Let $\Omega$ be a topological space, $\tau:\Omega\to\Omega$, $A\subseteq\Omega$, $\mathcal N(A):=\{N\subseteq\Omega:N\text{ is a neighborhood of }A\}$ and $$B(A):=\left\{x\in\Omega\mid\forall N\in\mathcal N(A):\exists n_0\in\mathbb N:\forall n\ge n_0:\tau^n(x)\in N\right\}.$$
$U\in\mathcal N(A)$ is called fundamental neighborhood of $A$ if $$\forall N\in\mathcal N(A):\exists n_0\in\mathbb N:\forall n\ge n_0:\tau^n(U)\subseteq V\tag1.$$
It is easy to show that if $U$ is a fundamental neighborhood of $A$, then $$U\subseteq B(A)\tag2$$ and hence $$B(A)\in\mathcal N(A).$$ Are we able to show the converse, i.e. does $B(A)\in\mathcal N(A)$ imply that $A$ has a fundamental neighborhood?
EDIT
I'm not sure if it is relevant or not, but please note that in the other direction (i.e. the direction where we assume that $A$ has a fundamental neighborhood $U$) we can show more than $(2)$, namely $$B(A)=\bigcup_{n\in\mathbb N_0}\tau^{-n}(U).\tag3$$
"$\subseteq$": Let $x\in B(A)$. Since $U\in\mathcal N(A)$, there is a $n_0\in\mathbb N$ with $\tau^n(x)\in U$ for all $n\ge n_0$ and hence it even holds $$x\in\bigcap_{n\ge n_0}\tau^{-n}(U)\tag4.$$
"$\supseteq$": Let $x$ belong to the right-hand side of $(3)$. Then there is a $n_1\in\mathbb N_0$ with $\tau^{n_1}(x)\in U$. Let $N\in\mathcal N(A)$. By $(1)$, there is a $n_2\in\mathbb N$ with $\tau^n(U)\subseteq N$ for all $n\ge n_2$. Let $n_0:=n_1+n_2$. Then it holds $$\tau^n(x)=\tau^{\overbrace{n-n_1}^{\ge\:n_2}}(\underbrace{\tau^{n_1}(x)}_{\in\:U})\;\;\;\text{for all }n\ge n_0\tag5.$$
You need stronger assumptions on the attractor. Take e.g. a circle homeomorphism that fixes precisely one point $p$. All other points then moves, say to the 'right'. Then $A=\{p\}$ is an attractor (with $B(A)=S^1$) but has no neighborhood which contracts uniformly towards $A$. The reason is that if $x_k$ converges to $p$ from the right and $N$ is a small neighborhood of $p$ then the time $n_k=n_0(x_k,N)$ it takes for $\tau^n(x_k)\in N$ for all $n\geq n_k$ will diverge with $k$.