Open/Closed Set Homework: Examples of \mathbb{R}^2

[Ted123]

Homework Statement

If f:\mathbb{R}\to\mathbb{R} and g:\mathbb{R}\to\mathbb{R} are continuous functions, give examples to show that the set \{ (f(x),g(x)) : x\in\mathbb{R} \} might or might not be closed in \mathbb{R}^2.

The Attempt at a Solution

Letting f(x)=g(x)=0 gives the set equal to \{ (0,0) \}, a singleton and singleton sets are closed.

What functions would make the set open?

 

[Dick]

That's a pretty trivial example. You should probably think about it a little more before someone just gives you the answer. Suppose f(x) and g(x) have a horizontal asymptote as x->inf. There's other ways it can fail to be closed as well, but that should get you started.

 

[Ted123]

That's a pretty trivial example. You should probably think about it a little more before someone just gives you the answer. Suppose f(x) and g(x) have a horizontal asymptote as x->inf. There's other ways it can fail to be closed as well, but that should get you started.

Well if the functions have horizontal asymptotes they won't be continuous on the real line for a start.

If we take f(x) a constant function, what condition must g(x) satisfy in order that the set isn't closed?

 

[micromass]

What functions would make the set open?

One VERY IMPORTANT THING:

Open is NOT the same as "not closed".

You don't need to find a set that is open, you need to find a set that is not closed. These are completely different questions!

 

[Dick]

Well if the functions have horizontal asymptotes they won't be continuous on the real line for a start.

If we take f(x) a constant function, what condition must g(x) satisfy in order that the set isn't closed?

I said HORIZONTAL asymptote. Why do you think that means it wouldn't be continuous?

 

[Ted123]

Sorry, of course it can be continuous.

Is the set \{ (0,x):x\in\mathbb{R} \} not closed?

 

[Dick]

Sorry, of course it can be continuous.

Is the set \{ (0,x):x\in\mathbb{R} \} not closed?

It doesn't look closed to me. What do you think? Keep micromass's comment in mind too. A set that is 'not closed' doesn't have to be open.

 

[Dick]

It doesn't look closed to me. What do you think? Keep micromass's comment in mind too. A set that is 'not closed' doesn't have to be open.

I changed my mind. I thought you were trying to write an open subsegment of a straight line. You're aren't, are you? Your set is the x-axis. It's closed in R^2. How would you show that?

 

[Ted123]

I changed my mind. I thought you were trying to write an open subsegment of a straight line. You're aren't, are you? Your set is the x-axis. It's closed in R^2. How would you show that?

I've just realized that I'm confusing 'open' and 'not closed'.

With f a constant function, I need to find g such that g(\mathbb{R}) is not closed. What is an example a function from R to R that isn't closed?

 

[Dick]

I've just realized that I'm confusing 'open' and 'not closed'.

With f a constant function, I need to find g such that g(\mathbb{R}) is not closed. What is an example a function from R to R that isn't closed?

So you aren't going to try the horizontal asymptote suggestion?

 

[Ted123]

So you aren't going to try the horizontal asymptote suggestion?

g(x)=e^x.

Then \{(0,e^x) :x\in\mathbb{R} \} is not closed is it?

 

[Dick]

g(x)=e^x.

Then \{(0,e^x) :x\in\mathbb{R} \} is not closed is it?

No, it isn't. Why isn't it closed?

 

[Ted123]

No, it isn't. Why isn't it closed?

The boundary of A = \{(0,e^x) : x\in \mathbb{R} \} is \partial A =\{(0,0)\} and (0,0) \notin A so A doesn't contain all its boundary and is therefore closed.

 

[Dick]

The boundary of A = \{(0,e^x) : x\in \mathbb{R} \} is \partial A =\{(0,0)\} and (0,0) \notin A so A doesn't contain all its boundary and is therefore closed.

Good answer! Except that the boundary of A is actually all of A AND (0,0). Every point in A is also a boundary point of A. But the critical thing is that (0,0) is on the boundary of A but not in A.

 

[Ted123]

Would you express the boundary as \partial A = A\cap \{(0,0)\}?

 

[Dick]

Would you express the boundary as \partial A = A\cap \{(0,0)\}?

Certainly not, it's a union not an intersection.