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If $\{A_n: n \in \mathbb{N} \}$ is a sequence of subsets of $\mathbb{R}$ and $|A_n| < \mathfrak{c}$ for all $n$. Prove $| \cup_n A_n| <\mathfrak{c}$

with $\mathfrak{c}$ the cardinality of $\mathbb{R}$.

I think we want to make a bijection from one $A_n$ to $ \cup_n A_n$. Because the first has cardinality smaller than $\mathfrak{c}$, the second must have too.

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  • 2
    $\begingroup$ Did you try anything on your own to solve the problem? $\endgroup$ – Asaf Karagila Oct 7 '13 at 14:06
  • $\begingroup$ You're assuming $|A_n| = |\cup A_k|$ for some $n\in\mathbb{N}$ which is not necessarily true if $\mathfrak{c} > 2^{\aleph_\omega}$. Assume $|A_n| = \aleph_n$, then $|\cup A_n|=\aleph{\omega}$. $\endgroup$ – kedrigern Oct 7 '13 at 14:33
  • $\begingroup$ @kedrigern: It is impossible that $2^{\aleph_\omega}<\frak c$. We can have, however, $\aleph_\omega<\frak c$ or $2^{\aleph_\omega}=\frak c$. $\endgroup$ – Asaf Karagila Oct 7 '13 at 15:21
  • $\begingroup$ Sure, I meant $\frak{c} > \aleph_{\omega}$. $\endgroup$ – kedrigern Oct 7 '13 at 15:29
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Clearly if there is some $k$ such that $|A_n|\leq|A_k|$ for all $k$, then the solution you propose works just fine. However your strategy may fail. It is consistent that there are such $A_n$'s such that $|A_n|<|A_{n+1}|$, and so $\bigcup A_n$ has cardinality greater than any single $A_n$.

Suppose that is the case, and let $\kappa_n=|A_n|$, then $\kappa=\sup\kappa_n=2^{\aleph_0}=\frak c$. We have now that $\kappa^\omega>\kappa$ by Koenig's theorem, however $\frak c^\omega=c$, which is a contradiction.

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  • $\begingroup$ Ah, Koenig's theorem does it! $\endgroup$ – 1234aaa Oct 7 '13 at 15:17
  • $\begingroup$ Question: $| \cup_n A_n|=\sum_{n \in \mathbb{N}} \mathfrak{a}_n < \prod_{n \in \mathbb{N}} \mathfrak{b}_n= \mathfrak{c}^k= \mathfrak{c}$, for $k \to \infty$. Is this true? $\endgroup$ – 1234aaa Oct 7 '13 at 15:38
  • $\begingroup$ I'm not sure what are $\frak a_n,b_n$ in this notation. Moreover the cardinal exponentiation is discontinuous (usually); and $k\to\infty$ is sorta vague in the context of cardinal (and ordinal) arithmetic anyway. $\endgroup$ – Asaf Karagila Oct 7 '13 at 17:37
  • $\begingroup$ Simpler: because there are possibly infinite $A_n$ shouldn't this lead to $\mathfrak{c}^\infty=\mathfrak{c}$? $\endgroup$ – 1234aaa Oct 7 '13 at 18:01
  • $\begingroup$ What's $\frak c^\infty$? $\endgroup$ – Asaf Karagila Oct 7 '13 at 18:06

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