# more examples needed...

• Oct 30th 2009, 04:35 PM
dannyboycurtis
more examples needed...
Hi I was wondering if anyone could point me in the right direction here:
I need an example of a function f:[a,b] --> R that is not continuous but whose range is:
(a) an open and bounded interval
(b) an open and UNbounded interval
(c) a closed and unbounded interval.
Thanks for any help
• Oct 30th 2009, 05:17 PM
Bruno J.
Well a function which is continuous is a function which is continuous everywhere. A non-continuous function is a function with at least one discontinuity. Is this what you want, or do you want a function which is nowehere continuous?
• Oct 30th 2009, 05:54 PM
redsoxfan325
Quote:

Originally Posted by dannyboycurtis
Hi I was wondering if anyone could point me in the right direction here:
I need an example of a function f:[a,b] --> R that is not continuous but whose range is:
(a) an open and bounded interval
(b) an open and UNbounded interval
(c) a closed and unbounded interval.
Thanks for any help

For a, take $f:[0,1]\to\mathbb{R}=\left\{\begin{array}{lr}1/2:&x=0~or~x=1\\x:& 0

For b and c, $f:[-\pi/2,\pi/2]\to\mathbb{R}=\left\{\begin{array}{lr}0:&x=-\pi/2~or~x=\pi/2\\ \tan x:&-\pi/2
• Oct 31st 2009, 06:39 PM
dannyboycurtis
Regarding:
$f:[-\pi/2,\pi/2]\to\mathbb{R}=\left\{\begin{array}{lr}0:&x=-\pi/2~or~x=\pi/2\\ \tan x:&-\pi/2<\pi/2\end{array}\right\}$
This function is closed and bounded isnt it? I was just wondering because you said it worked for (b) and (c). I can see that f(x)=tan(x) by itself has a range which is open and bounded, but I dont see how including the f(x)=0 at x=-pi/2 and x=pi/2 would make it an open interval...
• Oct 31st 2009, 06:44 PM
redsoxfan325
I had to pick arbitrary values for $\pm\pi/2$ because tan(x) isn't defined there, but f(x) had to be defined on a closed interval (so excluding the endpoints from the domain wasn't an option).

The range is $(-\infty,\infty)$, which is both open and closed (sometimes called "clopen") and clearly unbounded.