If $bigcup_{i=1}^infty A_i$ has cardinality $kappa$, then some $A_i$ has cardinality $kappa$?












2












$begingroup$


Is the following claim true?




If $bigcup_{i=1}^infty A_i$ has cardinality $kappa$, then some $A_i$ has cardinality $kappa$. Here $kappa$ is uncountable.




I'm interested in the particular case $kappa=c:=|mathbb{R}|$. In other words, if each $A_i$ has cardinality strictly less than $kappa$, can we deduce that $bigcup_{i=1}^infty A_i$ has cardinality less than $kappa$?



I know very little cardinal arithmetic beyond the usual rules for sums and products. But I came across this in an analysis text, so I wonder whether it's true.






set-theory cardinals







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$endgroup$












  • $begingroup$
    What text was this #*!% in?
    $endgroup$
    – DanielWainfleet
    Jan 14 at 1:55


















2












$begingroup$


Is the following claim true?




If $bigcup_{i=1}^infty A_i$ has cardinality $kappa$, then some $A_i$ has cardinality $kappa$. Here $kappa$ is uncountable.




I'm interested in the particular case $kappa=c:=|mathbb{R}|$. In other words, if each $A_i$ has cardinality strictly less than $kappa$, can we deduce that $bigcup_{i=1}^infty A_i$ has cardinality less than $kappa$?



I know very little cardinal arithmetic beyond the usual rules for sums and products. But I came across this in an analysis text, so I wonder whether it's true.






set-theory cardinals







share|cite|improve this question









$endgroup$












  • $begingroup$
    What text was this #*!% in?
    $endgroup$
    – DanielWainfleet
    Jan 14 at 1:55
















2












2








2





$begingroup$


Is the following claim true?




If $bigcup_{i=1}^infty A_i$ has cardinality $kappa$, then some $A_i$ has cardinality $kappa$. Here $kappa$ is uncountable.




I'm interested in the particular case $kappa=c:=|mathbb{R}|$. In other words, if each $A_i$ has cardinality strictly less than $kappa$, can we deduce that $bigcup_{i=1}^infty A_i$ has cardinality less than $kappa$?



I know very little cardinal arithmetic beyond the usual rules for sums and products. But I came across this in an analysis text, so I wonder whether it's true.






set-theory cardinals







share|cite|improve this question









$endgroup$




Is the following claim true?




If $bigcup_{i=1}^infty A_i$ has cardinality $kappa$, then some $A_i$ has cardinality $kappa$. Here $kappa$ is uncountable.




I'm interested in the particular case $kappa=c:=|mathbb{R}|$. In other words, if each $A_i$ has cardinality strictly less than $kappa$, can we deduce that $bigcup_{i=1}^infty A_i$ has cardinality less than $kappa$?



I know very little cardinal arithmetic beyond the usual rules for sums and products. But I came across this in an analysis text, so I wonder whether it's true.






set-theory cardinals




set-theory cardinals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 31 '18 at 10:15









ColescuColescu

3,20011136




3,20011136












  • $begingroup$
    What text was this #*!% in?
    $endgroup$
    – DanielWainfleet
    Jan 14 at 1:55




















  • $begingroup$
    What text was this #*!% in?
    $endgroup$
    – DanielWainfleet
    Jan 14 at 1:55


















$begingroup$
What text was this #*!% in?
$endgroup$
– DanielWainfleet
Jan 14 at 1:55






$begingroup$
What text was this #*!% in?
$endgroup$
– DanielWainfleet
Jan 14 at 1:55












1 Answer
1






active

oldest

votes


















1












$begingroup$

Counter-example:



$$left| bigcup_{i=1}^infty omega_i right| = omega_omega$$




We have:



$$kappa = left| bigcup_{i=1}^infty A_i right| le omega cdot sup_{i=1..infty} |A_i| = sup_{i=1..infty} |A_i|$$



So in order to conclude that $kappa = |A_i|$ for some $i$, we need $operatorname{cof}(kappa) = omega$, otherwise there are coutner-examples. For $kappa = mathfrak c$ though, you are lucky, since $operatorname{cof}(mathfrak c) > omega$, so you can always conclude that $kappa = |A_i|$ for some $i$.




