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In the ODE where $y'=f(y(t))$ and $y(0)=y_0$.

The omega limit set $\omega(y_0)$ is positively invariant and also negatively invariant.

I want to prove first that its positively invariant and then prove its negatively invariant.

But how do I show that using a flow function $(\phi(y,t)$) given that I know only the definition and the identity of flow function and very little understanding of the concept of flow function.

Thankyou for the help!

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    See this link for a proof: http://books.google.dz/books?id=yfY2uGROVrUC&pg=PA92&lpg=PA92&dq=Omega+limit+set+is+invariant,+proof&source=bl&ots=dXUvpOvds2&sig=JlubYmfAcrlo-VfizYPBw25lGTk&hl=fr&sa=X&ei=EujNUNq_JcKUtAaykoGICQ&redir_esc=y#v=onepage&q=Omega%20limit%20set%20is%20invariant%2C%20proof&f=false This is a google book.2012-12-16

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Consider the system $ \dot x=f(x),\quad x\in U\subset\mathbf{R}^k,\,f\colon U\to\mathbf R^k, $ with the solution $ x=x(t;x_0). $ By definition point $\bar x$ belongs to omega limit set is there exists a sequence $\{t_n\}$ such that $x(t_n;x_0)\to \bar x$ when $n\to\infty,\,t_n\to\infty$. This means that for any fixed $t$ one has (using the properties of the flow) $ x(t_n+t;x_0)=x(t;x(t_n;x_0))\to x(t;\bar x), $ which means that if $\bar x$ belongs to omega limit set then the whole orbit containing this point belongs to this set, and since the orbits are invariant this implies that omega limit set is both positive and negative invariant.

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    thanx a lot for explaining. i really appreciate ur help!2012-12-16