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Let $f$ be Cauchy continuous. $f$ is Cauchy continuous if for any Cauchy sequence $\{x_{n}\}$ in $(X,d_{X})$, $\{f(x_{n})\}$ is a Cauchy sequence in $(Y,d_{Y})$. Show that Cauchy continuous $\implies$ continuous.

Since $f$ is Cauchy continuous, we have that $\{f(x_{n})\}$ is a Cauchy sequence in $(Y,d_{Y})$. By definition, we have that for all $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that $d_{Y}(f(x_{n}),f(x_{m})) < \varepsilon$ for all $m,n \geqslant N$. We want to show the existence of $\delta > 0$ such that $d_{X}(x_{n},x_{m}) < \delta$. Since $\{f(x_{n})\}$ is Cauchy in $(Y,d_{Y})$, which was obtained from a Cauchy sequence in $(X,d_{X})$, let $\delta = \epsilon$. Thus, we have that $d_{Y}(f(x_{n}),f(x_{m})) < \varepsilon$ when $d_{X}(x_{n},x_{m}) < \delta$. So, $f$ is continuous.

This seems trivial but a little sketchy to me. Are there any glaring errors in this proof?

clocktower
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  • Here is a solution http://math.stackexchange.com/questions/344452/is-a-cauchy-sequence-preserving-continuous-function-is-uniformly-continuou – Svetoslav Feb 10 '16 at 22:17

1 Answers1

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 We want to show the existence of $\delta > 0$ such that $d_{X}(x_{n},x_{m}) < \delta$.

No, that's not what you want to show. In fact, let's go back to the definition of continuity. It starts with "Given $\epsilon > 0$ ..." That should tell you what you want to show, and it also shows what you need to assume as initial hypothesis.

Chan-Ho Suh
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