Can I choose $k+1$ hypersurfaces to avoid a fiber of dimension $k$ in projective space? Let $X$ be a closed subscheme of dimension $k$ in $\mathbb{P}^n_A$, where $A$ is a Noetherian ring. In Exercise 11.3.C of Ravi Vakil's notes, it is shown that one may choose $k+1$ hypersurfaces such that the intersection of these hypersurfaces avoids $X$. This uses Krull's Principal Ideal Theorem, which is why $A$ must be Noetherian. 
Let $\pi: X\rightarrow \text{Spec}(A)$ be the structure morphism. 
My question is this: If $p$ is a point of $\text{Spec}(A)$, and $\pi^{-1}(p) \subset X$ is the fiber of $p$ in $X$ of dimension $r$, can I still find $r+1$ hypersurfaces whose intersection avoids the fiber? If $p$ is closed this would be immediate as then the fiber is closed, but in general this will not be true. 
Note that my question is inspired from this Upper semicontinuity of fibre dimension on the target, where showing one can avoid a fiber of a given dimension is crucial to showing upper semicontinuity of fiber dimension.
 A: This is more or less a copy of part of my answer here, which I copied over rather than just linked to since that answer contains a lot of material irrelevant to this question. I also don't think that either question is a duplicate of the other, this question is interesting even without applying it to the linked one.
Since $\pi: X \rightarrow \operatorname{Spec}(A)$ factors through $\mathbb{P}_A^n$, the map $\pi^{-1}(p)\rightarrow \operatorname{Spec(\kappa(p))}$ factors through $\mathbb{P}_{\kappa(p)}^n$, and since $X \rightarrow \mathbb{P}_A^n$ is a closed embedding, so is $\pi^{-1}(p)\rightarrow \mathbb{P}_{\kappa(p)}^n$:
$$
\require{AMScd} \begin{CD}
    \pi^{-1}(p) @>>>\mathbb{P}_{\kappa(p)}^n@>>>\operatorname{Spec}(\kappa(p))
    \\ @VVV @VVV @VVV\\
    X @>>> \mathbb{P}_A^n @>>> \operatorname{Spec}(A)
    \end{CD}
$$
But now $\pi^{-1}(p)$ is a closed subscheme of $\mathbb{P}_{\kappa(p)}^n$, of dimension $\leq r$, and so we may find $r+1$ hypersurfaces $H'_i$ in $\mathbb{P}_{\kappa(p)}^n$ such that $\pi^{-1}(p)\cap_i H'_i = \phi$. We can now take hyperpsurfaces $H_i \subset \mathbb{P}_A^n$ that have pre-image $H'_i$ in $\mathbb{P}_{\kappa(p)}^n$ (We can choice some lift of the defining equation of the hyperplanes as polynomials in $\kappa(p)$ to polynomials in $A$). Then it must be the case that $\pi^{-1}(p)\cap_i H_i = \phi$.
