Arbitrary intersection in Zariski topology I tried to prove the closure under the arbitrary intersection of Zariski topology. I am aware that the intersection should become a union. Although I can't quite see what flaw I am making in the logical progression provided below. Can somebody point it out?
Given $\tau_{Zar} = \{V(S) \vert S = P_n\}$ such that $V(S) = \{x \in \mathbb{R}^n \vert (\forall f \in S)f(x) =0\}$
We take an arbitrary collection $\{{V(S_a)\} \in \tau_{Zar}}$. We will prove that $\bigcap_a V(S_a) =  V(\bigcup_a S_a)  $
\begin{align*}
    x \in \bigcap_a V(S_a) &\iff \forall a: x\in V(S_a) \iff (\forall a)( \forall f\in P_n)\left[f\in S_a \implies f(x)=0 \right]\\
    &\iff ( \forall f\in P_n)(\forall a)\left[f\in S_a \implies f(x)=0 \right]\\ 
    &\iff ( \forall f\in P_n)\left[f\in \bigcap_a S_a \implies f(x)=0 \right]\\
    &\iff x \in V\left(\cap_a S_a \right) \in \tau_{Zar}\text{ since } \cap_a S_a \in P_n
\end{align*}
 A: Well, if $P$ is some property,
$$\forall i \in I \ \bigl(x \in S_i \implies P\bigr)$$
is equivalent to
$$
x \in \bigcup_{i\in I} S_i \implies P
$$
and not to 
$$
x \in \bigcap_{i\in I} S_i \implies P
$$
A: Don't reduce to logical laws, but think about the definitions:
Suppose we have some collection $V(S_a)$, $a \in A$ of Zariski closed sets, so $S_a$ is some set of $n$-variable polynomials (so defined on $\Bbb R^n$) and $V(S_a)=\{x \mid \forall f \in S_a: f(x)=0\}$.
Now what does it mean for $x$ to be in $\bigcap_a V(S_a)$? It means that whatever $f$ I pick, from any $S_a$, I know that $f(x)=0$ (from $x \in V(S_a)$ for that $a$). So for any $f \in \bigcup_a S_a$, $f(x)=0$ (I don't need an $f$ that is in all $S_a$, just in one of them is enough), and so by definition $x \in V(\bigcup_a S_a)$.
And if $x \in V(\bigcup_a S_a)$, then pick any $a$ and any $f \in S_a$. Then $f \in \bigcup_a S_a$ for sure and so by definition $f(x)=0$ and $x \in V(S_a)$. As this holds for any $a$ I choose, $x \in \bigcap_a V(S_a)$. 
I did the other axioms in detail here but skipped over the above step with a simple "by definition" remark, which is, as you see, in fact true..
