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Can the real line be written as a disjoint union of sets with cardinality 5?

I tried using to write it as a set of sets.. but I didn't end up to a good position.

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    $\begingroup$ Can you write $\mathbb Z$ as a disjoint union of sets with cardinality $5$? $\endgroup$ – bof Sep 3 '15 at 9:04
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    $\begingroup$ If $S$ is a set, can you write $\mathbb Z\times S$ as a disjoint union of $5$-element sets? Can you write $\mathbb Z\times[0,1)$ as a disjoint union of $5$-element sets? $\endgroup$ – bof Sep 3 '15 at 9:06
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Sure, why not.

$$\mathbb{R} = \bigcup_{k \in \mathbb{Z}, \ \alpha \in [0,1)} \{5k+\alpha, 5k+1+\alpha, 5k+2+\alpha, 5k+3+\alpha, 5k+4+\alpha\}$$

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If we allow the axiom of Choice, then one can show that the cardinality of $\mathbb R \times 5$ is equal to the cardinality of $\mathbb R$.

Now, $card(\mathbb R \times 5)$ is the same as $card(\sum_{i \in \mathbb R}5) $. This indicates that there exists a correspondence between $\mathbb R$ and $\mathbb R$ disjoint sets of cardinality $5$.

Using this correspondence we can thus find an infinite collection of sets of cardinality 5, whose union is $\mathbb R$ and pairwise disjoint.

References: Naive Set Theory, by P. Halmos.

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    $\begingroup$ The axiom of choice is needed to show that the cardinality of $S\times5$ is equal to the cardinality of $S$ for an arbitrary infinite set $S$, but it is not needed to show that the cardinality of $\mathbb R\times5$ is equal to the cardinality of $\mathbb R.$ $\endgroup$ – bof Sep 3 '15 at 9:23
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Let $\mathfrak c=|\mathbb R|$. Let $S_0=\{1,2,3,4,5\}$. We may recursively define $S_\alpha$ for $\alpha<\mathfrak c$ such that $S_\alpha$ has $5$ real numbers not in $\bigcup _{\beta<\alpha} {S_\beta}$, using the fact that $|\bigcup _{\beta<\alpha} {S_\beta}|=|\alpha\times 5|\leq \max(\omega,|\alpha|)<\mathfrak c.$

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