Let I be an interval in $R$ and suppose that $f:I\to R$ is uniformly continuous. Show that $f(x_n)$ is Cauchy whenever $x_n$ is Cauchy.

Question

I believe that the answer might be somewhat trivial, but I am not sure because I am very confused by the concept of Cauchy sequences and uniform continuity.

My answer is as such:

If $x_n$ is a Cauchy sequence, then it converges to an arbitrary point $x$. Similarly, this will cause $f(x_n)$ to be convergent to $f(x)$. Because all convergent sequences are Cauchy, $f(x)$ is a Cauchy sequence.

I feel that this answer is not only trivial, but also partially incorrect; however, I don't know enough about either topic to provide further analysis. If someone could explain these topics and/or point me in the right direction for the problem that would be excellent!

Answer 1

Let $\epsilon>0$ be given.

Since $f$ is uniformly continuous, so there exists $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ for $|x-y|<\delta$.

Since $x_n$ is Cauchy, there exists $N>0$ such that $|x_n-x_m|<\delta$ for $m,n>N$

Hence $|f(x_n)-f(x_m)|<\epsilon$ for $m,n>N$.(Why?) So $f(x_n)$ is a Cauchy sequence.

You may also note that converse does not hole in general i.e. suppose $f$ maps cauchy sequences to cauchy sequences even $f$ may not be uniformly continuous.

The standard counterexample is $f : \mathbb R \to \mathbb R$ defined by $f(x)=x^2$

Answer 2

Your answer is valid if $I$ is a closed interval. In that case, you don't even need uniform continuity, just regular continuity. If $I$ is open, however, then $x_n$ may not converge to a point in $I$.

$x_n$ being a Cauchy sequence means that sufficiently far down the sequence, all points are very close to each other. Formally, for all $\epsilon > 0$ there exists an $N$ sufficiently large such that $n, m > N$ implies $|x_n - x_m| < \epsilon$.

$f$ being uniformly continuous means that the images of close together points are still close together, and how close the original points need to be doesn't depend on location. Formally, for all $\epsilon > 0$ there exists a $\delta > 0$, independent of $x$, such that $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon$.

See if you can combine these two definitions to show that $f(x_n)$ is Cauchy.