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



On the constructability of a particular set.
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.


