# more counterexamples needed

• October 22nd 2009, 11:15 PM
dannyboycurtis
more counterexamples needed
I know that the following are false, I just cant come up with counterexamples:
[f(x)]^2 continuous on (a,b) implies taht f is continuous on (a,b)

f and g are not continuous on (a,b) implies that fg is not continuous on (a,b)

f and g are not continuous on (a,b) implies that $f\circ g$ is not continuous on (a,b)
• October 22nd 2009, 11:49 PM
redsoxfan325
Quote:

Originally Posted by dannyboycurtis
I know that the following are false, I just cant come up with counterexamples:
[f(x)]^2 continuous on (a,b) implies taht f is continuous on (a,b)

f and g are not continuous on (a,b) implies that fg is not continuous on (a,b)

f and g are not continuous on (a,b) implies that $f\circ g$ is not continuous on (a,b)

This example works for the first two. Take $f(x)=g(x)=\left\{\begin{array}{lr}1:&x\in\mathbb{Q }\\-1:&x\notin\mathbb{Q}\end{array}\right\}$

This is obviously discontinuous, but $f^2(x)=f(x)g(x)=1$, which is continuous.

For the third one, let $f(x)=\left\{\begin{array}{lr}x:&x\in\mathbb{Q}\\0: &x\notin\mathbb{Q}\end{array}\right\}$ and $g(x)=\left\{\begin{array}{lr}\pi:&x\in\mathbb{Q}\\ 0:&x\notin\mathbb{Q}\end{array}\right\}$

Clearly both $f(x)$ and $g(x)$ are discontinuous, but $f(g(x))=0$.
• October 22nd 2009, 11:51 PM
chisigma
I think taht the following function pair is a countexample in all the cases You have proposed...

$f(x)= \left\{\begin{array}{cc}-1, &\mbox {if } -1\le x \le 0\\1, & \mbox {if } 0< x \le 1\end{array}\right.$

$g(x)= -f(x)$

$x \in [-1,1]$

Kind regards

$\chi$ $\sigma$
• October 25th 2009, 10:49 PM
binkypoo
Where you put a pi, any number, including x would work, right?
• October 25th 2009, 10:55 PM
redsoxfan325
Quote:

Originally Posted by binkypoo
Where you put a pi, any number, including x would work, right?

Any irrational number will work in place of $\pi$.
• October 25th 2009, 10:59 PM
binkypoo
how about a counterexample to this:
if f is continuous at x=a, then f(a+)=f(a-)
• October 25th 2009, 11:11 PM
redsoxfan325
That's true. If a function $f$ is continuous at $x=a$, then $\lim_{x\to a^-}f(x)=f(a)=\lim_{x\to a^+}f(x)$
• October 25th 2009, 11:17 PM
binkypoo
wow, that one has been boggling me for a while, because I was sure it was true, but my book says its false! Thanks
• October 25th 2009, 11:20 PM
binkypoo
last question on this thread, Ive been trying to find a counterexample for this one for a while, Im sure its simple but I cant seem to see it:
if |f| is cont. on (a,b) then f is cont. on (a,b).
Again my book claims its false...
• October 25th 2009, 11:29 PM
redsoxfan325
Quote:

Originally Posted by binkypoo
last question on this thread, Ive been trying to find a counterexample for this one for a while, Im sure its simple but I cant seem to see it:
if |f| is cont. on (a,b) then f is cont. on (a,b).
Again my book claims its false...

Consider $f(x)=\left\{\begin{array}{lr}1:&x\in\mathbb{Q}\\-1:&x\notin\mathbb{Q}\end{array}\right\}$