Decomposition of an $\aleph_1$ set into $\aleph_1$ sets $\aleph_1$

Question

Can we justify the claim

Any set of cardinality $\aleph_1$ can be expressed as the disjoint union of $\aleph_1$ sets of cardinality $\aleph_1$

in a simple yet reasonably-correct way?

As written? Trivially. What you probably meant to write? Please add context like the close votes are suggesting. — Mark S., Feb 21 '19 at 10:50

score 1 · Accepted Answer · answered Feb 21 '19 at 11:50

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I'm assuming you mean a disjoint union. In this cas here's a way to do it :

There is a bijection $f: \aleph_1\times \aleph_1 \to \aleph_1$.

Thus $\aleph_1 = \displaystyle\coprod_{a\in \aleph_1} f(\aleph_1\times\{a\})$.

Of course this partition can be transferred to any set of cardinality $\aleph_1$.

answered Feb 21 '19 at 11:50

Maxime Ramzi

Thanks for the answer. Before accepting it lets see if someone else come up with another idea. – Filburt Feb 21 '19 at 12:19
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Isn't the question just the same thing as asking to show that there is such a bijection? – Andrés E. Caicedo Feb 21 '19 at 12:27
@AndrésE.Caicedo yes somehow... – Filburt Feb 21 '19 at 12:41
@Filburt : if this is what you're asking for, then here is a proof of that result : https://math.stackexchange.com/questions/2153102/cardinals-arithmetic/2153155#2153155 – Maxime Ramzi Feb 21 '19 at 13:52
Yeah... maybe the answer is simply "no" (the adjective "simple" is pretty subjective though) – Filburt Feb 21 '19 at 18:57

1 Answers1