While learning the projective n-space, $\mathbb{P}_{k}^{n}$ , I cast doubt on the following mention during the lecture
suppose $[(a_0,....a_n)] \in \mathbb{P}_{k}^{n}$. If $a_0 \neq 0$, then $[(a_0,....a_n)] \in \mathbb{P}_{k}^{n}$ is an open subset in $\mathbb{P}_{k}^{n}$
Observation : I can mimic the zero set under the projective $n$-space, $Z_{\mathbb{P}}(S)$, like the zero set under the affine $n$-space, $\mathbb{A}^n$, and the element of $Z_{\mathbb{P}}(S)~$ consists of...
$$Z_{\mathbb{P}}(S):= \left \{ [(a_0,....a_n)] \in \mathbb{P}_{k}^{n} : f([(a_0,....a_n)])=0, \forall f \in S \in \bar{k}[x_1,....,x_n] \right \}$$
And such set , like an affine n-space, satisfies the condition of topology axiom (as closed version). $Z_{\mathbb{P}}(S)$ is a closed set in $\mathbb{P}_{k}^{n}$ is equivalent to $\mathbb{P}_{k}^{n}- Z_{\mathbb{P}}(S)$ is open. So, I know how the open set is given under the projective $n$-space. Hence, I think that if $[(a_0,....a_n)] $ is open in $\mathbb{P}_{k}^{n}$ iff there exists some polynomial $f \in \bar{k}[x_1,....,x_n] $ such that $f([(a_0,....a_n)] )\neq0$.
Hence, my claim is the following,
Claim: Suppose $a_0 \neq 0$, then there exists some polynomial $f \in \bar{k}[x_1,....,x_n]$ such that $f([(a_0,....a_n)] )\neq 0$.
However, I have no idea how to prove Claim. I want to get a contradiction by denying the conclusion. But I cannot get any contradiction. The following is my trial If I negate the given conclusion ,
for any polynomial $f \in \bar{k}[x_1,....,x_n] $, $f([(a_0,....a_n)] )=0$. Next, I expected the definition of the projective n-space and modifying the polynomial. Since, $f([(a_0,....a_n)] )=0$, clearly, $f(a_0,....a_n)=0=a_0 f(1, a_1/a_0, ..., a_n/a_0)$. But that's all I have tried. First, since $a_0 \neq 0$,
$$f(1, a_1/a_0, ..., a_n/a_0)=0............ (★)$$
I wished to occur in contradiction from the result, (★), but (★) holds well clearly because $(1, a_1/a_0,....a_n/a_0) \sim (a_0,a_1,....a_n)$ i.e $(1, a_1/a_0,....a_n/a_0) \in [(a_0,....a_n)] $
Answer: - By definition: The zariski topology on $\mathbb{P}^n$ has the following closed sets: Let $I \subseteq k[x_0,..,x_n]$ be an ideal generated by a set of homogeneous polynomials. It follows $Z(I)$ is a closed set and all closed sets arise in this way. The set $D(x_i):=Z((x_i))^c$ is the complement of the closed set $Z((x_i))$ hence by definition it follows $D(x_i)$ is open for all $i=0,,.,n$.