Why a finite set is bounded in a topological ring?

Question

I'm almost sure that this question is really idiot and I'm missing something very trivial. In page 36 of Wedhorn's "Adic Spaces" (https://www2.math.uni-paderborn.de/fileadmin/Mathematik/People/wedhorn/Lehre/AdicSpaces.pdf), he claims that every finite set $S \subset A$ is bounded.

Let $A$ be a topological ring (in my case, I just need $A$ to be an $f$-adic ring). Then a set $S \subset A$ is bounded if, for every open neighborhood $U$ of $0$, there exists an open neighborhood $V$ of zero such that $VS \subset U$.

If $S$ is finite and $A$ is non-archimedean, then it suffices to prove that a singleton $\{x \}$ is bounded. However even in this case it's not clear why given $U$, an open neighborhood of $0$, there exists some $V$, again a neighborhood of $0$, such that $xV \subset U$.

Thanks in advance.

Answer 1

Let $x$ be an arbitrary element of a topological ring $A$ and $U$ be an arbitrary neighborhood of $0$. Then $x0=0\in U$. Since the multiplication on a topological ring is continuous, there exists a neighborhood $V$ of $0$ such that $xV\subset U$.

Answer 2

Multiplication by a fixed element is continuous, so given $U$ we can find for every $x_i\in S$ an open neighborhood $V_i$ of $0$ such that $V_ix_i\subset U$. Now $V:=\bigcap_i V_i$ is open (as $S$ is finite) and satisfies $VS\subset U$.

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