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I am reading this paper on Atsuji metric space. A metric space $X$ is said to be Atsuji if every real-valued continuous function on $X$ is uniformly continuous.

In theorem 3.7 , it is proved that [see 3.7(e)], if $X$ is an Atsuji space then for any real-valued continuous function $f$ on $X$, there is some natural number $n$ such that every point of $A=\{x:|f(x)|\ge n\}$ is isolated and $\inf\{I(x):x\in A\}>0.$

Now $[0,1]$ is an Atsuji space (being compact) and $f:[0,1]\to\mathbb R:x\mapsto0.5$ is continuous. But I cannot find any $n$ such that $A=\{x:|f(x)|\ge n\}$ is isolated. How is it possible that there is no such $n$ that would make $A$ isolated?

Jave
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    Take $n=1$, in this case $A = \emptyset$ which is trivially isolated and $\inf {I (x) : x \in A} = \infty > 0$ – Kroki Apr 17 '18 at 01:55
  • @Youem Can't we consider $\emptyset$ to be a set where every point is an accumulation point? – Jave Apr 17 '18 at 02:13
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    I think you can. Because when you have an logic assertion $\mathcal P (x)$, then the assertion $\mathcal Q : {\forall x \in \emptyset, , \mathcal P(x) \text{ is true}}$ is always true. – Kroki Apr 17 '18 at 02:16
  • okay. Thanks so much. – Jave Apr 17 '18 at 02:18

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