This is a problem from revision material for a functional analysis class. Let $(X,d)$ and $(C,p)$ be two metric spaces and let $f:X\rightarrow C$ be a continuous function with $f(X)=C$.

Assuming $(X,d)$ is complete and for arbitrary $x,y\in X$ that $d(x,y)\leq p(f(x),f(y))$ give a brief proof that $(C,p)$ is complete.

So far I know the criteria of completeness is that all Cauchy sequences in the space converge to a limit point in that space. And I know that a uniformly continuous function will map a Cauchy sequence to a Cauchy sequence. But since this function is merely continuous so somehow the extra condition must make it preserve Cauchy sequences.

Since $X$ is complete a Cauchy sequence ${x_n}$ converges to a point $x\in X$.
Since $f$ is a continuous function $f(x_n)$ must converge to $f(x)$.
Since $x \in X$, $f(x)\in f(X)=C$
Since $f(x_n)$ is convergent then it should also be Cauchy.

My issue is I haven't used the extra condition and I know that continuity alone isn't enough to preserve Cauchy sequences. Where have I gone wrong and how can I proceed?


Let $(t_n)_{n=1}^\infty$ be a Cauchy sequence in $(C,p)$.

Pick $(x_n)_{n=1}^\infty \subseteq (X,d)$ such that $f(x_n) = t_n$.

Then $d(x_m, x_n) \leq p(t_m, t_n)$ and hence $(x_n)_{n=1}^\infty$ is a Cauchy sequence in $(X,d)$.

By completeness of $(X,d)$, there exists $x \in X$ such that $x_n \rightarrow x$.

Let $t = f(x) \in f(X) = C$.

By the continuity of $f$, we have $t = f(x) = f(lim_{n \rightarrow \infty }x_n) = lim_{n \rightarrow \infty }f(x_n) = lim_{n \rightarrow \infty } t_n$

We have the conclusion $t_n \rightarrow t \in C$, i.e. $(t_n)_{n=1}^\infty$ is a convergent sequence in $(C,p)$.

Hence $(C,p)$ is complete.

  • $\begingroup$ Thank you, this immediately makes sense. I feel like starting from the wrong point had me wanting to use the inequality in a different more convoluted (and probably incorrect) way. Cheers. $\endgroup$ – Jackamoto Nov 15 '14 at 8:52

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