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Consider a $p$-adic field $K$ with the standard topology inherited from the usual $p$-adic norm $\mid \cdot \mid$. Consider the ultrametric space $X=K^n$ with the topology inherited from the norm $\| \cdot \|$ defined as $\|x\|=\max_{i=1}^n (|x_1|,\dots,|x_n|)$ with $x=(x_1,\dots,x_n) \in K^n$. Now we have two questions:

  1. Is the topology on $X$ the same as the product topology of the $K$'s?

  2. Can we partition every open compact set in $X$ into a finite number of balls (defined using $\|\cdot \|$)?

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up vote 1 down vote accepted

The answers are yes and yes.

  1. is true because the sets $U_{\epsilon} = \{(x_1,\dots,x_n),\ |x_i| \leq \epsilon\}$ for a basis of neighborhood of $0$ in $X$ for your norm by definition, but also a basis of neighborhood of $X$ for the product topology since any neighborhood of $0$ in the product topology contains a product of neighborhoods of $0$ which each contains a ball $|x_i| \leq \epsilon_i$ and it suffices to take $\epsilon=\min(\epsilon_i)$.

  2. is true because every compact open metric space is obviously covered by finitely many open balls, and because in an ultrametric space, every two open balls either are disjoint, or are such that one of them contains the other. So remove from your finite cover all the balls that are contained in a larger ball, and you get a partition of your compact open space.

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Thank you Joel. I think your answer will help me to settle down another question on $p$-adic semilagebraic isomorphisms. – user16974 Oct 27 '11 at 15:13

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