Specializing on the case $kappa = mathfrak c$:



Let $|A_i| < mathfrak c$ for all $i$. By Konig's lemma:
$$left| bigcup_{i=1}^infty A_i right| < left| prod_{i=1}^infty mathfrak c right| = mathfrak c^omega = 2^{omega cdot omega} = 2^omega = mathfrak c$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for answering! Could you please elaborate on why this is true for $c$? I don't understand your notation cof and all that.. (I'm practically blind about set theory)
    $endgroup$
    – Colescu
    Dec 31 '18 at 10:25












  • $begingroup$
    The last part is not exactly true in ZF, it is consistent that the reals are countable union of countable sets. In ZFC it is true(the proof that $mbox{cof}(2^λ)>λ$ require choice)
    $endgroup$
    – Holo
    Dec 31 '18 at 10:33












  • $begingroup$
    @Colescu I've included some explanation
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:38










  • $begingroup$
    It might be better to wriite $aleph_i$ instead of $omega_i$, just to make sure we''re using cardinal operations. (For example: $omega=aleph_0$; one thinks of $omega$ as an ordinal annd $aleph_0$ as a cardinal. So, with cardinal multiplication, $aleph_0aleph_0=aleph_0$. But if the notation $omegaomega$ is interppreted as an ordinal product then $omegaomeganeomega$.)
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 15:27













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Counter-example:



$$left| bigcup_{i=1}^infty omega_i right| = omega_omega$$




We have:



$$kappa = left| bigcup_{i=1}^infty A_i right| le omega cdot sup_{i=1..infty} |A_i| = sup_{i=1..infty} |A_i|$$



So in order to conclude that $kappa = |A_i|$ for some $i$, we need $operatorname{cof}(kappa) = omega$, otherwise there are coutner-examples. For $kappa = mathfrak c$ though, you are lucky, since $operatorname{cof}(mathfrak c) > omega$, so you can always conclude that $kappa = |A_i|$ for some $i$.




Specializing on the case $kappa = mathfrak c$:



Let $|A_i| < mathfrak c$ for all $i$. By Konig's lemma:
$$left| bigcup_{i=1}^infty A_i right| < left| prod_{i=1}^infty mathfrak c right| = mathfrak c^omega = 2^{omega cdot omega} = 2^omega = mathfrak c$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for answering! Could you please elaborate on why this is true for $c$? I don't understand your notation cof and all that.. (I'm practically blind about set theory)
    $endgroup$
    – Colescu
    Dec 31 '18 at 10:25












  • $begingroup$
    The last part is not exactly true in ZF, it is consistent that the reals are countable union of countable sets. In ZFC it is true(the proof that $mbox{cof}(2^λ)>λ$ require choice)
    $endgroup$
    – Holo
    Dec 31 '18 at 10:33












  • $begingroup$
    @Colescu I've included some explanation
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:38










  • $begingroup$
    It might be better to wriite $aleph_i$ instead of $omega_i$, just to make sure we''re using cardinal operations. (For example: $omega=aleph_0$; one thinks of $omega$ as an ordinal annd $aleph_0$ as a cardinal. So, with cardinal multiplication, $aleph_0aleph_0=aleph_0$. But if the notation $omegaomega$ is interppreted as an ordinal product then $omegaomeganeomega$.)
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 15:27


















1












$begingroup$

Counter-example:



$$left| bigcup_{i=1}^infty omega_i right| = omega_omega$$




We have:



$$kappa = left| bigcup_{i=1}^infty A_i right| le omega cdot sup_{i=1..infty} |A_i| = sup_{i=1..infty} |A_i|$$



So in order to conclude that $kappa = |A_i|$ for some $i$, we need $operatorname{cof}(kappa) = omega$, otherwise there are coutner-examples. For $kappa = mathfrak c$ though, you are lucky, since $operatorname{cof}(mathfrak c) > omega$, so you can always conclude that $kappa = |A_i|$ for some $i$.




