I'm wondering whether it is possible to Partition the closed interval $[0,1]$ into closed, countably infinite sets.
The only observations I could make were as follows:
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When we remove the endpoints, we effectively end up with $\mathbb R$, and can consider the image of
$$\mathscr S :=\{r+\mathbb N\mid r\in [0,1)\}$$-
Because every $S\in \mathscr S$ is bounded, it must contain at least one limit point. Because our partition itself must be uncountable, $\bigcup_{S\in \mathscr S} L(S)$ ist uncountable, and by second countability, has uncountably many limit points.
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A natural example for a partition into countably infinite sets is $\{r+\mathbb Q\}$, but these elements are not closed – on the contrary, every sets closure is $\mathbb R$ already.
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I'm not really sure where to go from there. My intuition says that this won't work, because it would get “too crowded” near the boundary points, but I'm not sure how to make this precise.
Best Answer
Yes. You can build such a partition by transfinite recursion.
Specifically, we construct a sequence of disjoint closed countably infinite subsets $A_\alpha\subset[0,1]$ where $\alpha$ ranges over the ordinals. Having constructed $A_\beta$ for all $\beta<\alpha$, let $S=[0,1]\setminus\bigcup_{\beta<\alpha}A_\beta$. If $S$ is uncountable, then there is some sequence of distinct elements of $S$ which converges to an element of $S$ (if not, every element of $S$ would be isolated in $S$ and $S$ would be discrete, but there is no uncountable discrete subset of $[0,1]$). Pick such a sequence and let $A_\alpha$ consist of the terms of the sequence together with their limit.
If $S$ is countable, we instead halt the recursion, and put each element of $S$ into a different one of the $A_\beta$ for $\beta<\alpha$ (there must have been uncountably many $\beta<\alpha$ so we can do this, and the sets will remain closed after adding a single point).
Since $[0,1]$ is a set, the recursion must halt eventually, and we obtain a partition of $[0,1]$ into countably infinite closed sets.