Families of linear spaces in $\mathbb{P}^n$. Let $\mathbb{P}^n$ be the projective space (over some algebraically closed field). And let $k\le n+1$ be a positive integer.
Let $U\subset(\mathbb{P}^n)^k$ be the subset consisting of $k$-tuples $(p_1,...,p_k)$ such that the points $p_1,...,p_k$ are linearly independent. We denote by $\overline{p_1...p_k}$ the linear space generated by these points in $\mathbb{P}^n$.
In Joe Harris' Algebraic Geometry Book there is an exercise concerning these things (Exercise 4.16).
(1) Show that $U$ is open.
(2) Show that $\Omega:=\{(p_1,...,p_k;r)\in U\times\mathbb{P}^n\;|\;r\in\overline{p_1...p_k}\,\}$ is closed in $U\times\mathbb{P}^n$.
(3) What is the closure of $\Omega$ in $(\mathbb{P}^n)^k\times\mathbb{P}^n$?
I am struggling to understand (3).
For (1), we can write $U$ as
$$U=\{(p_1,...,p_k)\in(\mathbb{P}^n)^k\;|\;\text{at least one of the $k$-minors of the matrix $[p_1\; ...\;p_k]_{n\times k}$ is $\neq0$}\}$$
so, this amounts a collections of equations such that the coordinates of the points $(p_1,...,p_k)$ do not vanish (in at least one of them). Furthermore, since the determinant is a (multi)homogeneous-polynomial, these equations define a variety in $(\mathbb{P}^n)^k$, with $U$ being its complement. Thus $U$ is open.
For (2), let $\{L_{\alpha,p_1,...,p_k}(X)\}_{\alpha\in\Lambda}$ be the collection of linear equations defining $\overline{p_1...p_k}\subset\mathbb{P}^n$. We can also think of these as being equations over the coordinates of the $p_i$, $i.e.,$ $L_{\alpha,p_1,...,p_k}(X)=L_\alpha(p_1,...,p_k,X)$. We can then verify that these $L_\alpha(p_1,...,p_k,X)$ are bi-homogeneous (of degree 1 in X, and degree 1 over the part coming from the coordinates of $(p_1,...,p_k)$, $i.e.$, over the coordinates of $U$). But by definition, we have
$$\Omega=\{(p_1,...,p_k;r)\in U\times\mathbb{P}^n\;|\;L_\alpha(p_1,...,p_k,r)=0\;\;\forall\, \alpha\in\Lambda\,\}.$$
Thus $\Omega$ is closed in $U\times\mathbb{P}^n$.
For (3), I have no idea. Can someone help me with that?
 A: Here is an answer to my own question, following the hints provided by Mindlack in the comments.
Let $\overline{\Omega}$ be the closure of $\Omega$ in $(\mathbb{P}^n)^{k+1}$ and let $D_{k+1}$ be the set of all $(k+1)$-tuples in $(\mathbb{P}^n)^{k+1}$ that are linearly dependent. Note that $D_{k+1}$ is closed in $(\mathbb{P}^n)^{k+1}$. Indeed, it can be described as
$$D_{k+1}=\{(p_1,\ldots,p_k,r)\in(\mathbb{P}^n)^{k+1}\;|\;\text{ all ${k+1}$-minors of the matrix $[\,p_1\;\,\ldots\;\,p_k\;\,r\,]$ vanish}\},$$ but these ${k+1}$-minors amount to a collection of multi-homogeneous equations on the coordinates of the points, thus $D_{k+1}$ is a closed. We will show that $D_{k+1}=\overline{\Omega}$.
The inclusion $D_{k+1}\supset\overline{\Omega}$ is immediate from the definition of $\Omega$. To show the other inclusion, take $x=(p_1,\ldots,p_k,r)\in D_{k+1}\setminus\Omega$.
Claim: There exists a line $L\subset(\mathbb{P}^n)^{k+1}$ passing through $x$ such that $$L\cap\Omega=L-\{\text{finitely many points}\}.$$
(we will prove this claim later, let us assume it is true for now.)
Let $L':=L\cap\Omega$. Then we have $L'\subset\Omega$, and therefore $\overline{L'}\subset\overline{\Omega}$. But $\overline{L'}=L$, which contains $x$. Thus we have shown that $x\in\overline{\Omega}$ as we wanted.$$\tag*{$\Box$}$$
Proof of the claim:
(This claim was easy to visualize, but writing down a rigorous proof was harder than I expected. I bet there is a nicer way to prove it).
Since $x\notin\Omega$, we know that the points $\{p_1,\ldots,p_k\}$ are linear dependent (Indeed, if these points were independent, then, the fact that $x\in D_{k+1}$ would force a linear dependency with $r$, thus $r$ would be in the linear space spanned by the $p_i$'s, and so we would have $x\in\Omega$).
Therefore, there exist scalars $a_i\in K$ such that:
$$a_1 p_1 +\ldots+a_k p_k=0.$$
Take any set of points $\{v_1,\ldots,v_k\}\subset\mathbb{P}^n$ such that
(i) $\,a_1v_1+\ldots+a_k v_k=r\,$ and
(ii) $\{p_1+v_1,\ldots,p_k+v_k\}$ is a linearly independent set.
Define
$$L:=\{(p_1+tv_1,\ldots,p_k+tv_k,r)\;|\;t\in K\}.$$
Obviously $L$ passes through $x$ (at $t=0$). We have to verify that it has the desired property.
On the one hand, from (i) we get
$$a_1(p_1+tv_1)+\ldots+a_k(p_k+tv_k)=\sum_{i=1}^k a_ip_i+t\sum_{i=1}^k a_iv_i=0+ tr,$$
therefore, for every $t\neq0$ we have $r\in\overline{(p_1+tv_1)\ldots(p_k+tv_k)}$.
On the other hand, if $\{p_1+tv_1,\ldots,p_k+tv_k\}$ are linearly dependent, then all the $k$-minors of the matrix $[\,p_1+tv_1\;\,\ldots\;\,p_k+tv_k\,]$ have to vanish. Since the $p_i$ and $v_i$ are fixed, we can view these minors as polynomials in $t$. Therefore, they are $\equiv0$ or they have finitely many roots. But by (ii), we know that at $t=1$ at least one of these polynomials is not $0$. Therefore at least one of them is not $\equiv0$, and so must have finitely many roots. In other words, we have shown that
$$\{p_1+tv_1,\ldots,p_k+tv_k\}\text{ is linearly independent for all but finitely many $t\in K$}.$$
Putting this together with the fact that $r\in\overline{(p_1+tv_1)\ldots(p_k+tv_k)}$ for all $t\neq0$ yields
$$(p_1+tv_1,\ldots,p_k+tv_k,r)\in\Omega\text{ for all but finitely many $t\in K$}.$$
Thus $L$ has the desired property.
