I have a home work question which is: " what is the cardinality of the union of ${{\aleph }_{0}}$ disjoint sets of cardinality $\mathfrak{c}$?"

I believe somehow we can get to: cardinality = $({{\aleph }_{0}} )(\mathfrak{c})$

but how would we get there and what does that make the cardinality? Does it make it $\mathfrak{c}$? if so how could this be proven?

Help is appreciated, Thanks

EDIT: forgot to write disjoint

  • $\begingroup$ I am starting to have this feeling that the more I venture into choiceless badlands, the more complicated I make things look in answers to relatively simple questions. If someone thinks that the added tag of [axiom-of-choice] is too much, feel free to remove it. $\endgroup$ – Asaf Karagila Aug 13 '11 at 1:18

Under the assumption that each set of the form $\mathbb R\times\{n\}$ (or can be made of this form in a definable way), then the union of $\aleph_0$ sets, each of cardinality $2^{\aleph_0}$ is again $2^{\aleph_0}$.

If the axiom of countable choice holds then we may choose a function from each disjoint set into $\mathbb R\times\{n\}$. Now we have


(The cardinal arithmetic does not require the axiom of choice to hold in any form)

If, however, we do not impose any of the above limitations (the sets are not of a form similar to $X\times\{n\}$, and the axiom of countable choice is missing), then problems may arise which makes the question ill-defined.

It is possible for the question to be interpreted as above, while on the other hand, let $V$ be a model of $ZFA+AC$ with continuum many atoms, write $A=\bigcup A_n$ where $A_n$ are a disjoint partition of $A$, each piece of cardinality continuum. Consider permutations of $A$ which preserve the partitioning, alongside the ideal of supports generated by finite unions of $A_n$'s.

In the resulting permutation model $U$ we have that each $A_n\in U$, and so is the partition (since it is fixed by all permutations). We also have that there is no choice function on the $A_n$'s.

Suppose $f\colon A\to\mathcal P(\omega)$ a bijection in $U$, let $E$ be a support for $f$, without loss of generality $E=A_0\cup\ldots\cup A_n$. Let $\pi$ be a permutation fixing $E$, and therefore $\pi f=f$, we have that $\langle x,\alpha\rangle\in f$ then $\langle \pi x,\pi \alpha\rangle\in\pi f=f$.

Take $a\in A$ such that $\pi a\neq a$. Since $\pi f = f$ we have that $(\pi f)(\pi a) = \pi (f(a))$, which contradicts the injectivity of $f$, since $\operatorname{Rng}(f)$ is a subset of ordinals, which is fixed by all permutations and in particular by $\pi$.

Now use Jech-Sochor to transfer this into a symmetric extension of a model of $ZFC$, up to $\omega$.

We have a model of $ZF$ in which $2^{\aleph_0}$ can be well ordered, but the sum of $\aleph_0$ many copies of a continuum does not make a continuum.

This is quite the necrobumping, but I came up with a much simpler example of a countable union of sets of size $\frak c$ which is not of size $\frak c$:

Suppose that $V$ is a model of $$ZF+\text{There is a countable family of disjoint pairs without a choice function}$$ such model is Cohen's second model. Let $\{P_n\mid n\in\omega\}$ be such countable family of pairs, and let $S$ be $\bigcup_n P_n$.

Fun fact: $S$ cannot be linearly ordered, if it could be then we could choose the minimal one of each $P_n$.

Without loss of generality $S\cap\mathbb R=\varnothing$. Let $A_n=\{n\}\times\mathbb R\cup P_n$, since we only add two elements to a size isomorphic to $\mathbb R$ we do not increase in cardinality.

Consider $A=\bigcup_n A_n=\mathbb N\times\mathbb R\cup S$. We observe that $A$ cannot be linearly ordered since $S$ cannot be linearly ordered, however if $A$ were of size $\frak c$ then it could be matched with the real numbers and therefore be linearly ordered.

Conclusion, $|A|\neq\frak c$. On the other hand, clearly there is an injective function from $\mathbb R$ into $A$ therefore ${\frak c}<|A|$.