Specializing on the case $kappa = mathfrak c$:



Let $|A_i| < mathfrak c$ for all $i$. By Konig's lemma:
$$left| bigcup_{i=1}^infty A_i right| < left| prod_{i=1}^infty mathfrak c right| = mathfrak c^omega = 2^{omega cdot omega} = 2^omega = mathfrak c$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for answering! Could you please elaborate on why this is true for $c$? I don't understand your notation cof and all that.. (I'm practically blind about set theory)
    $endgroup$
    – Colescu
    Dec 31 '18 at 10:25












  • $begingroup$
    The last part is not exactly true in ZF, it is consistent that the reals are countable union of countable sets. In ZFC it is true(the proof that $mbox{cof}(2^λ)>λ$ require choice)
    $endgroup$
    – Holo
    Dec 31 '18 at 10:33












  • $begingroup$
    @Colescu I've included some explanation
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:38










  • $begingroup$
    It might be better to wriite $aleph_i$ instead of $omega_i$, just to make sure we''re using cardinal operations. (For example: $omega=aleph_0$; one thinks of $omega$ as an ordinal annd $aleph_0$ as a cardinal. So, with cardinal multiplication, $aleph_0aleph_0=aleph_0$. But if the notation $omegaomega$ is interppreted as an ordinal product then $omegaomeganeomega$.)
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 15:27
















1












1








1





$begingroup$

Counter-example:



$$left| bigcup_{i=1}^infty omega_i right| = omega_omega$$




We have:



$$kappa = left| bigcup_{i=1}^infty A_i right| le omega cdot sup_{i=1..infty} |A_i| = sup_{i=1..infty} |A_i|$$



So in order to conclude that $kappa = |A_i|$ for some $i$, we need $operatorname{cof}(kappa) = omega$, otherwise there are coutner-examples. For $kappa = mathfrak c$ though, you are lucky, since $operatorname{cof}(mathfrak c) > omega$, so you can always conclude that $kappa = |A_i|$ for some $i$.




Specializing on the case $kappa = mathfrak c$:



Let $|A_i| < mathfrak c$ for all $i$. By Konig's lemma:
$$left| bigcup_{i=1}^infty A_i right| < left| prod_{i=1}^infty mathfrak c right| = mathfrak c^omega = 2^{omega cdot omega} = 2^omega = mathfrak c$$






share|cite|improve this answer











$endgroup$



Counter-example:



$$left| bigcup_{i=1}^infty omega_i right| = omega_omega$$




We have:



$$kappa = left| bigcup_{i=1}^infty A_i right| le omega cdot sup_{i=1..infty} |A_i| = sup_{i=1..infty} |A_i|$$



So in order to conclude that $kappa = |A_i|$ for some $i$, we need $operatorname{cof}(kappa) = omega$, otherwise there are coutner-examples. For $kappa = mathfrak c$ though, you are lucky, since $operatorname{cof}(mathfrak c) > omega$, so you can always conclude that $kappa = |A_i|$ for some $i$.




Specializing on the case $kappa = mathfrak c$:



Let $|A_i| < mathfrak c$ for all $i$. By Konig's lemma:
$$left| bigcup_{i=1}^infty A_i right| < left| prod_{i=1}^infty mathfrak c right| = mathfrak c^omega = 2^{omega cdot omega} = 2^omega = mathfrak c$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 10:36

























answered Dec 31 '18 at 10:19









Kenny LauKenny Lau

19.9k2160




19.9k2160












  • $begingroup$
    Thanks for answering! Could you please elaborate on why this is true for $c$? I don't understand your notation cof and all that.. (I'm practically blind about set theory)
    $endgroup$
    – Colescu
    Dec 31 '18 at 10:25












