# Closed set proof

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## Homework Statement

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?

## Homework 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.

## 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.

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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.

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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.

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

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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?

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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?

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

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

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

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

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

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

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

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 $x \in N_\delta(x) \subset I$.

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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 $x \in N_\delta(x) \subset I$.

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 :)

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

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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?

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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.
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.

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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.

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?

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

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?
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.

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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 $j \neq i$.

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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 $j \neq i$.
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?

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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?
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.

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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.

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?

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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?
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.

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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 $i$.

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.

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Since every neighborhood of x is contained within some Ui...
No, this isn't true. All you know is that each $U_i$ contains SOME neighborhood of x, with some radius $\delta_i$. Of course, $U_i$ will also contain all neighborhoods of $x$ with radius smaller than $\delta_i$, but not necessarily those with larger radius.

And there may be neighborhoods of $x$ which are not contained in any of the $U_i$. In the concrete example I gave above, none of the $U_i$'s will contain a neighborhood of 0 with radius 5. That's larger than any of the sets.

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Let's continue for a moment with the concrete example above. Suppose I just have 3 sets: $U_1 = (-1, 1)$, $U_2 = (-1/2, 1/2)$, $U_3 = (-1/3, 1/3)$. And let's consider just $x = 0$.

What's the largest neighborhood of $x = 0$ which will fit inside $U_1$? How about $U_2$ and $U_3$? What's the intersection $U_1\cap U_2 \cap U_3$? What's the largest neighborhood that will fit inside $U_1 \cap U_2 \cap U_3$?

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No, this isn't true. All you know is that each $U_i$ contains SOME neighborhood of x, with some radius $\delta_i$. Of course, $U_i$ will also contain all neighborhoods of $x$ with radius smaller than $\delta_i$, but not necessarily those with larger radius.

Ahh I didn't see your example before, but now I get what you were getting at.

So since each Ui contains SOME neighborhood of x, would the rest of what I stated follow?

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Ahh I didn't see your example before, but now I get what you were getting at.

So since each Ui contains SOME neighborhood of x, would the rest of what I stated follow?

Well, the rest of what you stated is "it follows that every neighborhood of $x$ is also contained in $Q_o$". This is not true, but that's fine because you don't need that to be true. All you need is that SOME neighborhood of $x$ is contained in $Q_o$.

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Well, the rest of what you stated is "it follows that every neighborhood of $x$ is also contained in $Q_o$". This is not true, but that's fine because you don't need that to be true. All you need is that SOME neighborhood of $x$ is contained in $Q_o$.

'All you need is that SOME neighborhood of x is contained in Qo'

Does this not follow from the fact that we've found SOME Ui which contains SOME neighborhood of x? Since Qo is the intersection of these Ui, there will always be at least one neighborhood contained within Qo?

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'All you need is that SOME neighborhood of x is contained in Qo'

Does this not follow from the fact that we've found SOME Ui which contains SOME neighborhood of x? Since Qo is the intersection of these Ui, there will always be at least one neighborhood contained within Qo?

Each $U_i$ contains some neighborhood of $x$. However, these neighborhoods potentially all have different radii (values of $\delta$). How do you know that one of them will fit within $Q_o$?

This is not a fatuous question. In my earlier example, $U_i = (-1/i, 1/i)$, every one of these sets contains a neighborhood of $x = 0$, but if $Q_o = \cap_{i = 1}^\infty U_i$, then $Q_o = \{0\}$, so $Q_o$ does not contain ANY neighborhood of $x$. (And therefore $Q_o$ is not an open set.) Notice that I had to use an infinite number of sets to produce such an example. So the key is to explain why $Q_o$ DOES contain a neighborhood of $x$ if there are only finitely many sets in the intersection.

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Each $U_i$ contains some neighborhood of $x$. However, these neighborhoods potentially all have different radii (values of $\delta$). How do you know that one of them will fit within $Q_o$?

This is not a fatuous question. In my earlier example, $U_i = (-1/i, 1/i)$, every one of these sets contains a neighborhood of $x = 0$, but if $Q_o = \cap_{i = 1}^\infty U_i$, then $Q_o = \{0\}$, so $Q_o$ does not contain ANY neighborhood of $x$. (And therefore $Q_o$ is not an open set.) Notice that I had to use an infinite number of sets to produce such an example. So the key is to explain why $Q_o$ DOES contain a neighborhood of $x$ if there are only finitely many sets in the intersection.

Waaaait wait wait. So the intersection of your Ui example will converge to the set {0}. I understand why this happens. I'm a bit confused as to how to prove that a finite intersection will be different though. What if we had taken your example to p instead of ∞. Then would we be able to find a neighborhood inside its intersection?

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Waaaait wait wait. So the intersection of your Ui example will converge to the set {0}. I understand why this happens. I'm a bit confused as to how to prove that a finite intersection will be different though. What if we had taken your example to p instead of ∞. Then would we be able to find a neighborhood inside its intersection?
Yes, that's right. If there are only finitely many sets, then the intersection can't collapse to a single point like that. That's the intuitive explanation. Now, how to make the argument formal?

Once again I suggest thinking about a concrete case.

