Closed set proof

1. Sep 15, 2012

Zondrina

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

This question has two parts :

(a) Let F be a finite collection of closed sets in ℝn. Prove that UF ( The union of the sets in F ) is always a closed set in ℝn.

(b) Let F be a finite collection of closed intervals An = [$\frac{1}{n}, 1- \frac{1}{n}$] in ℝ for n = 1,2,3..... What do you notice about UF? Is it closed, open, both or niether?

2. Relevant equations

I know a set S is open ⇔ it is equal to its own interior, that is S = S0.
I know a set S is closed if its compliment ℝn-S is open.
I also know a set S is closed ⇔ B(S) $\subseteq$ S.
I know that the only open and closed sets are ℝn and ∅.
I know a set S is neither open or closed if there is at least one point which has a neighborhood containing points from both S and its compliment.

3. The attempt at a solution

(a) Suppose that F = {$S_1, S_2, ..., S_p | S_i\in ℝ^n, 1 ≤ i ≤ p$}

Suppose further that :

$$Q = \bigcup_{i=1}^{p} S_i$$

We want to show Q is always a closed set in ℝn, so let us show that the compliment of Q, ℝn - Q is open.

We know from F that : $S_i \subseteq B(S_i), \space 1≤ i ≤ p$. That is, each individual set contains its own boundary.

This implicates to us that : $ℝ^n - S_i$ is open since Si is closed for 1 ≤ i ≤ p now suppose that :

$$Q_o = \bigcup_{i=1}^{p} (ℝ^n - S_i)$$

So that Qo is composed of a finite collection of the compliments of F.

Am I on the right track here or have I missed something? Ill attempt (b) after I figure this one out.

2. Sep 15, 2012

HallsofIvy

It's impossible to answer this without knowing what definition of "closed set" you are using and what prior theorems you have. If your definition of "closed set" is "its complement is an open set" and you know that the finite union of open sets are open then it is easy.

3. Sep 15, 2012

Zondrina

Sadly all I have are these theorems and propositions. Is what I'm doing wrong? Would I use neighborhoods somehow?

4. Sep 15, 2012

jbunniii

You don't need to use boundaries at all.

I think you were heading in this direction: proving that the union of a finite number of closed sets is closed is equivalent to proving that the intersection of a finite number of open sets is open. This is correct and a useful observation. Thinking in terms of neighborhoods is also a good idea. What fact about neighborhoods characterizes an open set?

5. Sep 15, 2012

Zondrina

Uhm, if all the points of a set are interior points, then the set is open.

So : $\forall x \in S, \exists δ>0 | N_δ(x) \subseteq S$

Last edited: Sep 15, 2012
6. Sep 15, 2012

jbunniii

Correct. So what's an interior point? How would you define it in terms of neighborhoods?

7. Sep 15, 2012

Zondrina

I re - edited my last post, is that what you meant?

8. Sep 15, 2012

jbunniii

Yes, good. So here are a few steps toward a proof:

Suppose you have a finite collection of open sets, let's call them $U_1, \ldots, U_M$, and form their intersection:

$$I = \bigcap_{m=1}^{M} U_m$$.
If $I$ is empty then you're done. (Why?) If $I$ is not empty, then choose any $x \in I$. The goal is to prove that $x$ is an interior point of $I$. Thus you need to find a neighborhood $N_\delta(x)$ such that $x \in N_\delta(x) \subset I$.

9. Sep 15, 2012

Zondrina

So my proof should look more like :

(a) Suppose that Fc = {$S_1, S_2, ..., S_p | S_i\in ℝ^n, 1 ≤ i ≤ p$}

Suppose further that :
$$Q_c = \bigcup_{i=1}^{p} S_i$$

We want to show Qc which is composed of the union of finitely many closed sets is always a closed set in ℝn. This is equivalent to showing the intersection of finitely many open sets is open.

So suppose Fo = {$U_1, U_2, ..., U_p | U_i\in ℝ^n, 1 ≤ i ≤ p$} is such a set of finitely many open sets.

We want to show that :

$$Q_o = \bigcap_{i=1}^{p} U_i$$

Is an open set?

Sorry about not being to continue this, but I have a shift at work today. Perhaps you'll be on later.

Thanks for the help so far anyway :)

10. Sep 15, 2012

jbunniii

Yes, it looks fine so far. I'm sure if I'm not around when you continue, someone else will jump in and help.

11. Sep 16, 2012

Zondrina

Okay now continuing from before.

To prove Qo is an open set. Take x$\in$Qo. Since Qo is the intersection of finitely many open sets, we know that x is contained within some particular Ui. Since this particular Ui is open, $\exists δ>0 | N_δ(x) \subseteq U_i \Rightarrow N_δ(x) \subseteq Q_o$. So obviously, any x$\in$Qo has a neighborhood that is also in Qo, which means Qo is open.

