If $A$ is compact and $f : A \to \mathbb{R}$ is continuous and $(x_n) $ is Cauchy in $A$ show that $(f(x_n))$ is Cauchy and moreover show that if $A = \mathbb{R}$, $(f(x_n))$ is Cauchy

I can see this is leading to uniform continuity but we haven't covered that yet. Just making sure I have the correct idea. thanks.

Since $(x_n)$ is Cauchy in $A$, $\forall \delta > 0, \exists N \in \mathbb{N} \,\,s.t. n,m \ge N \implies |x_n-x_m| < \delta$ and if $A$ is compact then $A$ is closed and bounded, which gives that $(x_n)$ converges in $A$ (b/c Cauchy implies convergent)

since $f$ is continuous and $A$ is compact then $f(A)$ is compact. So it follows the $f(x_n), f(x_m) \in f(A)$, $(x_n \in A)(\forall \epsilon > 0) (\exists \delta > 0 \wedge x_m \in A) \implies|f(x_n)-f(x_m)| < \epsilon$ So $f(x_n)$ is Cauchy

And if $A = \mathbb{R}$ then this seems to follow from the above but since $\mathbb{R}$ is not compact, just use that $(x_n)$ is Cauchy implies it is convergent then use a $\delta - \epsilon$ proof and any convergent sequence in the reals is Cauchy.

  • $\begingroup$ are you assuming that $f$ is continuous? $\endgroup$
    – janmarqz
    Nov 14, 2015 at 18:42
  • $\begingroup$ oh yes, sorry I will edit the post $\endgroup$ Nov 14, 2015 at 18:43
  • $\begingroup$ Edit the title as well; also at the moment it doesn't really make sense (chose between 'prove' and 'show'). $\endgroup$ Nov 14, 2015 at 18:51
  • $\begingroup$ So it follows the $f(x_n), f(x_m) \in F(A)$, $(x_n \in A)(\forall \epsilon > 0) (\exists \delta > 0 \wedge x_m \in A) \implies|f(x_n)-f(x_m)| < \epsilon$ So $f(x_n)$ is Cauchy: It has absolutely no sense !! $\endgroup$
    – Surb
    Nov 14, 2015 at 18:55

1 Answer 1


$A$ is compact, so $(x_n)$ has a convergent subsequence $(x_{n_k})$ with limit $c$. Then $x_n \rightarrow c$ as well (this is a general fact for Cauchy sequences with a convergent subsequence):

Suppose $\varepsilon > 0$, then as the sequence is Cauchy, for some $N$ we have that $n,m \ge N$ implies $d(x_n, x_m) < \frac{\varepsilon}{2}$. Also some $K$ exists such that for $k > M$ we also have that $d(x_{n_k}, c) < \frac{\varepsilon}{2}$. So for $n \ge N$, we find $k > M$ such that $n_k > N$, so then $d(x_n, c) \le d(x_n, x_{n_k}) + d(x_{n_k},c) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$. This shows convergence of $(x_n)$ to $c$.

In fact this is the proof that a compact metric space is complete, which might already have been covered in your course. $A$ need not be a subset of the reals for this.

Now continuity shows that $f(x_n) \rightarrow f(c)$ (so the image sequence is not only Cauchy, it's already convergent).

For $A = \mathbb{R}$ the result is also true, but uninteresting: a Cauchy sequence is convergent in the reals, so its image is again convergent, so Cauchy.

  • $\begingroup$ this makes much more sense $\endgroup$ Nov 14, 2015 at 19:00

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