# Relatively open sets

1. Nov 16, 2008

### kittybobo1

1. The problem statement, all variables and given/known data

Let f be a continuous real-valued function of R^n. Show that [(x,y): x \in R^n, y > f(x)] is an open subset of R^{n+1}

2. Relevant equations

3. The attempt at a solution

If I am thinking about this right... Since f(x) goes from (-oo, y), this is an open subset, and as f is continuous, the domain of f(x) must be open, so x\in U is open.

Last edited: Nov 16, 2008
2. Nov 16, 2008

### VeeEight

I don't know if it is just me but I am having a hard time understanding what is going on in your set.

As for functions on open sets, perhaps you may want to consult the different definitions of a continuous function (besides the traditional episilon delta defn).

3. Nov 16, 2008

### kittybobo1

/Bump

4. Nov 16, 2008

### VeeEight

What is the complement of this set?

5. Nov 16, 2008

### kittybobo1

Would the complement be as all the x are open, then the complement would be closed. Then the complement of the y would be y <= f(x), and as both of these are closed, then in R^{n+1} they would be closed. Thus as the complement is closed, it must be open?

I guess one problem I am having is the y term. Are the f(x) chosen so that it is less than y? Thus y is one point. Or is it given an f(x), the y's go from (f(x), oo).

6. Nov 16, 2008

### VeeEight

It looks like y is chosen so it is less than f(x) for all f(x) in your range.
If you can show that the complement is closed then you are done.

7. Nov 16, 2008

### kittybobo1

I know if I show the complement is closed then it is open, but is there any reasonable way to show that the complement for this set is closed?

8. Nov 16, 2008

### kittybobo1

/bump

9. Nov 17, 2008

/bumpz