I read the book of Yuri. Kuznetsov, Elements of Applied Bifurcation Theory.
He claims, $W^s(x_0) = \{x : \phi^t x\rightarrow x_0, t \rightarrow \infty\}$, defined therein as a stable set, same (as much as I get) as Stable subspace as in Hirsch and Smale. The unstable counterpart is defined with negative infinity.
But, he also claims this to be an invariant. Now in an invariant set, would be $S = \{x \mid x \in S \implies S \ni \phi^t (x) \}$. My question is Why the former is also invariant.
To quench my question, I progressed so far as to :
Let $x_i = x^{\prime} \in W^s(x_0) \mid \phi^{t_1} (x^{\prime}) \not \in W^s, t < \infty$.
Thus , form assumption, $\phi^t(x^{\prime}) = x_0$
Furthermore $\phi^{t_2} (\phi^{t_1} (x^{\prime})) \neq x_0, t_2 \rightarrow \infty \implies \phi^{t_1+t_2}(x^{\prime}) \neq x_0, t_1 + t_2 = t \rightarrow \infty$. Thus Reductio ad absurdum.
Am I right to deduce so?
I have seen this already, but my question was not answered here.
Assume that $x$ is in $W^s(x_0)$ and let $y=\phi^T(x)$ for some $T$. Then $\phi^t(y)=\phi^{t+T}(x)$ hence, when $t\to+\infty$, $\phi^t(y)\to x_0$ because $t+T\to+\infty$ and $\phi^u(x)\to x_0$ when $u\to+\infty$. Thus, $y=\phi^T(x)$ is in $W^s(x_0)$. This holds for every $x$ in $W^s(x_0)$ and every $T$, hence $W^s(x_0)$ is stable by $\phi$, QED.