On the constructability of a particular set. Let $X\subset\mathbb{P}^N_{\mathbb{C}}$ be a projective irreducible variety and $p\in X$. I put 
$$
A= \{ x\in X:\langle x,p\rangle\subseteq X \},
$$
 where I denotes with $\langle x,p\rangle$ the line through $x$ and $p$. Is it true that $A$ is constructible?
Thank you
 A: Probably logicians like me should stay away from algebraic geometry, but isn't "constructible" (over algebraically closed fields) the same as "first-order definable"?  In that case, the answer is yes, because the question itself gives a first-order definition.  To make it look more like geometry and less like logic, you could argue as follows, using the fact that constructibility (if I understand it correctly) is preserved by complementation, intersection, and projection.  Since $X$ and therefore its complement are constructible, so is the set $B$ of pairs $\langle x,q\rangle$ where $q$ is on the line through $x$ and $p$ but is not in $X$.  Then the projection of $B$ along the $q$ coordinate is also constructible, and your $A$ is the intersection of $X$ with the complement of that projection.
A: Yes. Moreover, $A$ is closed. Indeed, assume that $X$ is given by equations $f_1,\dots,f_n$. First, consider the set $B$ of lines through $p$ contained in $X$. Note that the set of all lines through $p$ is $P^{N-1}$ and $B$ is a closed subset of $P^{N-1}$. Indeed, assume for example that $p = (1,0,\dots,0)$. Then a point $(y_1,\dots,y_N)$ is in $B$ if for any $t$ the point $(1,ty_1,\dots,ty_N)$ is in $X$. But
$$
f_i(1,ty_1,\dots,ty_N)
$$
is a polynomial in $t$ with coefficients being polynomials in $y_1,\dots,y_N$. The condition that for all $t$ this is zero is equivalent to all coefficients of this polynomial in $t$ being zero, that is a finite number of polynomial equations in $y_1,\dots,y_N$. 
Thus $B$ is closed. Finally, $A$ is the cone over $B$ (with vertex in $p$) hence is also closed.
By the way, note that irreducibility of $X$ is not required.
