Why Continuous Functions Don't Preserve Cauchy Sequences

[I like number]

Homework Statement

Why is it that continuous functions do not necessarily preserve cauchy sequences.

Homework Equations

Epsilon delta definition of continuity
Sequential Characterisation of continuity

The Attempt at a Solution

I can't see why the proof that uniformly continuous functions preserve cauchy sequences doesn't hold for 'normal' continuous functions.
In particular the example of f(x) = 1/x on (0,1)
I have worked through the examples
http://www.mathcs.org/analysis/reals/cont/answers/fcont3.html
and here
http://www.mathcs.org/analysis/reals/cont/answers/contuni4.html

where they address this issue directly, but I can't get my head around it.

I understand that if we have a cauchy sequence converging to 0, then f(x_n) is going to diverge to infinity, but I still can't see what the problem is.

Any explanation you can offer would be appreciated.

Kind regards

 

[jgens]

I understand that if we have a cauchy sequence converging to 0, then f(x_n) is going to diverge to infinity, but I still can't see what the problem is.

Recall that Cauchy sequences are bounded. So if $\{f(x_n)\}_{n \in \mathbb{N}}$ diverges, then the sequence cannot be Cauchy. In particular, $f$ does not take Cauchy sequences to Cauchy sequences.

 

[Robert1986]

The reason that we need uniform continuity is that we need to be able to find one $\delta$ for each $\epsilon$ that works for all $x$ in a certain interval. This is because in the proof, we do a "double triangle inequality." So, if $\{f(x_n)\}$ is a sequence of continuous functions that converges to $f(x)$ for each $x$ in the interval $(a,b)$ then we want to show that $\forall \epsilon \exists \delta$ such that $|f(x_0) - f(x)| < \epsilon$ whenever $|x_0 - x| < \delta$. We do this by writting:
$$|f(x_0) - f(x)| = |f(x_0) - f_n(x_0) + f_n(x_0) - f_n(x) + f_n(x_0)-f(x)| \leq<br /> |f(x_0) - f_n(x_0)| + |f_n(x_0) - f_n(x)| + |f_n(x_0)-f(x)|$$

Now, since the sequence is Cauchy, we can control the outer two terms with a big enough $n$ and make them both less than $\epsilon / 3$. So, we need to be able to ensure that $|f_n(x) - f_n(x_0)| \leq \epsilon / 3$ for every $x$ such that $|x_0-x|\leq \delta$. The only way we can do this is by making $f_n$ uniformly continuous.

As an example, consider the function $f_n(x) = x^n$ on $[0,1)$.

 

[I like number]

Thanks very much to you both.
I think I can see it more clearly now, (and a good nights sleep always helps too!).
I will continue to play around with these ideas and if I have any more questions I'll be back.

Thanks again