Fix a $p$-adic field $K$ with perfect residue field $k.$ Let $\mathcal{C}$ be the completion of the algebraic closure of $K,$ and let $$R = \varprojlim \mathcal{C}/p,$$ where the transition maps in the inverse limit are given by the Frobenius. One can then show that $R$ is a perfect valuation ring (of characteristic $p.$) We can then form $W(R),$ the ring of Witt vectors of $R.$

There is a natural topology on $W(R),$ making it into a topological ring, a basis of neighborhoods is given by $p^N W(R)+ W(I)$ for $N\geq 0$ and $I$ a non-zero ideal of $R.$

In this article, pg. 536, Fontaine uses the topology I just defined to define a topology on $W(R)[1/p] = K \otimes_{W(k)} W(R),$ by using what he calls the "tensor product topology". He then claims that the topology one get from the tensor product topology is the "same" (up to identifications under isomorphisms) as the one obtained by taking the topology coming from the inductive limit $$ \cdots \rightarrow W(R) \rightarrow W(R) \rightarrow \cdots$$ where transfer maps are multiplication by $p.$

My questions are the following:

1) What is precisely this tensor product topology? Naively, I would say that it is the topology on $K \otimes_{W(k)} W(R) \cong W(R)[1/p]$ where a basis of neighborhoods are given by $$(p^N W(R)+ W(I)) \otimes_{W(k)} K + W(R) \otimes_{W(k)} p^n \mathcal{O}_K$$ (where $\mathcal {O}_K$ is the valuation ring of $K).$ What makes me think this can not be a basis of neighborhoods comes from the fact that it seems to me (maybe erroneously) that $p^NW(R) \otimes_{W(k)} K \cong W(R)[1/p].$ Thus, it seems to me that Fontaine must have some other sort of topology in mind for this tensor product, or I am making a silly mistake. For example, what is a basis of neighborhoods for the tensor product topology? Is it part of a more general construction?

2. Why does the topology of the tensor product and inductive limit coincide?