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Let $X$ be a metric space and let $f : X → \mathbb{R}$ be a continuous function. Pick out the true statements.
(a) $f$ always maps Cauchy sequences into Cauchy sequences.
(b) If $X$ is compact, then $f$ always maps Cauchy sequences into Cauchy sequences.
(c) If $X = \mathbb{R}^n$ , then $f$ always maps Cauchy sequences into Cauchy sequences.

to preserve Cauchy sequence $f$ need to be uniformly continuous.
(a) is not true. as example $f(x)=1/x$ and the sequence {$1/n$}
(b) in compact set a continuous function becomes uniformly continuous.
(c) false by (a).

are my thinking correct or I am missing something because the given answer is (b) and (c)

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c> $\mathbb R^n$ is complete so... – jim Jan 5 '13 at 10:24
i did not understand. since $X = \mathbb{R}^n$ complete every Cauchy sequence is convergent.but i think the given condition is different – poton Jan 5 '13 at 10:36
isnt the image of the sequence $x_n$ in $\mathbb R^n$ cauchy in $\mathbb R$ ? use the definition of Cauchy sequence – jim Jan 5 '13 at 11:03
c is right, because continuous function is uniformly continuous on compact set. – lee Jan 5 '13 at 11:06
@lee in c> where did the compact set come from? – jim Jan 5 '13 at 11:09
up vote 2 down vote accepted

Concerning c): The counterexample $f(x) := \frac{1}{x}$ doesn't work, since you can't define it (as a continuous function) on $\mathbb{R}^n$, but only on $\mathbb{R}^n \backslash \{0\}$.

Let $(x_n)_{n} \subseteq X := \mathbb{R}^n$ a Cauchy sequence. Since $X$ is complete, we have $x_n \to x$ for some $x \in \mathbb{R}^n$. Since $f$ is (sequentially) continuous, we obtain $f(x_n) \to f(x)$, i.e. $(f(x_n))_{n \in \mathbb{N}}$ is convergent and in particular a Cauchy sequence.

Remark This proves also b) (since a compact metric space $X$ is in particular complete).

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any convergent sequence is cauchy? – Un Chien Andalou Jan 8 '13 at 6:53
@Panu In a metric space? Yes. – saz Jan 8 '13 at 6:58

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