A condition for irreducibility Let  $X$ be  a  closed   projective  set. Prove that $X$   is   an  irreducible  set if   and  only  if $X \cap U_i$ is irreducible  for every i=0,...,n; where $\cup U_i$ is an open cover of $\mathbb{P^n}$. For "$\Rightarrow $" I succeded.

It is a topological exercise!

Let $X$ be a topological space, let $\{U_i\}_{i\in I}$ be an open covering of $X$; a subset $Y$ of $X$ is irreducible only if $\forall i\in I,\,Y\cap U_i=Y_i$ is irreducible. Vice versa, if $Y_i$ is irreducible and $\forall i,j\in I,\,Y\cap U_i\cap U_j=Y_{ij}\neq\emptyset$ then $Y$ is irreducible.

Proof. Let $Y\subseteqq X$; if $Y$ is reducible, let $Z$ be an irreducible component of $Y$, then
\begin{equation*}
\exists h\in I\mid Y_h=Z\cap U_h\neq\emptyset
\end{equation*}
because $Y_h$ and $Z\cap U_h$ are irreducible. By hypothesis $\forall k\in I,\,Z\cap U_h\cap U_k=Z_{hk}$ is a non-empty open subset of $Z$ and so $Z_{hk}$ is dense in $Z$. By the same reasoning, $Z_{hk}$ is a non-empty, open and dense subset of $Y_k$; then
\begin{equation*}
\forall k\in I,\,Z_{hk}\subseteqq Y_k\Rightarrow Y_k\subseteqq Z\Rightarrow Z=Y.
\end{equation*}
The other implication is tautological! Q.E.D. $\Box$ 
