A subset $A\subset X$ is forward invariant if $f^{t}(A)\subset A$ for all $t\ge 0$ and backward invariant if $f^{-t}(A)\subset A$ for all $t\ge 0$.
I want to show that the complement of a forward invariant set is backward invariant (and vice-versa).
So the complement of $A$ is $X\setminus A$.
We have $f^{-t}(X\setminus A)=(f^{t})^{-1}(X\setminus A)...\subset X\setminus A$ for all $t\ge 0$.
How do I fill in the space between those dots?
Well, a simple reductio ad absurdum works here. Suppose that $X \setminus A$ is not backward invariant. It means that exist $x^{\ast} \in (X \setminus A)$ and $ t^{\ast} \in \mathbb{R_{+}}$ such that $f^{-t^{\ast}}(x^{\ast}) \notin (X\setminus A)$. From this follows that $f^{-t^{\ast}}(x^{\ast}) \in A$ and since $A$ is forward invariant $f^{t^{\ast}} \bigl ( f^{-t^{\ast}}(x^{\ast}) \bigr )$ is also in $A$. Therefore, $x^{\ast} \in A$, which is the contradiction.