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The Wikipedia article on Sarkovskii's theorem claims that the Sarkovskii ordering of the natural numbers is not a well-ordering, stating:

Note that this ordering is not a well-ordering, since the set $$\left\{ 2^k \mid k \in \mathbb{N} \right\}$$ doesn't have a least element.

I cannot see how this is true. Surely the element $2^0$ is a least element of this set?

Andrea
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$2^0$ is least in the usual ordering on the natural numbers, but not in the Sarkovskii ordering, which is a different ordering of the naturals. In this ordering, $2^0=1$ is actually the largest natural number, and the smallest is $3$.

$$3<5<\cdots<2^4<2^3<2^2<2^1<2^0$$

vadim123
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  • Oh, of course! That was stupid of me. I've always seen the Sarkovskii ordering defined in the opposite way, i.e. $3$ is the largest number and $1$ is the smallest. I didn't read the article carefully enough. Now it makes sense, thank you. – Andrea Jan 17 '15 at 21:08