Let $d>1, n>1$ and consider the vector space $V=\mathbb{C}[x\_0,\ldots,x\_n]\_d$. Let $\mathscr A \subseteq \mathbb{P}(V)$ be the set of all forms with the property that their zero locus in $\mathbb{P}^n$ has at least $k$ singularities. Let $p \in \mathbb{P}(V)$ be square free such that its zero locus has less than $k$ singularities and all of them are nodes. Is it true that in that case $p$ is not in the Zariski closure of $\mathscr A$?



Zariski closure of hypersurfaces with $k$ singularities
Yes, this is true.
This answer is essentially an expanded version of Jason Starr's comment, also including a reference.


The result that we need is the following, that can be easily deduced from [G. M. Greuel, C. Lossen, E. Schustin, *Introduction to singularities and deformations*, Theorem 2.6 Chapter I].



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> **Proposition.** Let $f \colon \mathcal{X} \to \Delta$ be a deformation of the affine hypersurface $X=f^{-1}(0)$ over a disk. Assume that $0$ is the only singular point for $X$ and that $X\_t$ has only isolated singularities. Then the total Milnor number of the fibres is upper semicontinuous, i.e. for all $t \in \Delta$ we have $$\mu(X, \, 0) \geq \sum\_{x \in \textrm{Sing} \, X\_t} \mu(X\_t, \, x).$$ In other words, the total Milnor number *cannot decrease under specialization.*
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In your case, the total Milnor number of every hypersurface in $\mathscr A$ is at least $k$, whereas the total Milnor number of the hypersurface $\{p=0\}$ is the number of its nodes, hence strictly less than $k$. 


Therefore, $\{p=0\}$ is not a specialization of hypersurfaces belonging to $\mathscr A$.


