# Closed set

## Homework Statement

How do I show that a simple set is closed?

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

## 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?

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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.

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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?

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

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.

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?

Anyone?

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.

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

Although proving that x -> (x, x^2) is an embedding is a lot more understandable and straightforward.
But being an embedding does not immediately imply closed.

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

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