TITLE: Is an open subscheme of a rationally connected variety, rationally connected?
QUESTION [8 upvotes]: Let $X$ be a projective, irreducible variety over an algebraically closed field (of characteristic zero) which is rationally connected. Is it true that any open dense subvariety of $X$ is rationally connected? If so, can we say that same for rationally chain connectedness?
By rationally (chain) connected we do not assume properness (similar to the definition in Kollar's "Rational curves on algebraic varieties"). We simply ask that two general points in the variety are connected by a (chain of) rational curve.
Any idea/reference is most welcome.

REPLY [10 votes]: If I have understood the OP's definitions correctly, the answer is no for rational chain connectedness. Let $E$ be a smooth genus $1$ curve in $\mathbb{P}^2$ and let $X \subset \mathbb{P}^3$ be the cone on $E$, with vertex $x_0$. For any $x_1$ and $x_2$ in $X$, the lines joining $x_1$ to $x_0$ and $x_0$ to $x_2$ are in $X$, so $x_1$ and $x_2$ are connected by a chain of rational curves. 
However, let $Y = X \setminus \{ x_0 \}$. Let $\pi : Y \to E$ be the projection. For any rational curve $C \subset Y$, the map $\pi : C \to E$ maps a genus $0$ curve to a genus $1$ curve, so it is constant. Thus, all rational curves of $Y$ lie in fibers of $\pi$, and points in different fibers of $\pi$ cannot be connected by a chain of rational curves.
Disclaimer: I'm not very familiar with the literature on rational connectedness, so I might have misguessed a definition.