# Thread: Cont. Func + Cauchy Seq

1. ## Cont. Func + Cauchy Seq

Give an example of the case or explain why it's not possible for each of the following:

1.) Continuous function $\displaystyle f : (0,1) \rightarrow \mathbb{R}$ and also a Cauchy sequence $\displaystyle (x_n)$ such that $\displaystyle f(x_n)$ is not a Cauchy sequence.

2.) Continuous function $\displaystyle f : [0,1] \rightarrow \mathbb{R}$ and also a Cauchy sequence $\displaystyle (x_n)$ such that $\displaystyle f(x_n)$ is not a Cauchy sequence.

3.) Continuous function $\displaystyle f : [0,\infty) \rightarrow \mathbb{R}$ and also a Cauchy sequence $\displaystyle (x_n)$ such that $\displaystyle f(x_n)$ is not a Cauchy sequence.

4.) Continuous function $\displaystyle f$ that's bounded on $\displaystyle (0,1)$ and that has a maximum value on the open interval, however, does not have a minimum value.

2. Originally Posted by fifthrapiers
Give an example of the case or explain why it's not possible for each of the following:
Good exam problems.

1.) Continuous function $\displaystyle f : (0,1) \rightarrow \mathbb{R}$ and also a Cauchy sequence $\displaystyle (x_n)$ such that $\displaystyle f(x_n)$ is not a Cauchy sequence.
Consider $\displaystyle f(x) = \frac{1}{x}$ and $\displaystyle x_n = \frac{1}{n}$.
2.) Continuous function $\displaystyle f : [0,1] \rightarrow \mathbb{R}$ and also a Cauchy sequence $\displaystyle (x_n)$ such that $\displaystyle f(x_n)$ is not a Cauchy sequence.
Impossible because continous functions on closed finite intervals are uniformly continous. But the property of uniformly continous functions is that their images of Cauchy sequences still remain Cauchy sequences.
3.) Continuous function $\displaystyle f : [0,\infty) \rightarrow \mathbb{R}$ and also a Cauchy sequence $\displaystyle (x_n)$ such that $\displaystyle f(x_n)$ is not a Cauchy sequence.
Impossible. Let $\displaystyle x_n$ be Cauchy sequence in $\displaystyle [0,\infty)$. Since it is Cauchy it is bounded. So the entire sequence is contained in $\displaystyle [0,M]$ where $\displaystyle M>0$ is its upper bound. But since $\displaystyle f(x)$ is uniformly continous on $\displaystyle [0,M]$ it means (as in 2) that $\displaystyle f(x_n)$ is Cauchy.

4.) Continuous function $\displaystyle f$ that's bounded on $\displaystyle (0,1)$ and that has a maximum value on the open interval, however, does not have a minimum value.
Draw a picture. Consider $\displaystyle f(x) = 1-|x|$ on $\displaystyle (-1,1)$.