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Theorem 24.1.15 says the following: "Let $M_1$ and $M_2$ be metric spaces, $S$ a subset of $M_1$, and $f : S \to M_2$. If $f$ is uniformly continuous and $M_2$ is complete, then there exists a unique continuous extension of $f$ to $\overline S$. Furthermore, this extension is uniformly continuous".

Let's concentrate on the special case where $M_1=M_2=:M$, and $S$ is dense in $M_1=M$, so we get: Let $M$ be a metric space, $S$ a dense subset of $M$, and $f : S \to M$. If $f$ is uniformly continuous and $M$ is complete, then there exists a unique continuous extension of $f$ to $M$. Furthermore, this extension is uniformly continuous.

My question: If we know that actually $f:S \to S$ and $f$ is an automorphism of finite order $d$, and if we also know that the unique continuous extension $F: M \to M$ is an automorphism, is it true that $F$ must be of order $d$?

Remarks: (1) See also this question. (2) Perhaps this question can serve as a counter-example.

user237522
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2 Answers2

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Yes, this is true. For $x \in M$, choose a sequence $x_i \in S$ such that $x_i \to x$ as $i \to \infty$. We have $f^d(x_i)=x_i$. Since $f^d$ is continuous, $$x = \lim_{i \to \infty} x_i = \lim_{i \to \infty} f^d(x_i) = f^d(\lim_{i \to \infty} x_i) = f^d(x) $$

Lee Mosher
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  • Thank you very much Eric Wofsey and Lee Mosher! (truly, I knew the argument with sequential continuity, but thought that it may happen that I am missing something). Actually, I wonder how much one can weaken the conditions on $M$ in order to get the same result. For example, I guess that it is enough for $M$ to be any topological space (am I right?). Please what is the most general form of the above quoted theorem? (where can I find it?). Thanks. – user237522 Apr 14 '17 at 02:34
  • Perhaps it is better that I will ask this as a new question. – user237522 Apr 14 '17 at 02:39
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Yes. Note that for any $d$, $F^d:M\to M$ is a continuous extension of $f^d:S\to S$. If $f^d$ is the identity, then the identity $M\to M$ is a continuous extension of $f^d$, and so by uniqueness of the continuous extension, $F^d$ must be the identity. It follows that if $f$ has order $d$, so does $F$.

Eric Wofsey
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