Does a global attractor contain every (forward) invariant set?

Question

In Remark 2.1.1 of this work, it is claimed that "the global attractor of a dynamical system contains every invariant set".

It is not completely transparent what their definition of a "(global) attractor" is, but I would like to understand under which assumptions this claim can be verified.

In order to do so, let $\Omega$ be a topological space and $\tau:\Omega\to\Omega$. Say that $A\subseteq\Omega$ is$^1$

1. forward invariant if $A=\tau(A)$;
2. backward invariant if $A=\tau^{-1}(A)$;
3. invariant if $A$ is forward and backward invariant.

Moreover, let $E(A)$ denote the basin of attraction for $A$, i.e. the set of all $x\in\Omega$ for which the orbit $$\operatorname{orb}x:\mathbb N_0\to\Omega\;,\;\;\;n\mapsto\tau^n(x)$$ is eventually in every neighborhood of $A$.

If $Q\subseteq\Omega$, let $A_Q:=\bigcap_{n\in\mathbb N_0}\tau^n(Q)$. It's then trivial to see that if $B\subseteq Q$ is forward invariant, then $$B=\bigcap_{n\in\mathbb N_0}\tau^n(\underbrace B_{\subseteq Q})\subseteq A_Q.\tag1$$ So, every forward invariant subset of $Q$ is contained in $A_Q$.

If $A\subseteq\Omega$ is forward invariant and $U$ is a fundamental neighborhood of $A$$^2$, then it is easy to show that $$\overline A=A_U\tag2.$$ Hence, from the former result, we can immediately conclude that every forward invariant subset of $U$ is contained in $\overline A$.

Now, assuming that this particular $A$ is "global" in the sense that $E(A)=\Omega$, are we somehow able to conclude that even every forward invariant subset of $\Omega$ is contained in $\overline A$?

If this is not possible in general, feel free to add further assumptions (e.g. $\tau$ being a homeomorphism as in the reference or $A$ being closed) or to rephrase the claim with "forward invariant" replaced by "backward invariant" (which might also be what the author meant).

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$^1$ Clearly, if $\tau$ is injective, then forward invariant implies backward invariant. And if $\tau$ is surjective, then backward invariant implies forward invariant. In particular, if $\tau$ is bijective, ten forward and backward invariance are equivalent.

$^2$ i.e. for every neighborhood $V$ of $A$, there is a $n_0\in\mathbb N_0$ with $\tau^n(U)\subseteq V$ for all $n\ge n_0$.

Answer 1

It's not true at all. Consider a very simple example: $\Omega = \mathbb R$ with $\tau(x) = x/2$. $\{0\}$ is a global attractor, but $\{2^n: n \in \mathbb Z\}$ is invariant and not contained in the global attractor.