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Is cardinality continuous?
Scott's trick without the Axiom of RegularityA special cofinal family in $(omega^omega,le)$Alternative argument - set theory problemDefining “structured sets”“Dual cardinality” in the graphs $(V_alpha,in_alpha)$Is this description of “sigma-algebra generated by collection of subsets” right?Equivalence classes of real sequences, an interesting concept of closenessFilter, which does not have the Baire propertyDoes existence of some (nice) non-trivial functionals in $ell_infty^*setminusell_1$ give a free ultrafilter on $omega$?Proving the set of non-increasing finite sequences of natural numbers is well-ordered
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Let the underlying set theory be ZFC. Let $x_1 subseteq x_2 subseteq dots$ and $y_1 subseteq y_2 subseteq dots$ be ascending sequences of sets such that, for every $n in 1,2,dots$, $|x_n| = |y_n|$. Is it the case that $big|cup_n =1^inftyx_nbig| = big|cup_n =1^inftyy_nbig|$? If this is not generally true, is it possible to characterize all those—or at least some interesting—cases for which this does hold? Is there a standard terminology for these cases? Can this be generalized to transfinite sequences? Does the answer change if we require that the sequences be strictly increasing, i.e. for every $n in 1,2,dots$, $x_n subsetneq x_n+1$ and $y_n subsetneq y_n+1$?
set-theory cardinals
asked 8 hours ago
Evan AadEvan Aad
5,8321 gold badge19 silver badges55 bronze badges
$endgroup$
|
show 10 more comments
$begingroup$
Let the underlying set theory be ZFC. Let $x_1 subseteq x_2 subseteq dots$ and $y_1 subseteq y_2 subseteq dots$ be ascending sequences of sets such that, for every $n in 1,2,dots$, $|x_n| = |y_n|$. Is it the case that $big|cup_n =1^inftyx_nbig| = big|cup_n =1^inftyy_nbig|$? If this is not generally true, is it possible to characterize all those—or at least some interesting—cases for which this does hold? Is there a standard terminology for these cases? Can this be generalized to transfinite sequences? Does the answer change if we require that the sequences be strictly increasing, i.e. for every $n in 1,2,dots$, $x_n subsetneq x_n+1$ and $y_n subsetneq y_n+1$?
set-theory cardinals
asked 8 hours ago
Evan AadEvan Aad
5,8321 gold badge19 silver badges55 bronze badges
$endgroup$
$begingroup$
Relevant: proofwiki.org/wiki/…
$endgroup$
– user658409
8 hours ago
$begingroup$
This can surely fail in ZF, if $Bbb R$ is a countable union of countable sets then you can pick $x_i$ all equal to $y_0$ and $y_i$ a sequence of countable sets whose union is $Bbb R$
$endgroup$
– Alessandro Codenotti
8 hours ago
$begingroup$
@AlessandroCodenotti: If $mathbbR$ were a countable union of countable sets, it would be countable, which it is not, since its cardinality is that of the powerset of the natural numbers. So I'm not sure what it is you intended to say.
$endgroup$
– Evan Aad
8 hours ago
1
$begingroup$
I know, that's why I wrote a comment rather than an answer, just to give some context
$endgroup$
– Alessandro Codenotti
8 hours ago
1
$begingroup$
In ZFC you can assume WLOG that every $x_i$ and $y_i$ is an ordinal. Then a counterexample would require the larger of the two unions to be a successor cardinal -- but it would have cofinality $omega$, which is impossible; successor cardinals are always regular.
