The definition of Lyapunov order number in Hartman ode book.

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In the Hartmann ode book: following are the definition of "Lyapunov order number"

consider a function $y(t)$ for $t\geq 0$.(Hartman use $y(t)$ as vector function in $\mathbb{R^d}$) A number $\tau$ is called a (Lyapunov) order number for y(t) if, for every $\epsilon>0$ there exist positive constants $C_0(\epsilon),C(\epsilon)$ such that

$$(1)\ \ \ \ \|y(t)\|\leq C(\epsilon)e^{(\tau+\epsilon)t}\ for\ all\ large\ t$$ $$(2)\ \ \ \ \|y(t)\|\geq C_0(\epsilon)e^{(\tau-\epsilon)}t\ for\ some\ arbitrarily\ large\ t.$$

here $\|\cdot\|$ is "Euclidean norm" for vector.

When $y(t)\neq 0$ for large $t$, an equivalent formulation of (1) and (2) is

$$(3)\ \ \ \ \limsup_{t\rightarrow\infty}t^{-1}\log\|y(t)\|=\tau;$$

actually Hale's ode book use only (3) as the defintion of "Lyapunov characteristic number".

actaully (3) is clear;

(3) hold if and only if

$$(4)\ \ \forall \epsilon>0, \exists t_0(\epsilon)\ s.t \ t\geq t_0\ imply\ t^{-1}log\|y(t)\|\leq \tau+\epsilon\ (i.e\ \|y(t)\|\leq e^{(\tau+\epsilon)t})$$ and $$(5)\ \ \exists\ sequence\ (t_n)_{n\in\mathbb{N}}\ \ s.t\ \ t_n\rightarrow \infty\ as\ n\rightarrow \infty\ and\ t_n^{-1}\log\|y(t_n)\|\geq\tau-n^{-1}\ (i.e\ \|y(t_n)\|\geq e^{(t-1/n)t_n})$$

I don't know why (1),(2) is equivalent (4),(5).

Actually I don't know the meaning of "for all large t" and "for some arbitrarily large t" in (1) and (2)

If we think (1) as $$(6)\ \ \exists T>0\ s.t\ \forall \epsilon>0, \exists C(\epsilon)>0\ s.t\ \ t>T\ imply\ \|y(t)\|\leq C(\epsilon)e^{(\tau+\epsilon)t}$$

If we assume $y(t)$ is continuous on $t>0$, then (4) imply (6).

if we let $C(\epsilon)=\max\{\sup_{t \in [T,t_0(\epsilon)]} \|y(t)\|e^{-(\tau+\epsilon)t}\ ,1\}$. it is clear (4) imply (6).

However I don't know how to verify (6) imply (4).

I want to know the meaning of "for all large t" and "for some arbitrary large t" in here. and I want to know why (3) is equivalent (1) and (2) under the condition $y(t)\neq 0$.

Please give me some help, If you have some idea.