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According to the classroom notes "Uniformly Continuous Linear Set" in American Mathematical Monthly, Vol. 62. No. 8(Oct., 1955) pp. 579-580, Author: Norman Levine link.

I'm usually confused about the state "without loss of generality".

Let see the proof of theorem 1. From the proof of theorem 1, how does "WLOG" work?

Thanks for all answers.

user90402
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2 Answers2

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Hint: Uniformly continuous functions can be extended continuously to their completions. If $p$ is a limit point of $X$, consider the function

$$f(x)=\frac{1}{d(x,p)}$$

Alex Youcis
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  • Sorry for asking again, in the proof of theorem 1 how to verify that f(x) is continuous but it's not uniformly continuous. – user90402 Aug 15 '13 at 08:53
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Perhaps you need the statement:

If a convergent sequence is not eventually constant, then it has a strictly monotone subsequence.

You read it in$\,$ C.G.Denlinger Elements of Real Analysis $\,$pag. 113.

Addendum (referring to my comment to prove the not uniformly continuity of $f$) the convergent sequence is a Cauchy sequence; therefore there exist always two terms $p_r$ and $p_s$, whatever near, such that $|f(p_r)-f(p_s)|=1$. Take $\epsilon<1$ ...

Tony Piccolo
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  • thanks it's very useful but now in the proof of theorem 1) I cannot find the way to verify that f(x) is continuous but it's not uniformly continuous. Thank you for your help. – user90402 Aug 15 '13 at 09:04
  • WLOG is related only to the existence of the sequence ... Have better edit your question. – Tony Piccolo Aug 15 '13 at 09:16
  • You need to find an $\epsilon>0$ such that for every $\delta>0$ there are points $u$ and $v$ in $E$ such that $|u-v|<\delta$ and $|f(u)-f(v)| \ge \epsilon$. – Tony Piccolo Aug 15 '13 at 09:37