|f(\mathbb{Z}p)| has at most 0 as a limit point in

Painevg

Painevg

Answered question

2022-01-13

|f(Zp)| has at most 0 as a limit point in the non-zero reals

Answer & Explanation

Debbie Moore

Debbie Moore

Beginner2022-01-14Added 43 answers

Step 1
The key concept is this: the non-zero accumulation points are deceptive, in the sense that if xnx in Cp, then either |xn| is eventually constant, or |xn|0; in other words, if |xn| tends to say 1, and |xn| is strictly increasing, then there is no chance that xn actually converges.
With this in mind, it is now a trivial application of the ''Bolzano–Weierstrass theorem'':
Assume yR is a non-zero accumulation point, so there are xnZp such that |f(xn)|y and |f(xn)| are all distinct. By ''Bolzano–Weierstrass theorem'', xn has a convergent subsequence, so we might WLOG assume that xnx; since f(xn)f(x) and |f(xn)| are all distinct, we must have |f(xn)|=y=0, contradiction.

autormtak0w

autormtak0w

Beginner2022-01-15Added 31 answers

Hint 1:
First, let SCp be any subset and assume some pr is an accumulation point of
|S|ppQ{0}.
That means there is a sequence of mutually distinct rational numbers rn such that
limnrn=r
and corresponding elements snS with |sn|p=prn.
Hint 2:
Show that a sequence sn as constructed above does not have any convergent subsequence. (Use that in Cp,limncn=c0 implies that |cn|p=|c|p for high enough n.
Apply to S=f(Zp).

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