# Is there a partition $\pi$ of $\mathbb{R}$ such that $\lvert \pi \rvert = \mathfrak{c}$ and $\forall X \in \pi~ \lvert X \rvert = \mathfrak{c}$?

I was wondering if there's a partition $$\pi$$ of $$\mathbb{R}$$ such that $$\lvert \pi \rvert = \mathfrak{c}$$ and $$\forall X \in \pi~ \lvert X \rvert = \mathfrak{c}$$. Now, I know that there exist partitions of $$\mathbb{N}$$ whose cardinality is $$\aleph_0$$ and every set in them is also of cardinality $$\aleph_0$$. I was wondering if that is the case for $$\mathbb{R}$$, too?
If so, what's an example of such partition? I think such partition exists because $$\mathfrak{c} \cdot \mathfrak{c} = \mathfrak{c}$$, but I can't think of a specific one.

For each $$\alpha$$ in the interval $$[0,9],$$ let $$S_{\alpha}$$ be the set of all real numbers which, when written in decimal expansion $$k_1\ k_2\ \ldots\ k_{n-1}\ k_n\ .\ k_{n+1}\ k_{n+2}\ldots,$$ have the property that $$\displaystyle\lim_{n\to\infty} \left(\frac{\displaystyle\sum_{i=1}^{n} k_i}{n}\right) = \alpha.$$

Further, let $$T$$ be the set of all real numbers whose decimal expansion has the property that $$\displaystyle\lim_{n\to\infty} \left(\frac{\displaystyle\sum_{i=1}^{n} k_i}{n}\right)$$ does not converge.

Then, $$\vert T \vert = 1,\ \left \vert\ \{\ S_{\alpha}: \alpha \in [0,9]\ \}\ \right \vert = \mathfrak{c},\$$ and

$$\ T\cup \left(\bigcup\limits_{\alpha\in [0,9]} S_{\alpha}\right) = \mathbb{R}$$ is an uncountable partition of $$\mathbb{R}$$ into uncountable sets.

• The limit doesn’t exist for all real numbers. For instance $0.00111100\dots$ where the blocks have length $2^{2^k}$, $k=0,1,….$. Thus the union of the $S_{\alpha}$ is not all of $\Bbb R$. Aug 25, 2022 at 13:29
• Yes you are right, but I think this is fairly easy to rectify. Please see my answer now. Aug 25, 2022 at 14:38

Such a partition does exist. If one has a bijection $$f:\mathbb R\to \mathbb R\times\mathbb R$$, one can define $$X_r=\{t\in \mathbb R\colon f(t)=(r,y)\text{ for some }y\in\mathbb R\},$$ i.e. the preimage of the line $$x=r$$ in $$\mathbb R^2$$. Then, since these lines in $$\mathbb R^2$$ are disjoint and have union all of $$\mathbb R^2$$, the sets $$\{X_r\colon r\in\mathbb R\}$$ are disjoint and have union $$\mathbb R$$. These sets are clearly indexed by $$\mathbb R$$, and so the set of them has cardinality $$\mathfrak c$$, but each set $$X_r$$ is also indexed by $$\mathbb R$$ (via $$y$$), so each such set has cardinality $$\mathfrak c$$.

• To be a bit more explicit, you could partition the real numbers according to the value of $\ldots x_2 x_0 . x_{-2}x_{-4}\ldots$ where the decimal expansion of $x$ is $x_d x_{d-1} \ldots x_0 . x_{-1} x_{-2} \ldots$ (choosing the expansion ending in 0's instead of the expansion ending in 9's where the expansion is ambiguous). Aug 24, 2022 at 22:55