Classical affine zero loci form the Zariski closed sets

For any indexed family $(S_{\lambda })$, a tuple vanishes on ${\bigcup }_{\lambda }{S}_{\lambda }$ exactly when it vanishes on every $S_{\lambda }$. Hence ${\bigcap }_{\lambda }V(S_{\lambda })=V({\bigcup }_{\lambda }{S}_{\lambda })$; the empty intersection is $k^n$.

If $a\in V(I)\cup V(J)$, every product of an element of $I$ with one of $J$ vanishes at $a$, hence every element of $IJ$ does. If $a$ belongs to neither locus, there are $f\in I,g\in J$ with $f(a),g(a)\mathrm{\ne }0$; then $(fg)(a)\mathrm{\ne }0$, so $a\mathrm{\notin }V(IJ)$. This proves both inclusions.

Replace arbitrary equation sets by generated ideals and iterate the two-set union identity. The zero-set identities for 0 and 1 supply the empty union and the full space. Intersecting these identities with $X$ verifies the induced closed-set axioms.