# 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:RR 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.

Homework Helper

## Homework Statement

I'm trying to do a problem, and in order to do it I need to find a function f:RR 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?

Homework Helper
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?

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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)).

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Gold Member
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.

1 person
Homework Helper
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.

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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!