I know that the union of countably many countable sets is countable. Is there an equivalent statement for uncountable sets, such as the union of uncountably many uncountable sets is uncountable? Furthermore, how does this generalize to other cardinals?
[Math] Union of Uncountably Many Uncountable Sets
cardinalsset-theory
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Let $x\in\mathbb{R}$. Define $$S_{\leq x}=S~\cap~(-\infty,x]~~~\mathrm{and}~~~S_{>x}=S~\cap~ (x,+\infty)$$ and define $S_{\leq}=\{$ the set of all $x$ such that $S_{\leq x}$ is uncountable $\}$, and $S_>=\{$ the set of all $x$ such that $S_{>x}$ is uncountable $\}$.
$S_{\leq}$ is a non empty$^{(*)}$ interval that extends to infinity on the right, and is therefore of the form $[a,+\infty)$ or $(a,+\infty)$ for some $a\in\mathbb{R}$ or is equal to the whole of $\mathbb{R}$.
Similarly, $S_>$ is a non empty$^{(*)}$ interval that extends to infinity on the left, and is thus of the form $(-\infty,b]$ or $(-\infty,b)$ for some $b\in\mathbb{R}$ or is equal to the whole of $\mathbb{R}$.
$^{(*)}$ : I have used some form of choice I believe when I say that $S_>$ and $S_{\leq}$ are non empty, because I use the fact that with some form of choice, the countable union of countable sets is still countable. Is it countable choice? Since $$S=\cup_{n\in\mathbb{N}}S_{\leq n}$$ is uncountable, at least one of the $S_{<n}$ must be uncountable, and so for some natural number $n$ you have $n\in S_{<}\neq\emptyset$.
Let's show that $S_>\cap S_{\leq}$ is non empty. If either one of $S_>$ or $S_{\leq}$ is the whole real line, there is no problem. So let's suppose none of them is the whole of $\mathbb{R}$.
Suppose $x\in S_{\leq}$ that is $S_{\leq x}=S~\cap~(-\infty,x]$ is uncountable. Since $$\bigcup_{n\in\mathbb{N}^*}S_{\leq x-\frac{1}{n}}\subset S_{\leq x}=S~\cap~(-\infty,x]\subset \{x\}\cup \bigcup_{n\in\mathbb{N}^*}S_{\leq x-\frac{1}{n}}$$ the already used fact that countable unions of countable sets are countable with some form of choice implies that at least one of the $S_{\leq x -\frac{1}{n}}$ is uncountable, which means that for some $n\in\mathbb{N}^*,~x -\frac{1}{n}\in L$ and thus $S_{\leq}$ is open, that is, $$\mathbf{S_{\leq}=(b,+\infty)}$$ for some real number $b$. Similarly, $$\mathbf{S_{>}=(-\infty,a)}$$ for some real number $a$.
Then $S_>\cap S_{\leq}$ is empty iff $a\leq b$. But this would entail that $$S=S_{\leq a}\cup S_{> a}$$ is countable as the union of two countable sets, since $a \notin (-\infty,a)=S_{\leq}$ means $S_{\leq a}$ is countable, and $a \notin (b,+\infty)=S_{>}$ means $S_{> a}$ countable.
You have $\mathfrak c=2^{\aleph_0}$ classes, each of them is of the size $\aleph_0$.
You get the set $V$ by choosing one element from each class. How many possibilities are for such choice? Precisely $$\aleph_0^{\mathfrak c}=2^{\mathfrak c}$$ possibilities. (Size of each class raised to the number of classes - you are basically counting the maps from the set of classes, for each class you have countably many possibilities.)
This cardinal number is definitely uncountable: $2^{\mathfrak c}>\mathfrak c>\aleph_0$.
The above cardinal equality holds since $$2^{\mathfrak c} \le \aleph_0^{\mathfrak c} \le \mathfrak c^{\mathfrak c} = (2^{\aleph_0})^{\mathfrak c} = 2^{\aleph_0\cdot\mathfrak c} = 2^{\mathfrak c},$$ so by Cantor-Bernstein Theorem all cardinal numbers in this chain of inequalities are the same.
Best Answer
It generalizes very naturally.
If $\kappa$ and $\lambda$ are cardinal numbers such that $\kappa$ is infinite and $0<\lambda\le\kappa$, the union of $\lambda$ sets of cardinality $\kappa$ has cardinality $\kappa$. In other words, the union of at least one and at most $\kappa$ sets of cardinality $\kappa$ has cardinality $\kappa$.
An alternative generalization is that if $\kappa$ is an infinite cardinal number, the union of at most $\kappa$ sets, each of cardinality at most $\kappa$, also has cardinality at most $\kappa$. In symbols, if $\lambda<\kappa$, and $\{A_\xi:\xi<\lambda\}$ is a family of sets such that $|A_\xi|\le\kappa$ for each $\xi<\lambda$, then $$\left|\bigcup_{\xi<\lambda}A_\xi\right|\le\kappa\;.$$