Is every fixed point in an attractor stable?

Question

Let $\Omega$ be a topological space and $\tau:\Omega\to\Omega$. We say that a fixed point $x_0\in\Omega$ of $\tau$ is stable if for every neighborhood $V$ of $x$, there is a neigheiborhood $U$ of $x$ with $$\tau^n(U)\subseteq V\;\;\;\text{for all }n\in\mathbb N_0\tag1.$$

On the other hand, $A\subseteq\Omega$ is called stable if for every neighborhood $V$ of $A$, there is a neigheiborhood $U$ of $A$ with $$\tau^n(U)\subseteq V\;\;\;\text{for all }n\in\mathbb N_0\tag2.$$

If $A$ is stable, can we conclude that every fixed point of $\tau$ in $A$ is stable?

If not, does the implication hold if $A$ is an attractor, i.e. $A$ is

1. forward invariant, i.e. $A=\tau(A)$;
2. stable;
3. there is a neighborhood $U_0$ of $A$ so that every $x\in U_0$ is attracted to $A$, i.e. $x\in E(A)$,

where $E(A)$ is 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 $A$?

Answer 1

We cannot conclude that if $A$ is stable, every fixed point of $A$ is stable. Consider $\tau : \mathbb{R} \to \mathbb{R}$ defined by $\tau(x) = 2x$. Then $0$ is not stable, but $\mathbb{R}$ is stable.

In fact, we see that $\mathbb{R}$ is an attractor. So we can't conclude the statement even with the stronger condition of $A$ being an attractor.