# can real line be written as a disjoint unions of set with cardinality 5

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.

• Can you write $\mathbb Z$ as a disjoint union of sets with cardinality $5$? – bof Sep 3 '15 at 9:04
• 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? – bof Sep 3 '15 at 9:06

Sure, why not.

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

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.

• 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 $\mathbb R.$ – bof Sep 3 '15 at 9:23

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 $|\bigcup _{\beta<\alpha} {S_\beta}|=|\alpha\times 5|\leq \max(\omega,|\alpha|)<\mathfrak c.$