleph_omegaimpliesaleph_{alpha+1}<aleph_omega$ - this is true simply because $omega$ is a limit ordinal - but "$aleph_alpha$ injects into $aleph_omega$" does not imply $aleph_alpha<aleph_omega$ (e.g. $aleph_omega$ injects into $aleph_omega$, obviously).
answered Nov 24 at 15:16
Noah Schweber
118k10146278
Ah yes, I see. How would you find an example of a cardinal with this property?
– MMR
Nov 24 at 15:32
@Matthew No cardinal has that property: $kappa$ always injects into $kappa$.
– Noah Schweber
Nov 24 at 15:35
Would it be true instead for the well ordering on ordinals i.e. If $kappa<aleph_omega$, then $(kappa)^+<aleph_omega$? Since in this case $aleph_omega$ is not strictly less than itself?
– MMR
Nov 24 at 15:49
1
@Matthew Yes. You can also phrase it in terms of cardinalities the same way.
– Andrés E. Caicedo
Nov 24 at 15:53
@Matthew And note that you can turn that into an embedding-phrased condition: if $kappa$ injects as a proper initial segment of $aleph_omega$, then $kappa^+<aleph_omega$.
– Noah Schweber
Nov 24 at 16:14
|
show 1 more comment
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It is true that $aleph_alpha<aleph_omegaimpliesaleph_{alpha+1}<aleph_omega$ - this is true simply because $omega$ is a limit ordinal - but "$aleph_alpha$ injects into $aleph_omega$" does not imply $aleph_alpha<aleph_omega$ (e.g. $aleph_omega$ injects into $aleph_omega$, obviously).
answered Nov 24 at 15:16
Noah Schweber
118k10146278
It is true that $aleph_alpha<aleph_omegaimpliesaleph_{alpha+1}<aleph_omega$ - this is true simply because $omega$ is a limit ordinal - but "$aleph_alpha$ injects into $aleph_omega$" does not imply $aleph_alpha<aleph_omega$ (e.g. $aleph_omega$ injects into $aleph_omega$, obviously).
answered Nov 24 at 15:16
Noah Schweber
118k10146278
answered Nov 24 at 15:16
Noah Schweber
118k10146278
answered Nov 24 at 15:16
Noah Schweber
118k10146278
answered Nov 24 at 15:16
Noah Schweber
118k10146278
118k10146278
Ah yes, I see. How would you find an example of a cardinal with this property?
– MMR
Nov 24 at 15:32
@Matthew No cardinal has that property: $kappa$ always injects into $kappa$.
– Noah Schweber
Nov 24 at 15:35
Would it be true instead for the well ordering on ordinals i.e. If $kappa<aleph_omega$, then $(kappa)^+<aleph_omega$? Since in this case $aleph_omega$ is not strictly less than itself?
– MMR
Nov 24 at 15:49
1
@Matthew Yes. You can also phrase it in terms of cardinalities the same way.
– Andrés E. Caicedo
Nov 24 at 15:53
@Matthew And note that you can turn that into an embedding-phrased condition: if $kappa$ injects as a proper initial segment of $aleph_omega$, then $kappa^+<aleph_omega$.
– Noah Schweber
Nov 24 at 16:14
|
show 1 more comment
Ah yes, I see. How would you find an example of a cardinal with this property?
– MMR
Nov 24 at 15:32
@Matthew No cardinal has that property: $kappa$ always injects into $kappa$.
– Noah Schweber
Nov 24 at 15:35
Would it be true instead for the well ordering on ordinals i.e. If $kappa<aleph_omega$, then $(kappa)^+<aleph_omega$? Since in this case $aleph_omega$ is not strictly less than itself?
– MMR
Nov 24 at 15:49
1
@Matthew Yes. You can also phrase it in terms of cardinalities the same way.
– Andrés E. Caicedo
Nov 24 at 15:53
@Matthew And note that you can turn that into an embedding-phrased condition: if $kappa$ injects as a proper initial segment of $aleph_omega$, then $kappa^+<aleph_omega$.
– Noah Schweber
Nov 24 at 16:14
Ah yes, I see. How would you find an example of a cardinal with this property?
– MMR
Nov 24 at 15:32
Ah yes, I see. How would you find an example of a cardinal with this property?
– MMR
Nov 24 at 15:32
@Matthew No cardinal has that property: $kappa$ always injects into $kappa$.
– Noah Schweber
Nov 24 at 15:35
@Matthew No cardinal has that property: $kappa$ always injects into $kappa$.
– Noah Schweber
Nov 24 at 15:35
Would it be true instead for the well ordering on ordinals i.e. If $kappa<aleph_omega$, then $(kappa)^+<aleph_omega$? Since in this case $aleph_omega$ is not strictly less than itself?
– MMR
Nov 24 at 15:49
Would it be true instead for the well ordering on ordinals i.e. If $kappa<aleph_omega$, then $(kappa)^+<aleph_omega$? Since in this case $aleph_omega$ is not strictly less than itself?
– MMR
Nov 24 at 15:49
1
1
@Matthew Yes. You can also phrase it in terms of cardinalities the same way.
– Andrés E. Caicedo
Nov 24 at 15:53
@Matthew Yes. You can also phrase it in terms of cardinalities the same way.
– Andrés E. Caicedo
Nov 24 at 15:53
@Matthew And note that you can turn that into an embedding-phrased condition: if $kappa$ injects as a proper initial segment of $aleph_omega$, then $kappa^+<aleph_omega$.
– Noah Schweber
Nov 24 at 16:14
@Matthew And note that you can turn that into an embedding-phrased condition: if $kappa$ injects as a proper initial segment of $aleph_omega$, then $kappa^+<aleph_omega$.
– Noah Schweber
Nov 24 at 16:14
|
show 1 more comment
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Doesn't $aleph_omega$ inject into $aleph_omega$?
– Andrés E. Caicedo
Nov 24 at 15:14
Yes, you are right. Thank you
– MMR
Nov 24 at 15:30