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.)

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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 $S$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.