 By Andrew Baker

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Extra info for An Introduction to p-adic Numbers and p-adic Analysis [Lecture notes]

Sample text

Let β ∈ D (α; δ). Then D (β; δ) = D (α; δ) . Hence every element of D (α; δ) is a centre. Similarly, if β ′ ∈ D (α; δ), then D (β ′ ; δ) = D (α; δ). Proof. This is a consequence of the fact that the p-adic norm is non-Archimedean. Let γ ∈ D (α; δ); then |γ − β|p = |(γ − α) + (α − β)|p max{|γ − α|p , |α − β|p } < δ. Thus D (α; δ) ⊆ D (β; δ). Similarly we can show that D (β; δ) ⊆ D (α; δ) and therefore these two sets are equal. A similar argument deals with the case of closed discs. Let X ⊆ Qp (for example, X = Zp ).

Similarly, DX (α; δ) = D (α; δ) ∩ X is the closed ball in X of radius δ centred at α. We will now deﬁne a continuous function. Let f : X −→ Qp be a function. 4. We say that f is continuous at α ∈ X if ∀ε > 0∃δ > 0 such that γ ∈ DX (α; δ) =⇒ f (γ) ∈ D (f (α); ε) . If f is continuous at every point in X then we say that it is continuous on X. 5. Let f (x) = γ0 + γ1 x + · · · + γd xd with γk ∈ Qp be a polynomial function. Then as in real analysis, this function is continuous at every point. To see this, we can either use the old proof with | |p in place of | |, or the following p-adic version.

K=1 Hence, for one of the numbers 1 inﬁnitely many values of n. Then k p, say a1 , the disc D (a1 ; 1) has αn ∈ D (a1 ; 1) for 2 D (a1 ; 1) = p ∪ D (k; 1/p) k=1 and again for one of the numbers 1 k p2 , say a2 , we have αn ∈ D (a2 ; 1/p) for inﬁnitely many values of n. Continuing in this way we have a sequence of natural numbers an for which ( ) D an ; 1/pn−1 contains αm for inﬁnitely many values of m. Moreover, for each n, ( ) D an ; 1/pn−1 ⊆ D (an ; 1/pn ) . ( ) Now for each n 1, choose s(n) so that αs(n) ∈ D an ; 1/pn−1 .