If $A=\bigcup_{n=1}^{\infty}A_n$ and $A$ has cardinality $\mathfrak{c}$, where $\mathfrak{c}$ is the cardinal of the continuum, prove that at least one of the $A_n$ has cardinality $\mathfrak{c}$.
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It might be easier to show that a countable union of countable sets is countable. – Dylan Nov 11 '15 at 09:46
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3@Dylan Then the rest follows from the truth of the continuum hypothesis? – Hagen von Eitzen Nov 11 '15 at 09:46
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@HagenvonEitzen Ah. Right. I did not think of that. – Dylan Nov 11 '15 at 09:48
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@Dylan how to solve this without using the continuum hypothesis cause the question doesn't depend on it – Idele Nov 11 '15 at 09:50
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It's not true. Let $A_1={1}$, $A_2={2}$,..., $A_c={c}$ and $A_n=\emptyset$ for all $n>c$, then $|A|=c$ but not any $A_i$ has card $c$. – Surb Nov 11 '15 at 09:51
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@Surb but by c I mean the cardinal of the continuum – Idele Nov 11 '15 at 09:54
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@hctb: Should do we guess ? – Surb Nov 11 '15 at 09:55
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@Surb sorry I didn't make it clear , I will add that – Idele Nov 11 '15 at 09:56
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5@hctb I was able to find this question: http://math.stackexchange.com/questions/199274/the-cardinality-of-a-countable-union-of-sets-with-less-than-continuum-cardiality which has an answer. – Dylan Nov 11 '15 at 10:05
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This follows from König's theorem, which reads
Theorem (König). If for $i \in I$ we have cardinals $\def\a{\mathfrak a}\def\b{\mathfrak b}\a_i < \b_i$, then $$ \sum_i \a_i < \prod_i \b_i $$
Now suppose for $I = \omega$, we have sets $A_i$ with cardinals $\a_i := |A_i| < \def\c{\mathfrak c}\c$, then $$ \left|\bigcup_i A_i\right| \le \sum_i \a_i < \prod_i \c = \c^{\aleph_0} = \c $$ Hence: If $\bigcup_i A_i$ has cardinality $\c$, one of the $A_i$ must also have cardinality $\c$.
martini
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Note that König's theorem is equivalent to the axiom of choice. (The above proposition - that continuum is not a union of countably many sets, all with cardinality strictly lesser than the continuum - is strictly weaker than the axiom of choice.) – Mike Rosoft Sep 22 '19 at 13:49