1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ are you assuming that $f$ is continuous? $\endgroup$ – janmarqz Nov 14 '15 at 18:42
  • $\begingroup$ oh yes, sorry I will edit the post $\endgroup$ – oliverjones Nov 14 '15 at 18:43
  • $\begingroup$ Edit the title as well; also at the moment it doesn't really make sense (chose between 'prove' and 'show'). $\endgroup$ – Silvia Ghinassi Nov 14 '15 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 '15 at 18:55
3
$\begingroup$

$A$ 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.

$\endgroup$
1
  • $\begingroup$ this makes much more sense $\endgroup$ – oliverjones Nov 14 '15 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.