# Closed & bounded on D imply closed and bounded on f(D)?

• February 21st 2011, 06:41 PM
Dagger2006
Closed & bounded on D imply closed and bounded on f(D)?
The question is:

For D a metric space and f:D--->R a continuous function. If D is closed and bounded. Is f(D) closed and bounded?

I'm pretty sure this is true for the case of D belonging to the real numbers, but I think this is true for D just being a metric space. I think if I can somehow find the example of a metric space that is closed and bounded, but not compact that would be the key. Can't think of one, though. I've also had no luck proving it true. So if someone can give a a counter example or an idea of his to proves, I would greatly appreciate it! Thanks!
• February 21st 2011, 07:20 PM
hatsoff
Not generally, no. Just take the identity function $f:(0,1)\to\mathbb{R}$. As a space, $D=(0,1)$ is closed (all spaces are closed). But clearly $f((0,1))=(0,1)$ is not closed in $\mathbb{R}$.
• February 21st 2011, 08:43 PM
Dagger2006
How is it that you can say that D = (0, 1) is closed? Are you limiting the entire metric space to just the interval (0, 1) and thus it is open and closed? Regardless (0, 1) doesn't contain all if it's limit points which would make it not closed? I'm guessing I'm not exactly clear on being closed in metric spaces. Thanks a lot fir the feedback!
• February 22nd 2011, 02:44 AM
hatsoff
Quote:

Originally Posted by Dagger2006
How is it that you can say that D = (0, 1) is closed? Are you limiting the entire metric space to just the interval (0, 1) and thus it is open and closed? Regardless (0, 1) doesn't contain all if it's limit points which would make it not closed? I'm guessing I'm not exactly clear on being closed in metric spaces. Thanks a lot fir the feedback!

Recall that a limit point of a set $D$ is a point in the space such that every neighborhood of the point contains a point not in $D$. If $D$ is the space, then it has to contain all its limit points by definition.

A topological space is always closed with respect to itself. Always---even in the metric case.

EDIT: By the way, you can make the image unbounded by using the continuous function defined by $g(x)=1/x$. Then $g(D)=(1,\infty)$ which is neither bounded nor closed in $\mathbb{R}$.