Is the Lyapunov stability definition ambiguous? I understand the conceptual idea of Lyapunov stability.
My question is about the formal definition.
Wikipedia's definition is very much in line with others I found:

An equilibrium, $x_{e}$, is said to be Lyapunov stable, if, for every $\epsilon >0$, there exists a $\delta >0$ such that, if $\|x(0)-x_{e}\|<\delta $, then for every $t\geq 0$ we have $\|x(t)-x_{e}\|<\epsilon$.

The confusing part for me is the phrase "for every $\epsilon >0$".
Assume $\|x(0)-x_{e}\|>0$ and that some $\delta$ exists as required. If I choose $\epsilon = \frac{|x(0)-x_{e}|}2$ then $x(0)$ falls outside the circle ($\|x(t)-x_{e}\|<\epsilon$).  By this interpretation no fixed point can be Lyapunov stable. And this is how I know I am in trouble :-).
This makes me wonder why this definition does not draw some relation between $\epsilon$ and $x(0)$. For example, if it includes $\epsilon > \|x(0)-x_{e}\|$ it would aid in my understanding and remove the argument I just provided.
Alternatively, I could wonder why not use the phrase "for some $\epsilon$ where $0 < \epsilon < \infty$" instead of "for every $\epsilon >0$".
I am clearly missing something. Please help me to eliminate the ambiguity I read in this definition.
 A: You're not interpreting the logic correctly.
It doesn't say “there exists a single positive number $\delta$ which works for all positive numbers $\epsilon$”, it says “for any positive number $\epsilon$, you can find a corresponding positive number $\delta$ which works for that particular number $\epsilon$”. If you change $\epsilon$ to a smaller number, you will in general have to make $\delta$ smaller too. But the point is that no matter how small an $\epsilon$ you take, you can find a $\delta$ that works (and that's why it has to say “for every $\epsilon$”, not just “for some $\epsilon$”).
This is just as for the $\epsilon$-$\delta$ definition of limits.
A: $ϵ$ is primary, $δ$ is secondary depending on it.
As you have chosen $x(0)$, you already know the radius $δ$ in $\|x(0)-x_e\|<δ$ and thus also the $ϵ$ it is based upon. So there remains no freedom to now alter $ϵ$ to some other value.
Or in other words, if you now select a new $ϵ$, then the stability property results in a new $δ$, for which the selected $x(0)$ is likely no longer admissible.
