Closed Set Proof: Homework Statement

[STEMucator]

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 A_n = [\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 S_i is closed for 1 ≤ i ≤ p now suppose that :

Q_o = \bigcup_{i=1}^{p} (ℝ^n - S_i)

So that Q_o 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.

 

[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.

 

[STEMucator]

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?

 

[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?

 

[STEMucator]

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

 

[jbunniii]

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?

 

[STEMucator]

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?

 

[jbunniii]

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.

 

[STEMucator]

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 F_c = {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 Q_c 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 F_o = {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 :)

 

[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.

 

[STEMucator]

Okay now continuing from before.

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

 

[jbunniii]

Okay now continuing from before.

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

 

[STEMucator]

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∈Q_o has a neighborhood that is also in Q_o, which means Q_o 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 U_i. Since this particular U_i is open...'

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

 

[jbunniii]

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

To : Suppose that x \in U_i, 1 ≤ i ≤ p, since every individual U_i 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.

 

[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.

 

[STEMucator]

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 U_i'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 U_i. This same neighborhood is not necessarily contained in any U_j with j≠i'.

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

 

[jbunniii]

So I have to show some circular neighborhood of x with radius δ is contained within ALL of the U_i so that the intersection of these U_i 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.

 

[STEMucator]

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 U_i is open, we can find a N_δ(x) \subseteq U_i.

Would this be the direction I would be taking then?

 

[jbunniii]

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 U_i is open, we can find a N_δ(x) \subseteq U_i.

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.

 

[STEMucator]

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

 

[jbunniii]

Since every neighborhood of x is contained within some U_i...

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.

 

[jbunniii]

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?

 

[STEMucator]

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 U_i contains SOME neighborhood of x, would the rest of what I stated follow?

 

[jbunniii]

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

So since each U_i 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.

 

[STEMucator]

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 Q_o'

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

 

[jbunniii]

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

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

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.

 

[STEMucator]

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 U_i 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?

 

[jbunniii]

Waaaait wait wait. So the intersection of your U_i 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?

 

[STEMucator]

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.

 

[jbunniii]

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?

 

[STEMucator]

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?

 

[jbunniii]

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?

 

[STEMucator]

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,...,δ_p) = δ_q for some 1≤i≤q≤p.

So N_δq(x) will be contained in all of the other δ neighborhoods which implies that N_δq(x) \subseteq U_i \forall1≤i≤q≤p. So it follows that N_δq(x) \subseteq Q_o which tells us that Q_o is indeed an open set.

 

[jbunniii]

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,...,δ_p) = δ_q for some 1≤i≤q≤p.

So N_δq(x) will be contained in all of the other δ neighborhoods which implies that N_δq(x) \subseteq U_i \forall1≤i≤q≤p. So it follows that N_δq(x) \subseteq Q_o which tells us that Q_o 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?

 

[STEMucator]

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?

 

[jbunniii]

We would wind up with a singleton set... and we both know what happens when we only have a single dot lol.

Right, or at least you could end up with one, depending on the sets. Looking at it in terms of the deltas, the set \{\delta_1, \delta_2, \delta_3, \ldots\} doesn't necessarily have a minimum if there are infinitely many deltas. For example, you could have \delta_n = 1/n, so you could never find a single \delta that satisfies 0 &lt; \delta \leq \delta_n for all n. Therefore x would not have any neighborhood contained in Q_o.

Cool, so it looks like you have a handle on this now. Don't forget that your original problem talked about a finite union of closed sets, and we translated that to an equivalent problem involving a finite intersection of open sets. So you still have a bit of work to do to relate the two. (Or maybe you already did that earlier, I can't remember.)

 

[STEMucator]

Right, or at least you could end up with one, depending on the sets. Looking at it in terms of the deltas, the set \{\delta_1, \delta_2, \delta_3, \ldots\} doesn't necessarily have a minimum if there are infinitely many deltas. For example, you could have \delta_n = 1/n, so you could never find a single \delta that satisfies 0 &lt; \delta \leq \delta_n for all n. Therefore x would not have any neighborhood contained in Q_o.

Cool, so it looks like you have a handle on this now. Don't forget that your original problem talked about a finite union of closed sets, and we translated that to an equivalent problem involving a finite intersection of open sets. So you still have a bit of work to do to relate the two. (Or maybe you already did that earlier, I can't remember.)

Oh yeah, I would make it clear I was using De Morgan's laws here to simplify this problem, I'm just getting some practice being as formal as possible here.

Also there was that other question which should be a bit easier (b) since it's only over the real line. The only thing that bothers me is.. well ill just restate the question here for convenience.

'Let be F a collection of closed intervals A_n = [1/n, 1 - 1/n] in ℝ for n=1,2,3... What do you notice about UF. Is it closed, open, both or niether?'

The interval A_n... i mean it's clear that 1/n < (n-1)/n for 1<n<2, but after that it flips. Not sure how to approach this one?

 

[jbunniii]

Oh yeah, I would make it clear I was using De Morgan's laws here to simplify this problem, I'm just getting some practice being as formal as possible here.

Also there was that other question which should be a bit easier (b) since it's only over the real line. The only thing that bothers me is.. well ill just restate the question here for convenience.

'Let be F a collection of closed intervals A_n = [1/n, 1 - 1/n] in ℝ for n=1,2,3... What do you notice about UF. Is it closed, open, both or niether?'

The interval A_n... i mean it's clear that 1/n < (n-1)/n for 1<n<2, but after that it flips. Not sure how to approach this one?

