User blog:Ynought/Attempt at an ordinal catching type function

So if this definition doesn't work in any way then please tell me what is wrong with it.

Catching type function
\(f\) is the FGH

\(g\) is the SGH

when i say \(f_\alpha(n)\approx g_\alpha(n)\) i am saying that for that \(\alpha\) there exists an\(n\) for which holds \(g_\alpha(n+1)>f_\alpha(n))\)

\(\gamma(\beta)\) is the ordinal needed that \(g_\alpha(n)\approx f_\alpha(n)\) is the case for the \(\beta\)'th time

the way you get that for succesors is easy but if you come across a limit then :

\(\eta\) is the limit ordinal which you come acros

so when \(\alpha=\eta+\delta\) then \(\gamma(\alpha)=\gamma(\eta[n]+\delta)\)

Formal definition?
\(f_\alpha(n)\approx g_\alpha(n)\) means \(\exists(n|f_\alpha(n)\leqslant g_\alpha(n+1))\)

for \(\alpha=\delta+1\):

\(\gamma(\alpha)[n]=\beta|\beta \text{ is the }\alpha\text{'th time that }f_\beta(n)\approx g_\beta(n)\)

for countable non 0 limit \(\alpha\):

\(\gamma(\alpha)[n]=\gamma(\alpha[n])\) here \(\alpha[n]\) denotes the n-th term in the fundamental sequnce leading up to \(\alpha\)

for countable limit \(\alpha\) + succesor

\(\gamma(\alpha+\eta)[n]=\gamma(\alpha[n]+\eta)\)