Equivalent definition of an "attractor"

Question

Let $\Omega$ be a topological space, $\tau:\Omega\to\Omega$, $\operatorname{Orb}(\tau,U):=\bigcup_{n\in\mathbb N_0}\tau^n(U)$ for $U\subseteq\Omega$, $A\subseteq\Omega$ and$^1$ $$E(A):=\left\{x\in\Omega\mid\forall N\in\mathcal N(A):\exists n_0\in\mathbb N_0:\forall n\ge n_0:\tau^n(x)\in N\right\}.$$

$U\in\mathcal N(A)$ is called fundamenal neighbhorhood of $A$ if $$\forall N\in\mathcal N(A):\exists n_0\in\mathbb N_0:\operatorname{Orb}(\tau\circ\tau^{n_0},U)\subseteq N\tag1.$$ Moreover, $A$ is called

1. attracting if $\exists U\in\mathcal N(A):U\subseteq E(A)$;
2. stable if $$\forall N\in\mathcal N(A):\exists U\in\mathcal N(A):\operatorname{Orb}(\tau,U)\subseteq N\tag2;$$
3. asymptotic stable if $A$ is attracting and stable;
4. attractor if $A$ is forward invariant and asymptotic stable.

If $A$ is forward invariant, I've read that (assuming that $A$ is compact, but I actually don't see why this should matter) $A$ is an attractor if and only if $A$ has a fundamental neighborhood $U$.

Are we really able to show this?

The condition in $(1)$ ensures that $A$ is somehow "eventually stable". On the one hand, it is a stronger condition than $(2)$, since $U$ doesn't depend on $N$. On the other hand, it is weaker, since it only "becomes stable" at time $n_0$ (and as if this wasn't bad enough, $n_0$ depends on $N$).

So, if I'm not missing something neither should $(1)$ imply $(2)$ nor should $(2)$ imply $(1)$. Thus, if the claim is true, the crucial ingredient must be the forward invariance or the compactness.

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$^1$ $\mathcal N(A)$ denotes the set of neighborhoods of $A$.