User blog comment:B1mb0w/Fundamental Sequences/@comment-5529393-20160124015439/@comment-27173506-20160206062007

The thing is that if we care about the ordinals, the only important thing is that their fundamental sequences limit to that ordinal. Hell, we could have \(\varepsilon_0[n]=\omega \uparrow \uparrow \F_{n}\), where Fn is the nth Fibonacci number. How about \(\zeta_0 [0]\)? Should that be equal to 0, to 1, or to BB(1000) (Busy Beaver function)? Fact is, it doesn't matter. It will always limit to the correct ordinal.

Your Veblen function however, does not. As Deedlit said, tetrating by \(\omega\) will bring you to the next epsilon number after that. To see that, observe: \(\varepsilon_{\zeta_{0}+1} = \text{sup} \lbrace \varespilon_{\zeta_0}, \varepsilon_{\zeta_0}^{\varepsilon_{\zeta_0}} , \varepsilon_{\zeta_0}^{\varepsilon_{\zeta_0}^{\varepsilon_{\zeta_0}}} \cdots \rbrace = \text{sup} \lbrace \zeta_0 , \zeta_{0}^{\zeta_0} , \zeta_{0}^{\zeta_{0}^{\zeta_0}} \cdots \rbrace = \zeta_0 \uparrow \uparrow \omega\).