Are all pure entangled states `robust'?

Question

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 $epsilon_0 > 0$?

Norbert Schuch · Accepted Answer

The set of separable states is closed.
Thus, around any entangled state - not necessarily pure - there is an $epsilon$-ball which lies entirely within the entangled states.
Or, in the language of your question: All entangled states are "robust".
(As illustrated by DaftWullie's answer, the size of this ball can depend on the state: There are pure entangled states arbitrarily close to separable ones.)

DaftWullie · Answer

For a fixed $epsilon_0$, why not simply consider
$$
|psirangle=costheta|00rangle+sintheta|11rangle?
$$
Since it's a two-qubit state, entanglement can be determined using the PPT criterion. Hence,
$$
rho=(1-epsilon)|psiranglelanglepsi|+epsilon I/4
$$
is entangled if $epsilon<2sin(2theta)/(1+2sin(2theta))$. Any $epsilon$ you give me, and I just pick $0<theta<arcsin(epsilon_0/2)/2approxepsilon_0/4$, and the state is not $epsilon_0$-robust. Given there exists a state that is not $epsilon_0$-robust, entanglement is not completely $epsilon_0$-robust.
To prove things the other way around (for fixed $|psirangle$, is there always a non-zero $epsilon_0$ such that for all $epsilon<epsilon_0$, the mixed state is entangled?), we can consider entanglement witnesses. Let $W$ be an entanglement witness for $|psirangle$. We have $text{Tr}(Wrho)geq 0$ for all separable states $rho$ and $text{Tr}(W|psiranglelangle psi|)=-sigma<0$.
Now, the trace of $W$ will be some specific value $text{Tr}(W)=kgeq 0$ (this is positive since the maximally mixed state is separable). Consider
$$
text{Tr}(W(epsilon I+(1-epsilon)|psiranglelanglepsi|))=epsilon k-(1-epsilon)sigma.
$$
For any
$$
epsilon<frac{sigma}{k+sigma},
$$
the trace is negative and hence the state is entangled.

Are all pure entangled states `robust'?

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