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

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Debbie Moore · Accepted Answer

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

autormtak0w · Accepted Answer

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

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

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