Let $\mathcal{H}_A \otimes \mathcal{H}_B$ be the tensor product of two finite dimensional Hilbert spaces, let $d = \operatorname{dim}(\mathcal{H}_A \otimes \mathcal{H}_B)$ and let $| \psi \rangle \in \mathcal{H}_A \otimes \mathcal{H}_B$ be a pure entangled state.
We say the entanglement in $| \psi \rangle$ is $\epsilon_0$-robust, for some $\epsilon_0 \in [0,1]$, if $ (1-\epsilon) | \psi \rangle \langle \psi | + \epsilon \, \mathbb{I}/d$ is entangled for all $\epsilon \in [0, \epsilon_0]$. More generally we say the entanglement in $| \psi \rangle$ is completely $\epsilon_0$-robust if $ (1-\epsilon) | \psi \rangle \langle \psi | + \epsilon \, \tau$ is entangled for all $\epsilon \in [0, \epsilon_0]$ and all states $\tau$ on $\mathcal{H}_A \otimes \mathcal{H}_B$.
Are there any pure entangled states that are not $\epsilon_0$-robust (or completely $\epsilon_0$-robust) for all $\epsilon_0 > 0$?