1
$\begingroup$

I seek to show whether the following statements are true or false:

  1. If $f$ is continuous on $(a,b)$ and $\{x_n\}$ is a sequence in $(a,b)$, then the sequence $\{f(x_n)\}$ has a convergent subsequence.

  2. If $f$ is continuous on $[a,b]$ and $\{x_n\}$ is a sequence in $(a,b)$, then the sequence $\{f(x_n)\}$ has a convergent subsequence.

My gut tells me one of them is true and the other false, but I'm not sure which. I think 2 is false and 1 is true? I can't really seem to explain why...

Thanks in advance.

$\endgroup$
3
$\begingroup$

HINTS:

  1. Look at the sequence $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$ in $(0,1)$ and the function $f(x)=\frac1x$.

  2. If $f$ is continuous on $[a,b]$, then $f\big[[a,b]\big]$ is a closed interval $[c,d]$ in $\Bbb R$. What do you know about sequences in closed, bounded sets?

$\endgroup$
  • $\begingroup$ so here, why is $[c,d]$ necessarily a closed, bounded interval? $\endgroup$ – Peej Gerard Apr 3 '13 at 1:26
  • 2
    $\begingroup$ @Paul: Because continuous functions preserve compactness and connectedness, and the only compact, connected subsets of $\Bbb R$ are sets of the form $[c,d]$. All you really need to know, however, is that the image is compact (closed and bounded); you don’t need to know that it’s an interval. $\endgroup$ – Brian M. Scott Apr 3 '13 at 1:29
2
$\begingroup$

The first statement is false. Consider the case where $f(x)=\frac{1}{x}$, $(a,b)=(0,2)$, and $\{x_n\}=\{\frac{1}{n}\}$. Obviously, $\{f(x_n)\}=\{n\}$, which has no convergent subsequence.

$\endgroup$
1
$\begingroup$

It's the other way around: 1. is false and 2. is true.

It's all about the endpoints. Imagine any continuous function that would go to $\pm\infty$ both in $a$ and in $b$. This is allowed for 1. but not allowed for 2. Now, if $x_n\to a$, then $f(x_n)$ will tend to $\pm\infty$, so it's not convergent.

$\endgroup$
1
$\begingroup$

If you suppose $f(x)=\frac{1}{x-a}$, and consider the sequence $x_n=a+\frac{1}{n}$, what happens to $f(x_n)$? Does it have a convergent subsequence?

On the other hand, since $[a,b]$ is compact and the continuous image of a compact set is compact, we know $f([a,b])$ is compact. Now we know any bounded sequence will have a convergent sequence, therefore $f(x_n)$ will have a convergent subsequence.

$\endgroup$
1
$\begingroup$

The second statement is true. This is because $[a,b]$ is compact, and $f$ is continuous, which implies that $f([a,b])$ is also compact. Hence, $f(\{x_n\})$ is a sequence in a compact set, which indicates that it has a convergent subsequence (compactness implies sequential compactness.)

$\endgroup$
  • $\begingroup$ Your last statement is not true in general. See here for reasoning. You may want to explicitly say that compactness implies sequential compactness in metric spaces. $\endgroup$ – Clayton Apr 3 '13 at 1:22

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.