This is a sequel to the question Accumulation of algebraic subvarieties: Near one subvariety there are many others (?) .

Let $Y$ be some projective variety, over $\mathbb{C}$. Let $X\subset Y$ be some closed subvariety, and let $\tilde{X}$ be some small open neighborhood of $X$, in the complex topology.

Naive Question: Are there many varieties $\subset \tilde{X}$ of dimension equal to the dimension of $X$?

In the previous question, this was answered positively if $Y=\mathbb{P}^n$, using crucially the fact that the tangent sheaf of $\mathbb{P}^n$ is ample, which allows to construct deformations of infinitesimal thickenings of $X$.

On other hand, the answer in the general case is obviously no, because one might take as $Y$ the blow-up of $\mathbb{P}^2$ at a point, and for $X$ the exceptional divisor. In that case, any $X^\prime\subset \tilde{X}$ has to project to the point in $\mathbb{P}^2$, for otherwise it meets any line in $\mathbb{P}^2$, which however may be outside of the image of $\tilde{X}$. But then $X^\prime$ is set-theoretically equal to $X$.

But in this case, the obstruction already exists if we take $\tilde{X}$ to be a small neighborhood in the *Zariski* topology.

Question: Assume that within any Zariski neighborhood of $X$, there are many projective subvarieties of the same dimension. Then is the same true for $\tilde{X}$?

Finally, note that the condition that within Zariski neighborhoods, there are many subvarieties of the same dimension is e.g. guaranteed by the following numerical criterion.

Lemma: Assume that $X$ meets every subvariety of complementary dimension. Then within any Zariski neighborhood of $X$, there are many projective subvarieties of the same dimension.

For the proof, just note that the complement of any Zariski neighborhood of $X$ has codimension smaller than $\mathrm{dim} X$, and hence by taking an intersection of generic hyperplane sections, you get subvarieties which do not meet the boundary. Also note that the condition of the Lemma is trivially verified if $Y=\mathbb{P}^n$.

However, note that one can not in general expect to have complete intersections inside of $\tilde{X}$, already if $Y=\mathbb{P}^n$ and $X$ is smooth. For the proof, note that if we take $\tilde{X}$ small enough, it will have the same (singular, say) cohomology as $X$; in particular, for any $X^\prime\subset \tilde{X}$, we get a map $H^i(X,\mathbb{Q})\rightarrow H^i(X^\prime,\mathbb{Q})$. It is compatible with cup-product, and injective in top-degree (compare with cohomology of $\mathbb{P}^n$), thus injective. But $H^i(X^\prime,\mathbb{Q})$ is zero below the middle dimension if $X^\prime$ is a complete intersection.

One might also ask the following weaker question (say, $Y$ is smooth in order to have a nice cup-product):

Question: Assume that the intersection pairing of $X$ with any subvariety of complementary dimension is positive. Are there many subvarieties in $\tilde{X}$ of the same dimension?

Counterexamples to these questions are most welcome, on the other hand I would be particularly interested in any positive result or method to construct such subvarieties, so feel free to make additional assumptions.

EDIT: Both questions above are shown to have negative answers by the nice example of Dmitri. Therefore, let me ask the following, pretty vague, question:

Question: Is it possible to formulate any positive result in this direction, possibly assuming that 1) the tangent or normal sheaf is in some sense positive or ample, 2) some positivity condition as already required above + 3) whatever you need.