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Problem:

$f:(0,\infty )\rightarrow \mathbb{R}$ defined as: $f(x)=x^{2}$

Can anyone show me how to prove that $f$ transforms Cauchy sequences of elements of $(0,\infty )$ into Cauchy sequences, but $f$ is not uniformly continuous? The purpose of this problem is to show how essential the boundedness of the set on which $f$ is defined is.

$f$ is definitely not uniformly continuous because for the two sequences: $\left \{ x_{n} \right \},\left \{ y_{n} \right \}$ defined by: $x_{n}=n+\frac{1}{n}$ and $y_{n}=n$. We have: $\left | x_{n}-y_{n} \right | \to 0$ as $n \to \infty $, but $\left | f(x_{n})-f(y_{n}) \right | \to 2$ as $(n \to \infty )$. Can anyone, please, show me how to prove that $f$ transforms Cauchy sequences of elements of $(0,\infty )$ into Cauchy sequences? Thanks

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Hint: Cauchy sequences are bounded. –  Jose27 Aug 8 '12 at 19:47

1 Answer 1

up vote 5 down vote accepted

Let $\{x_n\}$ a Cauchy sequence. It's bounded, since for $n,m$ large enough, say, $n,m\geq n_0$ we have $|x_n-x_m|\leq 1$ hence $|x_n|\leq \max\{1,|x_1|,\dots,|x_n|\}=:M$. We have for $k,j$ integers that $$|x_k^2-x_j^2|=|x_k-x_j||x_k+x_j|\leq |x_k-x_j|(|x_k|+|x_j|)\leq 2M|x_k-x_j|,$$ and since $M$ doesn't depend on $j$ or $k$, the sequence $\{x_k^2\}$ is Cauchy.

The statement:

If $f\colon \Bbb R\to \Bbb R$ is uniformly continuous and $\{x_k\}$ is Cauchy, then $\{f(x_k)\}$ is Cauchy.

is true, and this counter-example shows that the converse doesn't hold.

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