# Why Continuous Functions Don't Preserve Cauchy Sequences

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In summary: Continuous functions do not necessarily preserve cauchy sequences if the sequence is diverging. The proof is done using a double triangle inequality and requires that the function be uniformly continuous.
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## 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
and here

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(xn) 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

I like number said:
I understand that if we have a cauchy sequence converging to 0, then f(xn) 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.

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:
$$|f(x_0) - f(x)| = |f(x_0) - f_n(x_0) + f_n(x_0) - f_n(x) + f_n(x_0)-f(x)| \leq |f(x_0) - f_n(x_0)| + |f_n(x_0) - f_n(x)| + |f_n(x_0)-f(x)|$$

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 $[0,1)$.

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

## 1. What is a continuous function?

A continuous function is a function where small changes in the input result in small changes in the output. This means that as the input values get closer together, the output values also get closer together.

## 2. What is a Cauchy sequence?

A Cauchy sequence is a sequence of numbers where the terms get closer and closer together as the sequence progresses. This means that the distance between any two terms in the sequence eventually becomes smaller than any given positive number.

## 3. Why don't continuous functions preserve Cauchy sequences?

Continuous functions do not preserve Cauchy sequences because the Cauchy sequence may converge to a point outside of the function's range. In other words, the output of the function may not get closer and closer to a single point as the input values get closer and closer.

## 4. Can a function still be continuous even if it doesn't preserve Cauchy sequences?

Yes, a function can still be continuous even if it doesn't preserve Cauchy sequences. This is because continuity only requires small changes in the input to result in small changes in the output, not that the output values get closer and closer to a single point as the input values get closer and closer.

## 5. What is an example of a continuous function that doesn't preserve Cauchy sequences?

An example of a continuous function that doesn't preserve Cauchy sequences is the function f(x) = 1/x. As the input values get closer and closer to 0, the output values get larger and larger, and do not converge to a single point. Therefore, this function does not preserve Cauchy sequences.

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