$endgroup$
– Henning Makholm
8 hours ago
|
show 10 more comments
$begingroup$
Let the underlying set theory be ZFC. Let $x_1 subseteq x_2 subseteq dots$ and $y_1 subseteq y_2 subseteq dots$ be ascending sequences of sets such that, for every $n in 1,2,dots$, $|x_n| = |y_n|$. Is it the case that $big|cup_n =1^inftyx_nbig| = big|cup_n =1^inftyy_nbig|$? If this is not generally true, is it possible to characterize all those—or at least some interesting—cases for which this does hold? Is there a standard terminology for these cases? Can this be generalized to transfinite sequences? Does the answer change if we require that the sequences be strictly increasing, i.e. for every $n in 1,2,dots$, $x_n subsetneq x_n+1$ and $y_n subsetneq y_n+1$?
set-theory cardinals
asked 8 hours ago
Evan AadEvan Aad
5,8321 gold badge19 silver badges55 bronze badges
$endgroup$
Let the underlying set theory be ZFC. Let $x_1 subseteq x_2 subseteq dots$ and $y_1 subseteq y_2 subseteq dots$ be ascending sequences of sets such that, for every $n in 1,2,dots$, $|x_n| = |y_n|$. Is it the case that $big|cup_n =1^inftyx_nbig| = big|cup_n =1^inftyy_nbig|$? If this is not generally true, is it possible to characterize all those—or at least some interesting—cases for which this does hold? Is there a standard terminology for these cases? Can this be generalized to transfinite sequences? Does the answer change if we require that the sequences be strictly increasing, i.e. for every $n in 1,2,dots$, $x_n subsetneq x_n+1$ and $y_n subsetneq y_n+1$?
set-theory cardinals
set-theory cardinals
asked 8 hours ago
Evan AadEvan Aad
5,8321 gold badge19 silver badges55 bronze badges
asked 8 hours ago
Evan AadEvan Aad
5,8321 gold badge19 silver badges55 bronze badges
edited 7 hours ago
Evan Aad
asked 8 hours ago
Evan AadEvan Aad
5,8321 gold badge19 silver badges55 bronze badges
asked 8 hours ago
Evan AadEvan Aad
5,8321 gold badge19 silver badges55 bronze badges
asked 8 hours ago
Evan AadEvan Aad
5,8321 gold badge19 silver badges55 bronze badges
5,8321 gold badge19 silver badges55 bronze badges
$begingroup$
Relevant: proofwiki.org/wiki/…
$endgroup$
– user658409
8 hours ago
$begingroup$
This can surely fail in ZF, if $Bbb R$ is a countable union of countable sets then you can pick $x_i$ all equal to $y_0$ and $y_i$ a sequence of countable sets whose union is $Bbb R$
$endgroup$
– Alessandro Codenotti
8 hours ago
$begingroup$
@AlessandroCodenotti: If $mathbbR$ were a countable union of countable sets, it would be countable, which it is not, since its cardinality is that of the powerset of the natural numbers. So I'm not sure what it is you intended to say.
$endgroup$
– Evan Aad
8 hours ago
1
$begingroup$
I know, that's why I wrote a comment rather than an answer, just to give some context
$endgroup$
– Alessandro Codenotti
8 hours ago
1
$begingroup$
In ZFC you can assume WLOG that every $x_i$ and $y_i$ is an ordinal. Then a counterexample would require the larger of the two unions to be a successor cardinal -- but it would have cofinality $omega$, which is impossible; successor cardinals are always regular.
$endgroup$
– Henning Makholm
8 hours ago
|
show 10 more comments
$begingroup$
Relevant: proofwiki.org/wiki/…
$endgroup$
– user658409
8 hours ago
$begingroup$
This can surely fail in ZF, if $Bbb R$ is a countable union of countable sets then you can pick $x_i$ all equal to $y_0$ and $y_i$ a sequence of countable sets whose union is $Bbb R$
$endgroup$
– Alessandro Codenotti
8 hours ago
$begingroup$
@AlessandroCodenotti: If $mathbbR$ were a countable union of countable sets, it would be countable, which it is not, since its cardinality is that of the powerset of the natural numbers. So I'm not sure what it is you intended to say.
$endgroup$
– Evan Aad
8 hours ago
1
$begingroup$
I know, that's why I wrote a comment rather than an answer, just to give some context
$endgroup$
– Alessandro Codenotti
8 hours ago
1
$begingroup$
In ZFC you can assume WLOG that every $x_i$ and $y_i$ is an ordinal. Then a counterexample would require the larger of the two unions to be a successor cardinal -- but it would have cofinality $omega$, which is impossible; successor cardinals are always regular.
