What is the meaning of 'sup' in this equation: $$s(n)= \frac1n\sup_{k\ge0}T^{(k)}(n)$$

The equation is from this paper on the Collatz conjecture.

  • $\begingroup$ Supremum? $\endgroup$
    – suomynonA
    Oct 2, 2016 at 16:37
  • $\begingroup$ $\sup$ is always the supremum. So in this case you have to consider the set of real numbers $ \{ T^{(0)}(n), T^{(1)}(n), T^{(2)}(n), \cdots \} $ and consider it's least upper bound $\endgroup$ Oct 2, 2016 at 16:38
  • $\begingroup$ $\sup A_n$ where $A_n=\{T^{(k)}(n)\mid k\geqslant0\}$. $\endgroup$
    – Did
    Oct 2, 2016 at 16:38

1 Answer 1


Consider the set $\{T^{(k)}(n):k=0,1,2,\ldots\}.$ Then, $\sup_{k\ge0}T^{(k)}(n)$ is exactly $\sup\{T^{(k)}(n):k=0,1,2,\ldots\}.$ The supremum $\sup A$, where $A$ is a subset of an ordered set $E$, is an element of $E$ such that $x\le \sup A$ for all $x\in A$ and if $b < \sup A$ then there is some $x\in A$ such that $b < x$.

In other words, $\sup A$ is an upper bound of $A$ and it is the smallest upper bound of $A$.


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