# I need some examples

• Oct 25th 2009, 08:36 PM
binkypoo
I need some examples
Is there a function which is defined on $\mathbb{R}$ but which is NOT continuous on ANY point of $\mathbb{R}$?

What about a function which is defined on $\mathbb{R}$ and is continuous at exactly TWO points of $\mathbb{R}$?
• Oct 25th 2009, 09:00 PM
redsoxfan325
Quote:

Originally Posted by binkypoo
Is there a function which is defined on $\mathbb{R}$ but which is NOT continuous on ANY point of $\mathbb{R}$?

What about a function which is defined on $\mathbb{R}$ and is continuous at exactly TWO points of $\mathbb{R}$?

$f(x)=\left\{\begin{array}{lr}1:&x\in\mathbb{Q}\\0: &x\notin\mathbb{Q}\end{array}\right\}$

$g(x)=\left\{\begin{array}{lr}x:&x\in\mathbb{Q}\cap (-\infty,1]\\x-2:&x\in\mathbb{Q}\cap(1,\infty)\\0:&x\notin\mathbb {Q}\end{array}\right\}$

$f(x)$ is continuous nowhere, and $g(x)$ is continuous only at $x=0$ and $x=2$.
• Oct 25th 2009, 09:26 PM
binkypoo
what about a function f defined on an interval [a,b] which is continuous at 2 points of [a,b]?
• Oct 25th 2009, 09:32 PM
redsoxfan325
Use the exact same function; just restrict the domain.
• Oct 25th 2009, 09:39 PM
binkypoo
$g(x)=\left\{\begin{array}{lr}x:&x\in\mathbb{Q}\cap (-\infty,1]\\x-b:&x\in\mathbb{Q}\cap(1,\infty)\\a:&x\notin\mathbb {Q}\end{array}\right\}
$

would this work>?
• Oct 25th 2009, 09:42 PM
redsoxfan325
It needs to be defined on $[a,b]$; that function there is still defined on $(-\infty,\infty)$.
• Oct 25th 2009, 09:55 PM
binkypoo
$g(x)=\left\{\begin{array}{lr}x:&x\in\mathbb{Q}\cap[a,a+\delta]\\x-b:&x\in\mathbb{Q}\cap[a+\delta,b]\\a:&x\notin\mathbb{Q}\end{array}\right\}
$

I'm shooting in the dark here, not too sure
• Oct 25th 2009, 10:10 PM
redsoxfan325
Quote:

Originally Posted by binkypoo
$g(x)=\left\{\begin{array}{lr}x:&x\in\mathbb{Q}\cap[a,a+\delta]\\x-b:&x\in\mathbb{Q}\cap[a+\delta,b]\\a:&x\notin\mathbb{Q}\end{array}\right\}
$

I'm shooting in the dark here, not too sure

Take any three constants $c,d,e\in[a,b]$ such that $a. Define a function:

$g(x)=\left\{\begin{array}{lr}x-c:&x\in\mathbb{Q}\cap[a,d]\\x-e:&x\in\mathbb{Q}\cap(d,b]\\0:&x\notin\mathbb{Q}\end{array}\right\}$

$g$ will be continuous at $x=c$ and $x=e$.