Union of a set and it's limit points

[powerovergame]

Homework Statement

Prove that:
The union of a set U and the set of its limit points is the closure of U.

Homework Equations

Definitions:
Closure: The closure of U is the smallest closed set that contains U.
Limit points: if z is a limit point in U, then any open circle around z intersects U at points other than z.

The Attempt at a Solution

My attempt was to prove that:
The Closure is contained in the union of U and L(U), and
The union of L(U) and U contains it's closure.

However my instructor did something that I was never able to understand.
He first tried to prove that the union of U and L(U) is closed. To do so, he tries to prove it's complement is open. To prove its complement is open, he used contradiction:
This is where I really got lost. I think this is what he said:

Suppose it's complement of U is closed, then you can pick a point z (like on the boundary) on the complement of U, such that the any open disk around z will have to intersect U.

Now suppose the complement of L(U) is closed, then you can pick a point w (once again on the boundary of this set) that is in a open ball of z, and then any disk around w will have to intersect L(U). But if any of w's open disk must intersect L(U), then w is defined to be the limit point of L(U). This is a contradiction, as we let L(U) to be the set of all of its limit points, yet w is limit point of L(U) that is not in L(U).

The question is: While this does prove that the complement of L(U) is open, how does it prove the complement of the union of L(U) and U is open?

Of course this does not even finish the proof. In the second part of the proof the instructor attempted to prove that one of these two statements:
-The Closure is contained in the union of U and L(U), and
-The union of L(U) and U contains it's closure.
I listed them both because I can't remember which one he was trying to prove.

Any suggestions for the second part?

Thank you for reading this long thread.

 

[Gib Z]

Every element of the set U is the limit of some sequence in the set U, so you just need to show the limit of every sequence in U is in the closure, and that every point in the closure is the limit of a sequence in U.

For the first one, proof by contradiction would be my first attempt. Suppose a sequence in U had a limit not in the closure, and a contradiction will come very quickly.

For the second one, for every point in the closure of U, I would outline the construction of a sequence in U whose limit is that point.

 

[Tedjn]

In addition, regarding the following statements:

Suppose it's complement of U is closed, then you can pick a point z (like on the boundary) on the complement of U, such that the any open disk around z will have to intersect U.

Now suppose the complement of L(U) is closed, then you can pick a point w (once again on the boundary of this set) that is in a open ball of z, and then any disk around w will have to intersect L(U).

You need to be careful with language. A set can be neither open nor closed. Denote the union of U and L(U) by C. You want to show that C^c is open by contradiction; if it is not open it contains a point on the boundary but is not necessarily closed. Intuitively, your closed set contains all of its boundary, but a set that is not open just needs to contain some of its boundary (just one point will do). Continue the contradiction from here. By the way, this turns out to be equivalent to what I think GibZ is suggesting.

 

[tmccullough]

I have to confess that what you wrote confused me quite a bit. This is not a fact particular to metric spaces at all, so I'm just going to use open. Let me sketch a really quick proof, using your notation.

First, note that if V is an open set and $$V\cap U = \emptyset$$ then it is immediate that $$V\cap L(U) = \emptyset$$.

Suppose that $$x\notin \big(U\cup L(U)\big),$$ then (x is not a limit point of U, so by definition) there must be an open set $$V_x$$ such that
$$x\in V_x,\; V_x\cap U = \emptyset$$
It must be the case that
$$V_x\cap \big(U\cup L(U)\big) =\emptyset$$

This is enough to conclude that the complement of $$U\cup L(U)$$ is open. Does that help?