$endgroup$
– Henning Makholm
8 hours ago
$begingroup$
Relevant: proofwiki.org/wiki/…
$endgroup$
– user658409
8 hours ago
$begingroup$
Relevant: proofwiki.org/wiki/…
$endgroup$
– user658409
8 hours ago
$begingroup$
This can surely fail in ZF, if $Bbb R$ is a countable union of countable sets then you can pick $x_i$ all equal to $y_0$ and $y_i$ a sequence of countable sets whose union is $Bbb R$
$endgroup$
– Alessandro Codenotti
8 hours ago
$begingroup$
This can surely fail in ZF, if $Bbb R$ is a countable union of countable sets then you can pick $x_i$ all equal to $y_0$ and $y_i$ a sequence of countable sets whose union is $Bbb R$
$endgroup$
– Alessandro Codenotti
8 hours ago
$begingroup$
@AlessandroCodenotti: If $mathbbR$ were a countable union of countable sets, it would be countable, which it is not, since its cardinality is that of the powerset of the natural numbers. So I'm not sure what it is you intended to say.
$endgroup$
– Evan Aad
8 hours ago
$begingroup$
@AlessandroCodenotti: If $mathbbR$ were a countable union of countable sets, it would be countable, which it is not, since its cardinality is that of the powerset of the natural numbers. So I'm not sure what it is you intended to say.
$endgroup$
– Evan Aad
8 hours ago
1
1
$begingroup$
I know, that's why I wrote a comment rather than an answer, just to give some context
$endgroup$
– Alessandro Codenotti
8 hours ago
$begingroup$
I know, that's why I wrote a comment rather than an answer, just to give some context
$endgroup$
– Alessandro Codenotti
8 hours ago
1
1
$begingroup$
In ZFC you can assume WLOG that every $x_i$ and $y_i$ is an ordinal. Then a counterexample would require the larger of the two unions to be a successor cardinal -- but it would have cofinality $omega$, which is impossible; successor cardinals are always regular.
$endgroup$
– Henning Makholm
8 hours ago
$begingroup$
In ZFC you can assume WLOG that every $x_i$ and $y_i$ is an ordinal. Then a counterexample would require the larger of the two unions to be a successor cardinal -- but it would have cofinality $omega$, which is impossible; successor cardinals are always regular.
$endgroup$
– Henning Makholm
8 hours ago
|
show 10 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Yes, this is always true. If $x_1subseteq x_2subseteqdots$ and $x=bigcup x_n$, then $|x|=sup_n|x_n|$, and in particular $|x|$ is uniquely determined by the sequence of cardinalities $|x_n|$. Clearly $|x|geq|x_n|$ for all $n$ so $|x|geqsup_n |x_n|$. Conversely, $|x_n|leq sup|x_n|$ for all $n$ so $|x|leq aleph_0cdot sup|x_n|=sup|x_n|$ as long as $sup |x_n|$ is infinite (and if it is finite then the result is trivial).
Note that if you consider increasing sequences with possibly uncountable index sets then this is no longer true. For instance, with index set $omega_1$, if you let $x_alpha=omega+alpha$ and $y_alpha=omega$ for all $alpha<omega_1$, then $|x_alpha|=|y_alpha|=aleph_0$ for each $alpha$ but $left|bigcup x_alpharight|=aleph_1$ while $left|bigcup y_alpharight|=aleph_0$. If you require the sequences to be strictly increasing then it is true though: letting $kappa$ be the cofinality of the index set, the argument above shows that $|x|geqsup|x_i|$ and $|x|leq kappacdotsup|x_i|=max(kappa,sup|x_i|)$ but also $|x|geqkappa$ if the $x_i$ are strictly increasing (since looking at a cofinal well-ordered subsequence gives at least one new element of $x$ for each term in the subsequence), so $|x|=max(kappa,sup|x_i|)$.
answered 7 hours ago
Eric WofseyEric Wofsey
206k14 gold badges241 silver badges374 bronze badges
$endgroup$
$begingroup$
My set theory is very rusty. Could you please explain what the supremum of a set of cardinalities means? And how we know that it exists?