  • $begingroup$
    The last part is not exactly true in ZF, it is consistent that the reals are countable union of countable sets. In ZFC it is true(the proof that $mbox{cof}(2^λ)>λ$ require choice)
    $endgroup$
    – Holo
    Dec 31 '18 at 10:33












  • $begingroup$
    @Colescu I've included some explanation
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:38










  • $begingroup$
    It might be better to wriite $aleph_i$ instead of $omega_i$, just to make sure we''re using cardinal operations. (For example: $omega=aleph_0$; one thinks of $omega$ as an ordinal annd $aleph_0$ as a cardinal. So, with cardinal multiplication, $aleph_0aleph_0=aleph_0$. But if the notation $omegaomega$ is interppreted as an ordinal product then $omegaomeganeomega$.)
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 15:27




















  • $begingroup$
    Thanks for answering! Could you please elaborate on why this is true for $c$? I don't understand your notation cof and all that.. (I'm practically blind about set theory)
    $endgroup$
    – Colescu
    Dec 31 '18 at 10:25












  • $begingroup$
    The last part is not exactly true in ZF, it is consistent that the reals are countable union of countable sets. In ZFC it is true(the proof that $mbox{cof}(2^λ)>λ$ require choice)
    $endgroup$
    – Holo
    Dec 31 '18 at 10:33












  • $begingroup$
    @Colescu I've included some explanation
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:38










  • $begingroup$
    It might be better to wriite $aleph_i$ instead of $omega_i$, just to make sure we''re using cardinal operations. (For example: $omega=aleph_0$; one thinks of $omega$ as an ordinal annd $aleph_0$ as a cardinal. So, with cardinal multiplication, $aleph_0aleph_0=aleph_0$. But if the notation $omegaomega$ is interppreted as an ordinal product then $omegaomeganeomega$.)
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 15:27


















$begingroup$
Thanks for answering! Could you please elaborate on why this is true for $c$? I don't understand your notation cof and all that.. (I'm practically blind about set theory)
$endgroup$
– Colescu
Dec 31 '18 at 10:25






$begingroup$
Thanks for answering! Could you please elaborate on why this is true for $c$? I don't understand your notation cof and all that.. (I'm practically blind about set theory)
$endgroup$
– Colescu
Dec 31 '18 at 10:25














$begingroup$
The last part is not exactly true in ZF, it is consistent that the reals are countable union of countable sets. In ZFC it is true(the proof that $mbox{cof}(2^λ)>λ$ require choice)
$endgroup$
– Holo
Dec 31 '18 at 10:33






$begingroup$
The last part is not exactly true in ZF, it is consistent that the reals are countable union of countable sets. In ZFC it is true(the proof that $mbox{cof}(2^λ)>λ$ require choice)
$endgroup$
– Holo
Dec 31 '18 at 10:33














$begingroup$
@Colescu I've included some explanation
$endgroup$
– Kenny Lau
Dec 31 '18 at 10:38




$begingroup$
@Colescu I've included some explanation
$endgroup$
– Kenny Lau
Dec 31 '18 at 10:38












$begingroup$
It might be better to wriite $aleph_i$ instead of $omega_i$, just to make sure we''re using cardinal operations. (For example: $omega=aleph_0$; one thinks of $omega$ as an ordinal annd $aleph_0$ as a cardinal. So, with cardinal multiplication, $aleph_0aleph_0=aleph_0$. But if the notation $omegaomega$ is interppreted as an ordinal product then $omegaomeganeomega$.)
$endgroup$
– David C. Ullrich
Dec 31 '18 at 15:27






$begingroup$
It might be better to wriite $aleph_i$ instead of $omega_i$, just to make sure we''re using cardinal operations. (For example: $omega=aleph_0$; one thinks of $omega$ as an ordinal annd $aleph_0$ as a cardinal. So, with cardinal multiplication, $aleph_0aleph_0=aleph_0$. But if the notation $omegaomega$ is interppreted as an ordinal product then $omegaomeganeomega$.)
$endgroup$
– David C. Ullrich
Dec 31 '18 at 15:27




















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