Infinite set as a countably infinite union of infinite disjoint subsets

Mathematics Asked by XuUserAC on December 19, 2020

I am struggliglng with an exercise as the title.

I have been able to get to a proof, if the set is infinitely countable, but I am having problems on infinite uncountable sets.

Any thoughts on how to think about it?

EDITED: Added the word ‘disjoint’ and "countably"

One Answer

Given an infinite set $S$, for any element $xin S$ the subset $Ssetminus{x}$ is also infinite. By assumption, $S$ has infinitely many elements, therefore the union $$bigcup_{xin S}(Ssetminus{x})=S$$ is an infinite union of infinite subsets of $S$.


With the added requirement of countably many disjoint sets, the following needs the axiom of choice (via the well-ordering theorem).

According to the well-ordering theorem, the set $S$ is in bijection with a von-Neumann ordinal. Be $phi:alphato S$ such a bijection.

Since you say you already have the proof for countable $S$, let's further assume that $S$ (and therefore $alpha$) is uncountable.

Now it is possible to write each ordinal uniquely as sum $lambda + n$ of a non-successor (i.e. zero or limit) ordinal $lambda$ and a finite ordinal $n$.

Define $$alpha_n = {rhoinalpha | rho = lambda + n text{ for some non-successor ordinal $lambda$}}$$

Clearly there are countably many such $alpha_n$, one for each $ninomega$. Moreover, quite obviously the different $alpha_n$ are disjoint, their union is $alpha$, and each of them is clearly infinite (because otherwise $alpha$ would be countable).

Now we get the desired union for $S$ by simply using the images of $alpha_n$ under the bijection $phi$: $$S = bigcup_{ninomega} {phi(x)|xin alpha_n}$$

Answered by celtschk on December 19, 2020

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