Notation for eventually less than - MathOverflow

Notation for eventually less than

2010-01-06T13:13:19Z

Is there some existing notation for

\[f(n)\leq g(n)\] for sufficiently large n

Apart from just writing that itself? I'm thinking of something compact like the Landau notation $f\ll g$.

(Apologies if this is too specific for MathOverflow - just close it if so. I was also unsure what tags to add, so just edit it accordingly).

Answer by Joel David Hamkins for Notation for eventually less than

2010-01-06T13:49:27Z

In logic, this relation is called almost less than or equal, and is denoted with an asterisks on the relation symbol, like this: $f \leq^* g$.

For example, the bounding number is the size of the smallest family of functions from N to N that is not bounded with respect to this relation. Under CH, the bounding number is the continuum, but it is consistent with the failure of CH that the bounding number is another intermediate value.

Answer by Matt Noonan for Notation for eventually less than

2010-01-06T15:53:48Z

Why not just overload $\leq$ when applied to sequences? I don't think there is any opportunity for confusion, and it fits with the notation you would use when extending $\leq$ to an ultraproduct.

This is what Jim Henle does in his "non-nonstandard analysis", which uses "eventually" as a replacement for an ultrafilter.

Answer by Peter Michor for Notation for eventually less than

2012-12-17T07:31:34Z

Good notation should be self-explaining and not require the reader to remember to much. I would write either: $$ f\le g\quad\text{eventually}$$ $$ f\le g\quad\text{ near }\infty$$ If you use it more than 100 times in a paper you could use something like $$ f \preccurlyeq g.$$

Answer by Andreas Blass for Notation for eventually less than

2012-12-17T15:29:53Z

I agree with Joel Hamkins's answer, but I don't entirely agree with his comment on that answer. I generally use asterisks to mean "with finitely many exceptions" or "modulo finite sets", so I'd use $f\leq^*g$ and $A\subseteq^*B$ as Joel says. But when working modulo some ideal $I$ other than the ideal of finite sets, I'd ordinarily avoid asterisks and instead write $f\leq_Ig$ and $A\subseteq_IB$.

I'd like to protest vigorously against the use of $\ll$ in this situation. To me, $f\ll g$ means that $f$ is a lot smaller than $g$ (at least eventually), whereas here you might have $f(n)=g(n)-1$ for all $n$.