# Closed set

1. Apr 30, 2012

### dracox

1. The problem statement, all variables and given/known data
How do I show that a simple set is closed?

Ex: the set of points defined by the parabola y=x^2

3. The attempt at a solution
Well, a set is closed iff it contains all of its limit points. So, I want to show that this is true for the given set. I'm not exactly sure how to do this. Any help?

2. Apr 30, 2012

### BrownianMan

It's often easier to use certain theorems to prove things like this than trying to prove them from the definition.

Couldn't you use the theorem that states that if f: X -> Y is a continuous function and V is contained in Y (where V is closed) then f^-1 (V) is closed?

EDIT: nevermind. I didn't see the example you provided in the OP. This theorem won't work.

Last edited: Apr 30, 2012
3. Apr 30, 2012

### dracox

Hmm, I'm not familiar with this theorem. Is there somewhere I could read about it on the web? Or, if you're inclined, you could explain it. What would V be in this case?

4. Apr 30, 2012

### dracox

Ah, nevermind, I am familiar with this theorem. I just found it in my textbook. How shall I apply it, though?

5. Apr 30, 2012

### hamsterman

In the example case it is easiest to say that there is a homomorphism from a curve to R, since being closed (or open) is a topological property.

Another way should be to show that its complement is open by taking a point outside, finding its (least) distance to the set and then showing that if that distance is 0, the point is in the set.

I'm not sure how to apply the definition directly.

6. Apr 30, 2012

### dracox

Ok, I think I'll try it your second way. Let (x_0, y_0) be a point in the complement of the set (call it A). How would I proceed from here?

7. Apr 30, 2012

Anyone?

8. May 1, 2012

### hamsterman

Well, since you have a curve, it is easy to write a function of distance from some point to (x0, y0). Then the least distance d(x, y) is the minimum of that function. Then show that d(x, y) = 0 means that y = x^2.

Although proving that x -> (x, x^2) is an embedding is a lot more understandable and straightforward.

9. May 1, 2012

### micromass

Can you find a function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ such that f(x,y)=0 if and only if y=x²??

10. May 1, 2012

### micromass

But being an embedding does not immediately imply closed.

11. May 1, 2012

### hamsterman

It should, if it's an embedding from R, I think.

12. May 1, 2012

### micromass

We have that

$$\mathbb{R}\rightarrow \mathbb{R}^2:x\rightarrow (e^x,\sin(\frac{1}{e^x}))$$

is an embedding that is not closed.