Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three vriables, where $\delta_d = \frac{d(d+3)}{2}$. Let $$ X \subset \mathcal{D} \times \mathbb{P}^2$$
be a smooth embedded complex submanifold, not necessarily closed. Given a point $p\in \mathbb{P}^2$, we get a hyperplane $$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$ Note that a point $p$ first of all gives a hyperplane $H_p$ in $\mathcal{D}$ (which is the space of degree $d$ polynomials passing through the point $p$). This gives us a hyperplane $$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Is it true that ifLet us further define $\tilde{H}_p$ is not transverse$H_p^* \subset H_p$ to $X$, then for every open neighborhood $U$ of $p$ in $\mathbb{P}^2$, there exists abe the pointspace of degree $q\in U$$d$ curves such that the hyperplane $\tilde{H}_q$ does intersect $X$ transversally? This seems to intuitively say that if something$p$ is not transversea smooth point of the then you can ``perturb'' it so that it is transversecurve. Similarly define More precisely, I want to claim$$ \tilde{H}_p^* := H_p^*\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$
Is it true that the setfor almost all choices of points $q$ where $\tilde{H}_q$ intersects $X$ transversally is an open$p \in \mathbb{P}^2$, dense subset of $\mathbb{P}^2$. It is certainly open. Please note that apriori it$\tilde{H}^*_p$ is possible that there does not exist any point $q$ such thattransverse to $\tilde{H}_q$ intersects $X$ transversally. This is the part I don't see how to prove, although intuitively it seems obvious.?