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    $\begingroup$ @William: I do not recall at any point above that I assumed that the continuum can be well ordered. $\endgroup$ – Asaf Karagila Aug 12 '11 at 17:04
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    $\begingroup$ What is your definition of a cardinal? $\endgroup$ – William Aug 12 '11 at 17:06
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    $\begingroup$ @Chris: but you can prove that $\cup_{i\in\omega}2\times\{i\}$ is bijectable with $2\times\omega$. If you define a sum by explicitly "disjointing", say, $\sum_{i\in I}X_i$ being the cardinality of $\cup_{i\in I}X_i\times\{i\}$, then $\sum_{i\in\omega}2$ is bijectable with $2\times\omega$. $\endgroup$ – Arturo Magidin Aug 12 '11 at 18:54
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    $\begingroup$ @Chris: Yes, I am aware, but I don't see how it relates. I distinguish in my answer between the sum of sets-equipped-with-bijections and just sets; in ZF, the countable union of sets-equipped-with-bijections-to-$\mathbb{N}$ is provably countable. I just don't get the point you are driving at. $\endgroup$ – Arturo Magidin Aug 12 '11 at 21:40
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    $\begingroup$ @Chris: The only place where choice might be hiding in Asaf's answer is in going from "disjoint union of sets, each of cardinality $\mathfrak{c}$" to $\mathfrak{c}\aleph_0$. To be honest, Asaf is far better at AC as it relates to cardinals than I am, so whether or not you need AC for that step, I'll defer to him. But the chain of equalities just before his final paragraph does not seem to me to require AC at all. $\endgroup$ – Arturo Magidin Aug 12 '11 at 21:46

The fact that the union of $\kappa$ sets, each of cardinality $\lambda$, is equal to $\kappa\lambda$ follows simply from the fact that $\kappa\lambda$ is defined to be the cardinality of $\kappa\times\lambda$, and there is an obvious bijection between the disjoint union of $\{X_i\mid i\in\kappa\}$ in which each $X_i$ comes equipped with a bijection $f_i\colon X_i\to\lambda$, and $\kappa\times\lambda$: map each $(a,i)\in \bigcup_{i\in\kappa}X_i\times\{i\}$ to $(f_i(a),i)$. This is easily seen to be a bijection.

If your $X_i$ don't "come equipped" with bijections to $\lambda$ then you may need the Axiom of Choice in order to select a bijection for each $X_i$. On the other hand, the cardinal equality $$\sum_{i\in\kappa}\lambda = \lambda\kappa$$ does not require the Axiom of Choice, since we can pick $f_i$ to be the identity for each $i$.

  • $\begingroup$ How are you even defining the sum $\sum_{i\in\kappa}\lambda$ without choice? I would define it as the cardinality of the union of $\kappa$-many disjoint sets, each of cardinality $\lambda$, but without choice, that might not be well-defined. $\endgroup$ – Chris Eagle Aug 12 '11 at 18:46
  • $\begingroup$ @Chris: As Asaf points out, you can define cardinality without choice; why would it not be well-defined? You don't need choice here, because the cardinality of $\sum_{i\in \kappa}\lambda$ is the cardinality of $\cup\lambda\times\{i\}$, and you can explicitly biject this with $\lambda\times\kappa$ as I note above. If there is a problem, I honestly don't see it... $\endgroup$ – Arturo Magidin Aug 12 '11 at 18:51
  • $\begingroup$ The sum as I've defined it is ill-defined because taking a different family of sets can give you a different cardinality of the union. Your definition of the sum gets around this (you do need to replace $\lambda$ in $\cup \lambda \times \{ i \}$ with some fixed set of cardinality $\lambda$, though). $\endgroup$ – Chris Eagle Aug 12 '11 at 20:10
  • $\begingroup$ @Chris: I was using $\lambda$ to represent a particular set, yes, as opposed to the previous paragraph in which I had $X_i$ with bijections to $\lambda$. $\endgroup$ – Arturo Magidin Aug 12 '11 at 21:16

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