I have extended the Bertini Theorem (Hartshorne II.8.18) for singular projective varieties to show this, this or that.

(I think they use Bertini for a singular variety, but Hartshorne shows that only for a not singular variety.)

Please check my proof.

To let $ k $ be an arbitrary algebraically closed field and $ X subseteq mathbb {P} ^ N $ a closed normal subvariety of dimension $ d $, (possibly singular)

(To use liner systems, I think we need $ X $ normal.)

Then we will show it $ mathfrak {d} = {H subseteq mathbb {P}: text {a hyperplane with} X not subseteq H text {and} X cap H text {is regular} } $ is open in $ | H | $, a complete linear system of a hyperplane.

Almost all parts are identical to those of Harthorne.

For a closed point $ x in X $, To let $ B_x = {H | X subseteq H text {or} (X not subseteq H text {and} x in X cap H text {and} X cap H text {is unique to} x) } $,

By the following lemma, if $ x $ is a singular point, $ B_x = {x in H } $,

Let A be a singular noetheric local domain of dimension $ n $. $ f in A $,

Then $ A / f $ is also unique.

(The proof is very simple, assuming that $ dim A / f = n -1 $ in this case )

So by the proof of Bertini in Hartshorne, $ dim B_x = N – d – 1 $ if $ x $ is a not singular point and is $ = N – 1 $ if $ x $ is a singular point.

Finally, look at the subset $ B = {(x, H) in X times | H | : H in B_x } $,

This is closed $ X times | H | $, (I can not show that …)

So, consider it as a closed subschema.

Then the first projection $ h: B to X $ is surjective.

So after the general proposition about morphisms and fibers for every closed point $ b in B $ and his picture $ x = h (b) $. $ dim mathscr {O} _ {h ^ {- 1} (x), b} ge dim mathscr {O} _ {B, b} – dim mathscr {O} _ {X, x } $,

Now for $ X $ is a strain over a field, $ dim mathscr {O} _ {X, x} = d $, and since $ dim mathscr {O} _ {h ^ {- 1} (x), b} le dim h ^ {- 1} (x) $, it is $ le N -d – 1 $,

That's why $ dim B le N – 1 $,

So under the second projection $ p: B to | H | $the picture is a correct subset.

Since $ p $ is a projective morphism, this picture is closed.

Because this picture is accurate $ mathfrak {d} $we have made the desired set.

That's the way it is $ B $ closed?

Any help is greatly appreciated !!