If a set $S$ has countably many elements $\{x_n\}$, it can be expressed as a union of a chain of finite sets

$$ \{x_1\} \subset \{x_1,x_2\} \subset $$

But what about a set of arbitrary cardinality $S$? Can we express it as a union : $$S=\bigcup_{i\in I}S_i$$

where $I$ is a totally ordered set of some cardinality and each $|S_i|$ has cardinality less than $|S|$?


Yes, assuming the axiom of choice. Suppose that $S$ is infinite. Then $|S|=\kappa$ for some infinite cardinal $\kappa$ so there is a bijection $f:\kappa\to S$. For each $\alpha<\kappa$ let $S_\alpha=f[\alpha]=\{f(\xi):\xi<\alpha\}$; since $\kappa=\bigcup_{\alpha<\kappa}\alpha$, clearly $S=\bigcup_{\alpha<\kappa}S_\alpha$, and $S_\alpha\subseteq S_\beta$ whenever $\alpha\le\beta<\kappa$.

Without the axiom of choice it need not be possible. In particular, it cannot be done with a sequence of proper subsets if $S$ is an amorphous set. (As Asaf points out in the comments, it’s possible with a two-element sequence if one is $S\setminus\{p\}$ for some $p\in S$, and the other is $S$ itself.)

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  • $\begingroup$ I just wanted to double check whether the equation $\kappa = \cup_{\alpha<\kappa}\alpha$ holds for all infinite cardinals (regular and otherwise). I'm sorry, I'm just really nervous about dealing with cardinals and I need this reasoning for an argument I'm trying to do in commutative algebra. Cheers! $\endgroup$ – abeliancats Jan 9 '16 at 10:30
  • $\begingroup$ @gocardinals: Yes, you’re safe: it does. It’s just that if $\kappa$ is singular, you can get a cofinal sequence that is shorter than $\kappa$. $\endgroup$ – Brian M. Scott Jan 9 '16 at 10:37
  • $\begingroup$ Great! Thanks :) $\endgroup$ – abeliancats Jan 9 '16 at 10:39
  • $\begingroup$ @gocardinals: You’re welcome! $\endgroup$ – Brian M. Scott Jan 9 '16 at 11:00
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    $\begingroup$ I think that I know what I have to do now. Take a nap. $\endgroup$ – Asaf Karagila Jan 9 '16 at 11:52

Yes, you can always do this (as long as $S$ is infinite). Choose a well-ordering $<$ of $S$ of minimal possible length. Since $S$ is infinite, it has no $<$-greatest element (otherwise, you could get a shorter well-ordering by moving the greatest element to the beginning of the ordering). Then for each $i\in S$, the set $S_i=\{j\in S:j<i\}$ must has smaller cardinality than $S$ (otherwise, choosing a bijection between $S_i$ and $S$, the ordering on $S_i$ would give a shorter well-ordering of $S$). Since $S$ has no greatest element, it is the union of the sets $S_i$, and these sets are totally ordered (with the same ordering as $S$).

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