Let's say I have three sets, $U_1$, $U_2$, and $U_3$. Let's say there is a point $x$ contained in all three, and let's further assume that for each set, we have found a neighborhood that works:

$$x \in N_1(x) \subseteq U_1$$
$$x \in N_{1/2}(x) \subseteq U_2$$
$$x \in N_{1/10}(x) \subseteq U_3$$

In other words, a neighborhood of $x$ with radius 1 fits inside $U_1$, whereas I had to make the neighborhood smaller, radius 1/2, to fit inside $U_2$, and even smaller, radius 1/10, to fit inside $U_3$.

Without knowing anything else about the sets except the above, what radius $\delta$ is guaranteed to give us $x \in N_\delta(x) \subseteq U_1 \cap U_2 \cap U_3$?

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Yes, that's right. If there are only finitely many sets, then the intersection can't collapse to a single point like that. That's the intuitive explanation. Now, how to make the argument formal?

Once again I suggest thinking about a concrete case.

Let's say I have three sets, $U_1$, $U_2$, and $U_3$. Let's say there is a point $x$ contained in all three, and let's further assume that for each set, we have found a neighborhood that works:

$$x \in N_1(x) \subseteq U_1$$
$$x \in N_{1/2}(x) \subseteq U_2$$
$$x \in N_{1/10}(x) \subseteq U_3$$

In other words, a neighborhood of $x$ with radius 1 fits inside $U_1$, whereas I had to make the neighborhood smaller, radius 1/2, to fit inside $U_2$, and even smaller, radius 1/10, to fit inside $U_3$.

Without knowing anything else about the sets except the above, what radius $\delta$ is guaranteed to give us $x \in N_\delta(x) \subseteq U_1 \cap U_2 \cap U_3$?

Obviously your smallest radius of 1/10 will ensure this I would presume.

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Obviously your smallest radius of 1/10 will ensure this I would presume.

Correct. So generalize this. If I have $U_1$, $U_2$, ..., $U_p$ and corresponding neighborhoods such that

$$x \in N_{\delta_1}(x) \subseteq U_1$$
$$x \in N_{\delta_2}(x) \subseteq U_2$$
...
$$x \in N_{\delta_p}(x) \subseteq U_p$$

then what radius $\delta$ will ensure $x \in N_\delta \in \cap_{i=1}^p U_i$?

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Correct. So generalize this. If I have $U_1$, $U_2$, ..., $U_p$ and corresponding neighborhoods such that

$$x \in N_{\delta_1}(x) \subseteq U_1$$
$$x \in N_{\delta_2}(x) \subseteq U_2$$
...
$$x \in N_{\delta_p}(x) \subseteq U_p$$

then what radius $\delta$ will ensure $x \in N_\delta \in \cap_{i=1}^p U_i$?

Sorry for the late response, My differential eqs TA somehow didn't understand the fundamental theorem of calc and I was talking to her for a moment ( lol ).

So I believe that... 0 < δ < 1 would work?

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Sorry for the late response, My differential eqs TA somehow didn't understand the fundamental theorem of calc and I was talking to her for a moment ( lol ).

So I believe that... 0 < δ < 1 would work?

No, your answer should be a function of $\delta_1$, $\delta_2$, ..., $\delta_p$.

Recall the concrete example, where we had $\delta_1 = 1$, $\delta_2 = 1/2$, and $\delta_3 = 1/10$. How did you choose $\delta = 1/10$?

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No, your answer should be a function of $\delta_1$, $\delta_2$, ..., $\delta_p$.

Recall the concrete example, where we had $\delta_1 = 1$, $\delta_2 = 1/2$, and $\delta_3 = 1/10$. How did you choose $\delta = 1/10$?

Sorry about that terrible answer, I was thinking about it and my laptop decided to die on me so I posted some random garbage.

What i figured was that all the δs over i form a set over the reals and what we want is the min(δ1,...,δp) = δq for some 1≤i≤q≤p.

So Nδq(x) will be contained in all of the other δ neighborhoods which implies that Nδq(x) $\subseteq$ Ui $\forall$1≤i≤q≤p. So it follows that Nδq(x) $\subseteq$ Qo which tells us that Qo is indeed an open set.

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Sorry about that terrible answer, I was thinking about it and my laptop decided to die on me so I posted some random garbage.

What i figured was that all the δs over i form a set over the reals and what we want is the min(δ1,...,δp) = δq for some 1≤i≤q≤p.

So Nδq(x) will be contained in all of the other δ neighborhoods which implies that Nδq(x) $\subseteq$ Ui $\forall$1≤i≤q≤p. So it follows that Nδq(x) $\subseteq$ Qo which tells us that Qo is indeed an open set.

Yes, that's right. You want to choose $\delta = \min(\delta_1, \delta_2, \ldots, \delta_p)$. This works precisely because $p$ is finite. Can you see what could go wrong if there were infinitely many deltas?

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Yes, that's right. You want to choose $\delta = \min(\delta_1, \delta_2, \ldots, \delta_p)$. This works precisely because $p$ is finite. Can you see what could go wrong if there were infinitely many deltas?

We would wind up with a singleton set... and we both know what happens when we only have a single dot lol. More precisely though, if we had infinitely many deltas, there's no guarantee that we could find a delta to suit our needs?