I have a question about the proof of Bertini' Theorem found in Harris' book Algebraic Geometry, on page 216/217:
Theorem 17.16. Bertini's Theorem. If $X$ is any quasi-projective variety over $\mathbb{C}$, $f: X \to \mathbb{P}^n$ a regular map, $H \subset \mathbb{P}^n$ a general hyperplane, and $Y = f^{-1}(H)$, then
$$Y_{\text{sing}}= X_{\text{sing}} \cap Y. $$
where $X_{\text{sing}}=X- X_{\text{sm}}$ is the singular locus.
The essence of the proof is to analyze the incidence correspondence
$$ \Gamma = \{(p,H) \ \vert f(p) \in H \} \subset X \times (\mathbb{P}^n)^* $$
If $X$ is $k$-dimensional, $\Gamma$ will have dimension $k + n - 1$. Now, let $p \in X$ be any smooth point and $H \subset \mathbb{P}^n$ a hyperplane containing $f(p)$; then it's easy to see that
$$ \dim(T_{(p,H)}(\Gamma))= \dim(T_{(p,H)}(X \times (\mathbb{P}^n)^*))-1 $$
in particular, the singular locus of $\Gamma$ is exactly the inverse image $\pi_1(X_{\text{sing}}$ of the singular locus of $X$ under $\pi_1: X \times (\mathbb{P}^n)^* \to X$.
Now look at the restricted map $\tilde{\pi}_2 : \Gamma_{\text{sm}} \to (\mathbb{P}^n)^*$. Bertini's theorem follows from applying Proposition 14.4 to $\tilde{\pi}_2$ ; or, more directly, by Exercise 14.6 the locus $U \subset (\mathbb{P}^n)^*$ of hyperplanes $H$ such that the fiber $\tilde{\pi}^{-1}_2(H) = X_{\text{sm}} \cap H$ is smooth is either contained in a proper subvariety of $(\mathbb{P}^n)^*$ or contains an open subset of $(\mathbb{P}^n)^*$, and we can simply apply Sard's theorem to $\tilde{\pi}_2$ deduce that the former must be the case.
Question: I would like to understand better the last part of the proof. Harris offers in the last sentence alongside to the crystal clear argument applying Proposition 14.4 another "more directly" argument by means of Exercise 14.6 which I not understand. This exercise asserts:
Exercise 14.6. Let $f: X \to Y$ be any regular map of affine varieties. Then the dimension of the kernel of the differential $df_p: T_pX \to T_{f(p)}Y$ is an upper-semicontinuous function of $p$.
In other words
$$X_{\ge k} :=\{p \in X \ \vert \ \dim \text{ Ker} (df_p)) \ge k\} \subset X$$ is closed for every $k \ge 0$. This implies also that every subset $X_k:= \{p \in X \ \vert \ \dim \text{ Ker} (df_p) = k\}$ is locally closed or constructible and therefore $f(X_k)$ is constructible too, since images of constructible sets are constructible.
Note that this statement does not imply that the locus of smooth points of fibers of $f$ - that is, the locus of $p \in X$ such that the fiber $X_{f(p)} = f^{-1} (f(p))$ is smooth at $p$ - is open or constructible in $X$, or that the dimension of $T_p(X_{f(p)})$ is an upper-semicontinuous function of $p$, since the Zariski tangent space to $X_{f(p)}$ may be smaller than $\text{Ker}(df_p)$.
And that's exactly the central problem. We know that the subsets $(\Gamma_{\text{sm}})_k$ and therefore their images under $\tilde{\pi}_2$ are locally closed / constructible. This implies that these have this mentioned property that they are either contained in a proper subvariety of $(\mathbb{P}^n)^*$ or contain an open subset.
But we know by means of this exercise 14.6 not enough about the set $U \subset (\mathbb{P}^n)^*$ of those hyperplanes $H$ whose fibers $\tilde{\pi}^{-1}_2(H) \cong X_{\text{sm}} \cap H$ are smooth, ie those $H$ such that for every $(p, H) \in \Gamma_{\text{sm}}$ (equivalently $p \in H \cap X_{\text{sm}}$) the dimension of the tangent space of $\tilde{\pi}^{-1}_2(H)$ at $(p,H)$ equals $\dim H \cap X_{\text{sm}} =k-1$.
We only know that on the domain side there is an inclusion
$$ \{(p,H) \in \Gamma_{\text{sm}} \ \vert \ \dim \text{Ker} (T_{(p,H)} \tilde{\pi}_2)= k-1\} \subset \{(p,H) \ \vert \ \tilde{\pi}^{-1}_2(H) \text{ smooth at } (p,H) \}$$
because obviously $T_{(p,H)}\tilde{\pi}^{-1}_2(H) \subset \text{Ker} (T_{(p,H)} \tilde{\pi}_2)$. Therefore we conclude that the set on the right hand side is dense in $\Gamma_{\text{sm}}$, because $(\Gamma_{\text{sm}})_{k-1}$ is open in $\Gamma_{\text{sm}}$. But clearly it could happen that $\tilde{\pi}_2((\Gamma_{\text{sm}})_{\ge k})= (\mathbb{P}^n)^*$, and so we would lose any control at the codomain side. Therefore I not see why $U$ should have the claimed property to be either contained in a proper subvariety of $(\mathbb{P}^n)^*$ or to contain an open subset of $(\mathbb{P}^n)^*$.