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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.

Thanks in advance!

tcb93
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2 Answers2

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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.

Adam Rubinson
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  • 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 $\Bbb R$. – Jamie Radcliffe Aug 25 '22 at 13:29
  • Yes you are right, but I think this is fairly easy to rectify. Please see my answer now. – Adam Rubinson Aug 25 '22 at 14:38
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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 $\mathfrak c$.

Carl Schildkraut
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    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). – Daniel Schepler Aug 24 '22 at 22:55