$endgroup$
– Evan Aad
7 hours ago
$begingroup$
It means the same thing supremum always means: the least cardinal greater or than equal to all of them. It exists because cardinals are well-ordered.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I'm accepting your answer based on the first paragraph, which is good enough for my purposes at the present time. The second paragraph I don't understand, but it is not due to you but to my deficient knowledge of set theory. Hopefully I'll be able to come back to this answer in the future and understand it in its entirety.
$endgroup$
– Evan Aad
7 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
Yes, this is always true. If $x_1subseteq x_2subseteqdots$ and $x=bigcup x_n$, then $|x|=sup_n|x_n|$, and in particular $|x|$ is uniquely determined by the sequence of cardinalities $|x_n|$. Clearly $|x|geq|x_n|$ for all $n$ so $|x|geqsup_n |x_n|$. Conversely, $|x_n|leq sup|x_n|$ for all $n$ so $|x|leq aleph_0cdot sup|x_n|=sup|x_n|$ as long as $sup |x_n|$ is infinite (and if it is finite then the result is trivial).
Note that if you consider increasing sequences with possibly uncountable index sets then this is no longer true. For instance, with index set $omega_1$, if you let $x_alpha=omega+alpha$ and $y_alpha=omega$ for all $alpha<omega_1$, then $|x_alpha|=|y_alpha|=aleph_0$ for each $alpha$ but $left|bigcup x_alpharight|=aleph_1$ while $left|bigcup y_alpharight|=aleph_0$. If you require the sequences to be strictly increasing then it is true though: letting $kappa$ be the cofinality of the index set, the argument above shows that $|x|geqsup|x_i|$ and $|x|leq kappacdotsup|x_i|=max(kappa,sup|x_i|)$ but also $|x|geqkappa$ if the $x_i$ are strictly increasing (since looking at a cofinal well-ordered subsequence gives at least one new element of $x$ for each term in the subsequence), so $|x|=max(kappa,sup|x_i|)$.
answered 7 hours ago
Eric WofseyEric Wofsey
206k14 gold badges241 silver badges374 bronze badges
$endgroup$
$begingroup$
My set theory is very rusty. Could you please explain what the supremum of a set of cardinalities means? And how we know that it exists?
$endgroup$
– Evan Aad
7 hours ago
$begingroup$
It means the same thing supremum always means: the least cardinal greater or than equal to all of them. It exists because cardinals are well-ordered.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I'm accepting your answer based on the first paragraph, which is good enough for my purposes at the present time. The second paragraph I don't understand, but it is not due to you but to my deficient knowledge of set theory. Hopefully I'll be able to come back to this answer in the future and understand it in its entirety.
$endgroup$
– Evan Aad
7 hours ago
add a comment |
$begingroup$
Yes, this is always true. If $x_1subseteq x_2subseteqdots$ and $x=bigcup x_n$, then $|x|=sup_n|x_n|$, and in particular $|x|$ is uniquely determined by the sequence of cardinalities $|x_n|$. Clearly $|x|geq|x_n|$ for all $n$ so $|x|geqsup_n |x_n|$. Conversely, $|x_n|leq sup|x_n|$ for all $n$ so $|x|leq aleph_0cdot sup|x_n|=sup|x_n|$ as long as $sup |x_n|$ is infinite (and if it is finite then the result is trivial).
Note that if you consider increasing sequences with possibly uncountable index sets then this is no longer true. For instance, with index set $omega_1$, if you let $x_alpha=omega+alpha$ and $y_alpha=omega$ for all $alpha<omega_1$, then $|x_alpha|=|y_alpha|=aleph_0$ for each $alpha$ but $left|bigcup x_alpharight|=aleph_1$ while $left|bigcup y_alpharight|=aleph_0$. If you require the sequences to be strictly increasing then it is true though: letting $kappa$ be the cofinality of the index set, the argument above shows that $|x|geqsup|x_i|$ and $|x|leq kappacdotsup|x_i|=max(kappa,sup|x_i|)$ but also $|x|geqkappa$ if the $x_i$ are strictly increasing (since looking at a cofinal well-ordered subsequence gives at least one new element of $x$ for each term in the subsequence), so $|x|=max(kappa,sup|x_i|)$.
answered 7 hours ago
Eric WofseyEric Wofsey
206k14 gold badges241 silver badges374 bronze badges
$endgroup$
$begingroup$
My set theory is very rusty. Could you please explain what the supremum of a set of cardinalities means? And how we know that it exists?
