Union of all the rational curves in a complex algebraic variety Can the union of all the rational curves in a complex algebraic variety contain a complex ball and not be the whole variety?
 A: I am just posting my comments as an answer.  The OP clarified that they want to use the "arithmetic genus" and not the "geometric genus" when working with irreducible singular curves in a projective variety.  With this clarification, the question has a negative answer (both the original question for genus $0$ and the question for genus $\leq 2$): the union of all irreducible curves in a projective variety having arithmetic genus $0$, respectively having arithmetic genus $\leq 2$, is typically not a union of countably many closed subvarieties.  Thus, even if the union contains a nonempty open subset, it can happen that the union is not the entire projective variety.
One negative example arises from the non-normal variety whose normalization equals $\mathbb{P}^1\times C$, where $C$ is a curve of genus $>2$ and where the normalization morphism is an isomorphism away from a length-$n$, finite subset of a fiber $\mathbb{P}^1\times \{t\}$, where $n>2$.  The curves of geometric genus $\leq 2$ in the non-normal variety are images of curves of geometric genus $\leq 2$ in the normalization, and these are each of the form $\mathbb{P}^1\times \{s\}$, for a choice of closed point $s$ of $C$.  The image of such a curve in the non-normal variety has arithmetic genus $0$ if $s\neq t$, but has arithmetic genus $>2$ if $s$ equals $t$.  Thus, the union of the irreducible curves of arithmetic genus $\leq 2$ equals the image in the non-normal variety of $\mathbb{P}^1 \times (C\setminus \{t\})$.
If one uses "geometric genus" instead of "arithmetic genus", the answer is positive.  The union of all irreducible curves in a projective variety having geometric genus $\leq g$ is a countable union of closed subvarieties of the projective variety.  Thus, if this union contains a Zariski dense subset (over an arbitrary field) or contains a nonempty open subset for the Hausdorff topology (over $\mathbb{C}$), then the union equals the entire projective variety.
