Continuous Function- Open Sets

[analysis001]

Homework Statement

I'm trying to do a problem, and in order to do it I need to find a function f:R→R which is continuous on all of R, where A$\subseteq$R is open but f(A) is not. Can anyone give an example of a function that satisfies these properties? I think once I have an example I'll understand things a lot better. Thanks.

 

[Dick]

Homework Statement

I'm trying to do a problem, and in order to do it I need to find a function f:R→R which is continuous on all of R, where A$\subseteq$R is open but f(A) is not. Can anyone give an example of a function that satisfies these properties? I think once I have an example I'll understand things a lot better. Thanks.

Try a function with a maximum or a minimum and think about it. A lot of functions work.

 

[analysis001]

Oh ok would f(A)=sin(A) work?

 

[Dick]

Oh ok would f(A)=sin(A) work?

It would. Now you have to give an example of an open set A such that f(A) isn't open.

 

[analysis001]

It would. Now you have to give an example of an open set A such that f(A) isn't open.

If I take A to be the interval (-$\pi$,$\pi$) which is open, then sin(A) would be on the interval [-1,1], which is closed. Am I understanding this correctly?

 

[Dick]

If I take A to be the interval (-$\pi$,$\pi$) which is open, then sin(A) would be on the interval [-1,1], which is closed. Am I understanding this correctly?

Almost, but sin((-pi,pi))=(-1,1) which is an open interval. The endpoints aren't included. Try sin((0,pi)).

 

[jbunniii]

Almost, but sin((-pi,pi))=(-1,1) which is an open interval. The endpoints aren't included. Try sin((0,pi)).

That's not true. $\sin(\pi/2) = 1$, $\sin(-\pi/2) = -1$

Any open interval containing $[-\pi/2,\pi/2]$ will work.

 

[STEMucator]

I believe something like:

$f: x → a^2 - (x-a)^2$ should fit those requirements.

 

[analysis001]

Almost, but sin((-pi,pi))=(-1,1) which is an open interval. The endpoints aren't included. Try sin((0,pi)).

I don't get why sin((-$\pi$,$\pi$)=(-1,1). I thought because sin($\pi$/2)=1 and since $\pi$/2$\in$ (-$\pi$,$\pi$) and also because sin(-$\pi$/2)=-1 and since -$\pi$/2$\in$ (-$\pi$,$\pi$) then sin((-$\pi$,$\pi$)=[-1,1].

If sin((0,$\pi$) then it equals (0,1] which is neither open nor closed. I'm not sure where my reasoning is wrong.

 

[Dick]

I don't get why sin((-$\pi$,$\pi$)=(-1,1). I thought because sin($\pi$/2)=1 and since $\pi$/2$\in$ (-$\pi$,$\pi$) and also because sin(-$\pi$/2)=-1 and since -$\pi$/2$\in$ (-$\pi$,$\pi$) then sin((-$\pi$,$\pi$)=[-1,1].

If sin((0,$\pi$) then it equals (0,1] which is neither open nor closed. I'm not sure where my reasoning is wrong.

jbunniii's correction is correct and you are correct. I slipped up. Sorry!

 

[analysis001]

Thanks everyone!