So since I've shown that the intersection of finitely many open sets is open, we can simply say P implies Q in this scenario and we know that the union of finitely many closed sets is closed.

Is this good?

Last edited: Sep 16, 2012
12. Sep 16, 2012

jbunniii

No, this doesn't work. It's true that $N_\delta(x) \subseteq U_i$, but it does not follow that $N_\delta(x) \subseteq Q_o$. This is because $Q_o$ is the intersection of $U_i$ with other sets, so in general $Q_o$ is smaller than $U_i$. You have to take advantage that $x$ is in all of the $U_i$'s, not just one of them.

13. Sep 16, 2012

Zondrina

I really don't understand what you said there. The reason being is my concluding statement here : 'So obviously, any x∈Qo has a neighborhood that is also in Qo, which means Qo is open'.

Presuming what you're saying to me is correct though, would I simply just change this statement : 'we know that x is contained within some particular Ui. Since this particular Ui is open.....'

To : Suppose that $x \in U_i, 1 ≤ i ≤ p$, since every individual Ui is open..... blah blah. Then I would finish it off with a P implies Q?

14. Sep 16, 2012

jbunniii

The problem is that this is not "obvious" and in fact it does not follow from what you have written so far.

Well, you need to carefully spell out the details. You are arguing that there is a neighborhood of x which is entirely contained in $Q_o$. What is the radius of that neighborhood? It may seem like I'm being pedantic here, but you need to be careful here. If there were infinitely many $U_i$'s, the theorem is actually false. So your proof needs to use the fact that there are only finitely many.

15. Sep 16, 2012

jbunniii

Here's the main thing wrong with your argument. If x is in $Q_o$, then given $i$, we have $x \in U_i$, and there is a neighborhood of $x$ contained in $U_i$. OK so far. However: it's not necessarily true that this neighborhood is contained in $Q_o$. Generally this will not be true. This is because all you know is that the neighborhood is contained in $U_i$. This same neighborhood is not necessarily contained in any $U_j$ with $j \neq i$.

16. Sep 16, 2012

Zondrina

I appreciate you being pedantic, otherwise what would I actually be learning right? My ultimatum is to understand so by all means show me what for.

Okay I understand your argument clearly now. So what struck me the most :

'You have to take advantage that x is in all of the Ui's, not just one of them'

As well as :

'What is the radius of that neighborhood?' ( δ ).

Finally :

'This is because all you know is that the neighborhood is contained in Ui. This same neighborhood is not necessarily contained in any Uj with j≠i'.

So I have to show some circular neighborhood of x with radius δ is contained within ALL of the Ui so that the intersection of these Ui will also contain this neighborhood?

17. Sep 17, 2012

jbunniii

Yes, that's exactly what you need to show. This will show that $x$ is an interior point of $Q_o$. Since x was an arbitrary point of $Q_o$, this implies that $Q_o$ is open.

18. Sep 17, 2012

Zondrina

So we must show : $\forall x \in Q_o, \exists δ > 0| N_δ(x) \subseteq Q_o$

Suppose $x \in U_i, \forall i≤p.$ Since every Ui is open, we can find a Nδ(x) $\subseteq$ Ui.

Would this be the direction I would be taking then?

19. Sep 17, 2012

jbunniii

Yes, it's the right direction. However, there's something important to note: the radius $\delta$ of the neighborhood for one of the $U_i$'s might not work for another (it might be too large). So I suggest making the following small but important change:

Suppose $x \in U_i, \forall i \leq p$. Since every $U_i$ is open, we can find a $N_{\delta_i}(x) \subseteq U_i$, where the radius $\delta_i$ depends on $i$.

It might be helpful to keep in mind a concrete example. Suppose $U_i = (-1/i, 1/i)$, in other words, the open interval from $-1/i$ to $1/i$. Now, $x = 0$ is contained in $U_i$ for every $i$. However, for larger values of $i$, the length of $U_i$ shrinks. So a neighborhood that works for a smaller $i$ might not work for a larger one.

Another important thing to note about this concrete example is that if $i$ is allowed to be arbitrarily large, then $x = 0$ is the ONLY point contained in all of the $U_i$'s, and therefore there is no neighborhood which will be contained in all of them. This is why the theorem fails if you don't include the restriction that there are only finitely many sets.

Last edited: Sep 17, 2012
20. Sep 17, 2012

Zondrina

Yes yes, duly noted makes sense obviously. So knowing that, I shall try to continue here.

Since every neighborhood of x is contained within some Ui, would it follow that every neighborhood of x is also contained within Qo? That would tell us that Qo is equal to its own interior and is therefore open.