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Suppose $f(x)$ defines on the bounded interval $I$. Prove that $f(x)$ is uniformly continuous on $I$ if and only if the image of each Cauchy sequence under $f$ is also a Cauchy sequence.

The "only if" part is easy. For the "if" part I can prove that $f$ is continuous on $I$. But how to prove it's uniformly continuous? I try to prove by contradiction, but could not get a Cauchy sequence.

Knt
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    I think no, for in my question, the interval $I$ is bounded. – Knt Oct 05 '19 at 11:33
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    If $I = [a,b]$ is closed, then $f$ is automatically uniformly continuous. If not, can you prove that $\lim_{x\to a}f(x)$ and $\lim_{x\to b}f(x)$ exist? Then you can extend $f$ to a continuous function on the closure of $I$. – amsmath Oct 05 '19 at 12:04
  • @KaviRamaMurthy Read the question. OP has already proved continuity. – amsmath Oct 05 '19 at 12:12

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Hint: The following can be used together with some basic theory of real analysis to complete the OP's analysis.

Given $a. b \in \Bbb R$ with $a \lt b$.

Let $f: (a,b) \to \Bbb R$ be a continuous function satisfying

$$\tag 1 {\displaystyle \bigcap_{n \ge 1}} \, \overline {f\big ( \,(a,a+\frac{b-a}{2^n}] \, \big )} = \{\alpha\}$$

and

$$\tag 2 {\displaystyle \bigcap_{n \ge 1}} \, \overline {f\big( [b-\frac{b-a}{2^n},b) \big )} = \{\beta\}$$

Then $f$ can be extended to a continuous function $f_♯: [a,b] \to \Bbb R$ by defining $f_♯(a) = \alpha$ and $f_♯(b) = \beta$.

CopyPasteIt
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