# Open sets and cartesian products

1. Mar 29, 2013

### bedi

Let f be a continuous function from R to R and let A be a subset of R^2. Define A={(x,y): y<f(x)}.

Can you express A as a cartesian product of two open sets? I tried RxU alpha_x where alpha_x = {y:y<f(x)}. But that didn't work, i need to change something about R.

2. Mar 29, 2013

### Staff: Mentor

If those open sets should be subsets of R, I don't see how this can be possible.
Two open sets B,C of R or R^2, where A={b+c|B in B, c in C} are possible, but that is not B x C.

3. Mar 29, 2013

### bedi

Let A=U_y {{f^-1(x):f(x)>y}x{y':y<maxf(R) is there is such a max and y'<y}}. Does that work?

4. Mar 29, 2013

### bedi

But this doesn't contain those y's. If we add them manually to A will that change something?

Edit:
If f has a max, say f(D), then add (f^-1(D), y) to A. If not then there exists an f(x') such that f(x')>y so add (f^-1(x'), y) to A. What do you think?

Last edited: Mar 29, 2013
5. Mar 29, 2013

### Staff: Mentor

What does "{f^-1(x):f(x)>y}" mean? You use "x" to denote values otherwise written as y, and get a set of values usually described as x. In addition, it is meaningless to use the same x in f and f^(-1). And how does y get defined here?

What is "{y':y<maxf(R) is there is such a max and y'<y}"? What happens if there is no such maximum? And if there is one, what does that mean?

6. Mar 29, 2013

### bedi

Actually you can ignore my last reply above.
Alright so write {x:f(x)>y} instead. We know this is open because f is continuous. y is defined as an arbitrary real number such that y<sup f so that y<f(x) could make sense. And if we put y'<y we get an open subset of real numbers consisting of y' s such that y'<y<f(x).

7. Mar 29, 2013

### Staff: Mentor

I don't see what you are doing.

Let's consider two examples:
(1) f(x)=-y^2
(2) f(x)=y

How do your sets look like?

8. Mar 29, 2013

### bedi

(1)
A={{x: f(x)=-y^2>-1}x{y'<-1: y' is a real}}U{{x: f(x)=-y^2>-2}x{y'<-2: y' is a real}}U{{x: f(x)=-y^2>-sqrt2}x{y'<-sqrt2: y' is a real}}U....

(2) is similar, I just substitute random real values for y.

9. Mar 29, 2013

### WannabeNewton

The assumption isn't even true in general so I'm not sure what you're trying to prove with all that work. Given $f:\mathbb{R}\rightarrow \mathbb{R}$, assume that $A = \mathbb{R}\times U$ is in fact true, where $U\subseteq \mathbb{R}$ is open. If $f$ isn't constant then there exist $x,y\in \mathbb{R}$ such that $f(x) < f(y)$. Therefore $(y,f(x))\in A$ but then $f(x)\in U$ implying $(x,f(x))\in A$ since $A = \mathbb{R}\times U$ by hypothesis, which is a contradiction.

10. Mar 29, 2013

### bedi

Yes, this is why I said R x {something} doesn't work. I think I should write these in latex, it may be the case that you think I wrote U as a set but it is the union operation.

Last edited: Mar 29, 2013