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I found in this post the following question:

Can one partition the plane $\mathbb{R}^2$ by closed intervals of equal length?

Then it is written: "The answer to the first question is "yes"". My question is: why is the answer 'yes'?

I think that the length should be $>0$, otherwise it is obviously true and uninteresting. According to Partitioning $[0,1]$ into pairwise disjoint nondegenerate closed intervals, Can $\mathbb R$ be written as the disjoint union of (uncountably many) closed intervals?, Is $[0,1]$ a countable disjoint union of closed sets?, this can't be done in dimension $1$.

Thank you!

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Alphonse
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1 Answers1

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A construction of such partition into unit closed closed segments is in:

J. Conway, H. Croft, Covering a sphere with congruent great-circle arcs. Proc. Cambridge Philos. Soc. 60 (1964) p. 787–800.

Moreover, they use only countably many different slopes in the construction.

A sketch of their construction you can also find in chapter 12 of

M.Gardner, Knotted Doughnuts and Other Mathematical Entertainments, W.H. Freeman & Company, 1986.

Edit. They also prove that the plane cannot be partitioned into open unit intervals. Lastly, it is easy to see that the plane can be partitioned into vertical/horizontal open intervals of lengths $1$ and $2$.

Moishe Kohan
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