$endgroup$
– Evan Aad
7 hours ago
$begingroup$
It means the same thing supremum always means: the least cardinal greater or than equal to all of them. It exists because cardinals are well-ordered.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I'm accepting your answer based on the first paragraph, which is good enough for my purposes at the present time. The second paragraph I don't understand, but it is not due to you but to my deficient knowledge of set theory. Hopefully I'll be able to come back to this answer in the future and understand it in its entirety.
$endgroup$
– Evan Aad
7 hours ago
add a comment |
$begingroup$
Yes, this is always true. If $x_1subseteq x_2subseteqdots$ and $x=bigcup x_n$, then $|x|=sup_n|x_n|$, and in particular $|x|$ is uniquely determined by the sequence of cardinalities $|x_n|$. Clearly $|x|geq|x_n|$ for all $n$ so $|x|geqsup_n |x_n|$. Conversely, $|x_n|leq sup|x_n|$ for all $n$ so $|x|leq aleph_0cdot sup|x_n|=sup|x_n|$ as long as $sup |x_n|$ is infinite (and if it is finite then the result is trivial).
Note that if you consider increasing sequences with possibly uncountable index sets then this is no longer true. For instance, with index set $omega_1$, if you let $x_alpha=omega+alpha$ and $y_alpha=omega$ for all $alpha<omega_1$, then $|x_alpha|=|y_alpha|=aleph_0$ for each $alpha$ but $left|bigcup x_alpharight|=aleph_1$ while $left|bigcup y_alpharight|=aleph_0$. If you require the sequences to be strictly increasing then it is true though: letting $kappa$ be the cofinality of the index set, the argument above shows that $|x|geqsup|x_i|$ and $|x|leq kappacdotsup|x_i|=max(kappa,sup|x_i|)$ but also $|x|geqkappa$ if the $x_i$ are strictly increasing (since looking at a cofinal well-ordered subsequence gives at least one new element of $x$ for each term in the subsequence), so $|x|=max(kappa,sup|x_i|)$.
answered 7 hours ago
Eric WofseyEric Wofsey
206k14 gold badges241 silver badges374 bronze badges
$endgroup$
Yes, this is always true. If $x_1subseteq x_2subseteqdots$ and $x=bigcup x_n$, then $|x|=sup_n|x_n|$, and in particular $|x|$ is uniquely determined by the sequence of cardinalities $|x_n|$. Clearly $|x|geq|x_n|$ for all $n$ so $|x|geqsup_n |x_n|$. Conversely, $|x_n|leq sup|x_n|$ for all $n$ so $|x|leq aleph_0cdot sup|x_n|=sup|x_n|$ as long as $sup |x_n|$ is infinite (and if it is finite then the result is trivial).
Note that if you consider increasing sequences with possibly uncountable index sets then this is no longer true. For instance, with index set $omega_1$, if you let $x_alpha=omega+alpha$ and $y_alpha=omega$ for all $alpha<omega_1$, then $|x_alpha|=|y_alpha|=aleph_0$ for each $alpha$ but $left|bigcup x_alpharight|=aleph_1$ while $left|bigcup y_alpharight|=aleph_0$. If you require the sequences to be strictly increasing then it is true though: letting $kappa$ be the cofinality of the index set, the argument above shows that $|x|geqsup|x_i|$ and $|x|leq kappacdotsup|x_i|=max(kappa,sup|x_i|)$ but also $|x|geqkappa$ if the $x_i$ are strictly increasing (since looking at a cofinal well-ordered subsequence gives at least one new element of $x$ for each term in the subsequence), so $|x|=max(kappa,sup|x_i|)$.
