Show that if $X$ is an irreducible variety, and if $U,V\subset X$ are open not empty, then its intersection $U\cap V$ is not empty

Question

Show that if $X$ is an irreducible variety, and if $U,V\subset X$ are open not empty, then its intersection $U\cap V$ is not empty

I have reasoned in the following way: Reasoning by the absurd suppose that $U\cap V$ is empty, with which $X=U\cup V\cup (U^c-V)$ would be a decomposition of $X$,Which is absurd because P is irreducible. I am thinking well? Many thanks.

Answer 1

If $U\cap V$ is empty, $(U\cap V)^c=U^c\cup V^c=X$. Since $U^c,V^c$ are closed, this shows that $X$ is reducible. Contradiction.

Answer 2

from the book "Algebraic Geometry I, Ulrich Görtz, Torsten Wedhorn" page 13.

Theorem: Let X be a non-empty topological space. The following assertions are equivalent.

(i) X is irreducible.

(ii) Any two non-empty open subsets of X have a non-empty intersection.

(iii) Every non-empty open subset is dense in X.

(iv) Every non-empty open subset is connected.

(v) Every non-empty open subset is irreducible.