# Thread: more counterexamples needed

1. ## 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)

2. 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 $f(g(x))=0$.

3. 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

$\chi$ $\sigma$

4. Where you put a pi, any number, including x would work, right?

5. Originally Posted by binkypoo
Where you put a pi, any number, including x would work, right?
Any irrational number will work in place of $\pi$.

6. how about a counterexample to this:
if f is continuous at x=a, then f(a+)=f(a-)

7. That's true. If a function $f$ is continuous at $x=a$, then $\lim_{x\to a^-}f(x)=f(a)=\lim_{x\to a^+}f(x)$

8. wow, that one has been boggling me for a while, because I was sure it was true, but my book says its false! Thanks

9. 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...

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