answered 7 hours ago
Eric WofseyEric Wofsey
206k14 gold badges241 silver badges374 bronze badges
edited 7 hours ago
answered 7 hours ago
Eric WofseyEric Wofsey
206k14 gold badges241 silver badges374 bronze badges
answered 7 hours ago
Eric WofseyEric Wofsey
206k14 gold badges241 silver badges374 bronze badges
answered 7 hours ago
Eric WofseyEric Wofsey
206k14 gold badges241 silver badges374 bronze badges
206k14 gold badges241 silver badges374 bronze badges
$begingroup$
My set theory is very rusty. Could you please explain what the supremum of a set of cardinalities means? And how we know that it exists?
$endgroup$
– Evan Aad
7 hours ago
$begingroup$
It means the same thing supremum always means: the least cardinal greater or than equal to all of them. It exists because cardinals are well-ordered.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I'm accepting your answer based on the first paragraph, which is good enough for my purposes at the present time. The second paragraph I don't understand, but it is not due to you but to my deficient knowledge of set theory. Hopefully I'll be able to come back to this answer in the future and understand it in its entirety.
$endgroup$
– Evan Aad
7 hours ago
add a comment |
$begingroup$
My set theory is very rusty. Could you please explain what the supremum of a set of cardinalities means? And how we know that it exists?
$endgroup$
– Evan Aad
7 hours ago
$begingroup$
It means the same thing supremum always means: the least cardinal greater or than equal to all of them. It exists because cardinals are well-ordered.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I'm accepting your answer based on the first paragraph, which is good enough for my purposes at the present time. The second paragraph I don't understand, but it is not due to you but to my deficient knowledge of set theory. Hopefully I'll be able to come back to this answer in the future and understand it in its entirety.
$endgroup$
– Evan Aad
7 hours ago
$begingroup$
My set theory is very rusty. Could you please explain what the supremum of a set of cardinalities means? And how we know that it exists?
$endgroup$
– Evan Aad
7 hours ago
$begingroup$
My set theory is very rusty. Could you please explain what the supremum of a set of cardinalities means? And how we know that it exists?
$endgroup$
– Evan Aad
7 hours ago
$begingroup$
It means the same thing supremum always means: the least cardinal greater or than equal to all of them. It exists because cardinals are well-ordered.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
It means the same thing supremum always means: the least cardinal greater or than equal to all of them. It exists because cardinals are well-ordered.
$endgroup$
– Eric Wofsey
7 hours ago
$begingroup$
I'm accepting your answer based on the first paragraph, which is good enough for my purposes at the present time. The second paragraph I don't understand, but it is not due to you but to my deficient knowledge of set theory. Hopefully I'll be able to come back to this answer in the future and understand it in its entirety.
$endgroup$
– Evan Aad
7 hours ago
$begingroup$
I'm accepting your answer based on the first paragraph, which is good enough for my purposes at the present time. The second paragraph I don't understand, but it is not due to you but to my deficient knowledge of set theory. Hopefully I'll be able to come back to this answer in the future and understand it in its entirety.
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– Evan Aad
7 hours ago
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Relevant: proofwiki.org/wiki/…
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– user658409
8 hours ago
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This can surely fail in ZF, if $Bbb R$ is a countable union of countable sets then you can pick $x_i$ all equal to $y_0$ and $y_i$ a sequence of countable sets whose union is $Bbb R$
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– Alessandro Codenotti
8 hours ago
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@AlessandroCodenotti: If $mathbbR$ were a countable union of countable sets, it would be countable, which it is not, since its cardinality is that of the powerset of the natural numbers. So I'm not sure what it is you intended to say.
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– Evan Aad
8 hours ago
1
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I know, that's why I wrote a comment rather than an answer, just to give some context
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– Alessandro Codenotti
8 hours ago
1
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In ZFC you can assume WLOG that every $x_i$ and $y_i$ is an ordinal. Then a counterexample would require the larger of the two unions to be a successor cardinal -- but it would have cofinality $omega$, which is impossible; successor cardinals are always regular.
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– Henning Makholm
8 hours ago