Showing the join of two disjoint projective varieties is a projective variety. I'm looking at the following proposition in Harris' Algebraic Geometry: a first course:

Proposition: Let $X,Y\subset\mathbb P^n$ be disjoint projective varieties. Then the join $J(X,Y)$ of $X$ and $Y$, i.e. the union of all lines in $\Bbb P^n$ which intersect $X$ and $Y$, is a projective variety.

What I'm having trouble with is determining where the disjoint assumption is used. The proof given is as follows (modulo some rewording):

Proof: By [previous example], the locus $\mathscr C_1(X)$ of lines in $\Bbb P^n$ that intersect $X$ is a closed subvariety of the Grassmannian $\Bbb G(1,n)$, and similarly $\mathscr C_1(Y)$ is a closed subvariety of $\Bbb G(1,n)$, so their intersection $Z:=\mathscr C_1(X)\cap\mathscr C_1(Y)$ is closed in $\Bbb G(1,n)$. But then by [previous proposition], the union in $\Bbb P^n$ of all elements of $Z$ is a closed subvariety of $\Bbb P^n$, and this is exactly equal to $J(X,Y)$.

Thanks in advance to anybody who can point out whatever it is I'm missing.
 A: From my reading, it appears that the assumption of disjointness is used in defining the "join" construction, not the proof of showing that it is projective. Indeed, if $X\cap Y$ is nonempty, then the union of all lines in $\Bbb P^n$ which intersect $X$ and $Y$ is exactly $\Bbb P^n$, which is a boring construction.
A: Per the discussion, the disjoint assumption is usually made to avoid the "boring" case allowed by the usual definition. However, we can, instead, modify the definition slightly.
Let us form the locus of triples $(\ell, P, Q)$ where $\ell$ is a line and $P$ and $Q$ are points on $\ell$ with $P$ in $X$ and $Q$ in $Y$. Let us now define the join of $X$ and $Y$ as the (Zariski) closure of the union in $\mathbb{P}^n$ of all these $\ell$.
For example, one important special case is the secant variety of $X$. By definition, it is the case with $Y := X$.
The above construction can be formalized as follows. Consider the incidence correspondence $\ell$ in $\text{G}(1, n) \times \mathbb{P}^n$ parameterizing the pairs $(\ell, P)$ with $P$ on $\ell$. Form the triple fiber product $F \times F \times F$ over $\text{G}(1, n)$. Denote by$$p_i: F \times F \times F \to \mathbb{P}^n$$the $i$th projection, and by $D$ the diagonal subvariety parameterizing the triples $(P, Q, R)$ with $P = Q$. Form the closure $C$ in $F \times F \times F$ of $p_1^{-1}(X) \cap p_2^{1}(Y) - D$. Finally, set$$J(X, Y) :=p_3(C).$$Update: Let's try a second explanation. The observation in the discussion is correct. If $X$ and $Y$ are disjoint there is a morphism from $X \times Y$ to $\text{G}(1, n)$ sending $(x, y)$ to the line defined by these $2$ points. As $X \times Y$ is projective the image is closed. Using the incidence correspondence$$\{(x, \ell) : x \in \ell\} \in \mathbb{P}^n \times \text{G}(1, n)$$we see that the union $J(X, Y)$ of the lines in the image is also closed.
If $X$ and $Y$ intersect, when $X = Y$ it is slightly more complicated. The morphism becomes a rational mapping defined outside the diagonal $= \{(x, x) : x \in X \cap Y\}$. The image is then typically not closed in $\text{G}(1, n)$. But, we can take the closure. This includes limiting lines. Think limit line as $x$ and $y$ approach each other.
If $X = Y$, we get the proper secant lines meeting $X$ in $2$ or more points, and for a nonsingular point the tangent lines. If the point is singular it is slightly more complicated, but we get the lines in the tangent star.
Another simple case is when $Y$ is a point. Then $J(X, Y)$ is a cone. If $Y$ is not in $X$ this is easy to visualize. If $Y$ is a point on $X$, we get the secants through this point and their limiting lines. Again the lines in the tangent star, I suppose.
