Given a topology on a field $K$ (in my case a non-archimedean local field with valuation $\nu$ and ring of integers $R$ with maximal Ideal $\mathfrak{m}$), the goal is to define a topology on $\mathbb{P}^n(K)$ that comes from $K$. I know that one way to do this is to look at the map
\begin{align}
K^{n+1} \setminus \lbrace 0 \rbrace &\longrightarrow \mathbb{P}^n(K) \\
(x_0, \dots, x_n) &\longmapsto [x_0, \dots,x_n]
\end{align}
which is surjective, so we can define the topology on the target to be the quotient topology from $K^{n+1}$. (The topology there being understood as product topology).
My question is the following: Since we know that $\mathbb{P}^n(K)$ admits an open cover by $n+1$ copies of $K^n$, is there a way to define a topology via this covering? Would this give us the same topology as the quotient?
Another question is, when we want to check the continuity of the specialization map
\begin{align}
\text{Sp }\colon \mathbb{P}^n(K) &\longrightarrow \mathbb{P}^n(k) \\
[x_0, \dots x_n] &\longmapsto [\lambda x_0 \text{ mod } \mathfrak{m}, \dots, \lambda x_0 \text{ mod } \mathfrak{m}]
\end{align}
where $k=R/\mathfrak{m}$ and $\lambda$ scales the coordinates so that all are in $R$ (but not all in $\mathfrak{m}$), with $\mathbb{P}^n(k)$ having discrete topology and $\mathbb{P}^n(K)$ the avove, can we check this using an open cover of $\mathbb{P}^n(K)$ by copies of $R^n$? In that case, the continuity of reduction mod $\mathfrak{m}$ is obvious via the metric on each space. For clarification: $\mathbb{P}^n(K)$ does admit an open cover by $n+1$ copies of $\mathbb{A}^n(R)=R^n$, because coordinates in $\mathbb{P}^n(K)$ can be scaled so that all of them are in the ring of integers R. This works for any local field, but thinking specifically of $K=\mathbb{Q}_p$, $R=\mathbb{Z}_p$ or even $K=\mathbb{Q}$ and $R=\mathbb{Z}$ should help intuition.
Best Answer
Sure: you can define a topology on $\mathbb{P}^n(K)$ by saying a subset is open iff its intersection with each of the copies of $K^n$ is open in the product topology of $K^n$. Or equivalently, we can take as a basis for our topology all the open subsets of any of the copies of $K^n$ that cover $\mathbb{P}^n(K)$.
This topology is in fact the same as the quotient topology. To prove this, let $U_i\subset K^{n+1}\setminus\{0\}$ be the set of points whose $i$th coordinate is nonzero. Each $U_i$ is open and saturated with respect to the quotient map $p:K^{n+1}\setminus\{0\}\to\mathbb{P}^n(K)$, so each $p(U_i)$ is open with respect to the quotient topology, and $p$ restricts to a quotient map $U_i\to p(U_i)$. So, all we need to show is that the quotient topology on each $p(U_i)$ the the same as the product topology on $K^n$ when we identify them in the usual way.
For convenience of notation, we show this in the case $i=0$. To show the quotient topology on $p(U_0)$ is the same as product topology on $K^n$, we need to show the map $q:U_0\to K^n$ given by $q(x_0,\dots,x_n)=(x_1/x_0,\dots,x_n/x_0)$ is a quotient map. Now note that the map $f:U_0\to U_0$ defined by $f(x_0,\dots,x_n)=(x_0,x_1x_0,\dots,x_nx_0)$ is a homeomorphism, and $qf$ is just the projection that drops the first coordinate. This projection is a quotient map, and thus so is $q$.