Oh right, I forgot there was a part (b). I'm not sure what you mean by "it flips" - although it does seem that the author of the question made a slight mistake. Here's what you get if you write out the first few intervals:

A_1 = [1, 0]
A_2 = [1/2, 1/2]
A_3 = [1/3, 2/3]
A_4 = [1/4, 3/4]
A_5 = [1/5, 4/5]

It's clear that the intent is to have the left side approach 0, and the right side approach 1. However, A_1 is problematic, unless you consider [1,0] to be the empty set, which is what I would recommend. (Or just ignore A_1 and start with A_2).

So what is the union of all these intervals? i.e. what is \cup_{n=1}^\infty A_n?

 

[STEMucator]

Oh right, I forgot there was a part (b). I'm not sure what you mean by "it flips" - although it does seem that the author of the question made a slight mistake. Here's what you get if you write out the first few intervals:

A_1 = [1, 0]
A_2 = [1/2, 1/2]
A_3 = [1/3, 2/3]
A_4 = [1/4, 3/4]
A_5 = [1/5, 4/5]

It's clear that the intent is to have the left side approach 0, and the right side approach 1. However, A_1 is problematic, unless you consider [1,0] to be the empty set, which is what I would recommend. (Or just ignore A_1 and start with A_2).

So what is the union of all these intervals? i.e. what is \cup_{n=1}^\infty A_n?

Yeah what I meant was we usually have : a<b when we see [a,b], but he pulled a fast one and said a>b for 2 > n ≥ 1. That's what I intended when i said flip for lack of better terminology.

Anyways, I digress. So we want the union of these sets over ℝ which is :

\bigcup_{n=1}^{∞} A_n = {[a,b] \in ℝ | a&gt;b \space for \space 2 &gt; n ≥ 1 \space and \space b&gt;a \space for \space n≥2}

Would this be it?

 

[jbunniii]

Yeah what I meant was we usually have : a<b when we see [a,b], but he pulled a fast one and said a>b for 2 > n ≥ 1. That's what I intended when i said flip for lack of better terminology.

Anyways, I digress. So we want the union of these sets over ℝ which is :

\bigcup_{n=1}^{∞} A_n = {[a,b] \in ℝ | a&gt;b \space for \space 2 &gt; n ≥ 1 \space and \space b&gt;a \space for \space n≥2}

Would this be it?

But what do you get when you evaluate the union? You should be able to express it as an interval.

 

[STEMucator]

But what do you get when you evaluate the union? You should be able to express it as an interval.

So would I get a finite interval? I've never been asked that before actually, but my hunch is that I would get [1,1]?

 

[jbunniii]

So would I get a finite interval? I've never been asked that before actually, but my hunch is that I would get [1,1]?

Take another look at the first 5 intervals: A_1, A_2, A_3, A_4, A_5 that I listed above. Let's take for granted that A_1 = [1, 0] is just the empty set, so it doesn't contribute anything to the union. (You can ignore it.) You can see that as n increases, the left endpoint of A_n is decreasing (toward what number?) and the right endpoint is increasing (toward what number?)

Keep in mind that the union consists of exactly those points that appear in at least one of the A_n's.

I have to take off for a while, but will check in again later this evening.

 

[STEMucator]

Take another look at the first 5 intervals: A_1, A_2, A_3, A_4, A_5 that I listed above. Let's take for granted that A_1 = [1, 0] is just the empty set, so it doesn't contribute anything to the union. (You can ignore it.) You can see that as n increases, the left endpoint of A_n is decreasing (toward what number?) and the right endpoint is increasing (toward what number?)

Keep in mind that the union consists of exactly those points that appear in at least one of the A_n's.

I have to take off for a while, but will check in again later this evening.

Ahhh so [0,1] would be my interval, and thanks for the help so far i'll catch you soon.

 

[Dick]

Ahhh so [0,1] would be my interval, and thanks for the help so far i'll catch you soon.

I don't think it's [0,1]. That contains both 0 and 1 and I don't think either of those are contained in any of your intervals.

 

[jbunniii]

Dick is right - if the union contains 0, then 0 must be in at least one of the A_n's. Same is true for 1. Can you find an A_n that contains either of these points?

 

[STEMucator]

Dick is right - if the union contains 0, then 0 must be in at least one of the A_n's. Same is true for 1. Can you find an A_n that contains either of these points?

Well, if we're not counting the first one then no I cannot find one.

 

[jbunniii]

Yes, I'm 99.99% sure that whoever wrote the question intended for A_1 to be empty. Assuming the interval notation has the usual meaning, [1,0] would be the set of all x such that 1 \leq x \leq 0, and there are obviously no values of x that satisfy this. Therefore the first non-empty interval is A_2 = [1/2, 1/2] = \{1/2\}, and from there, the intervals grow outward in both directions as n increases.

 

[jbunniii]

So, assuming what I just wrote is the correct interpretation, neither 0 nor 1 will be contained in any of the A_n's, and hence neither one is in the union. So what IS the union?

 

[STEMucator]

Yes, I'm 99.99% sure that whoever wrote the question intended for A_1 to be empty. Assuming the interval notation has the usual meaning, [1,0] would be the set of all x such that 1 \leq x \leq 0, and there are obviously no values of x that satisfy this. Therefore the first non-empty interval is A_2 = [1/2, 1/2] = \{1/2\}, and from there, the intervals grow outward in both directions as n increases.

If i represent what you're saying with [a,b]. Then as n->∞, a->0 and b->1. Why is this incorrect?

 

[jbunniii]

If i represent what you're saying with [a,b]. Then as n->∞, a->0 and b->1. Why is this incorrect?

Well, we just agreed that 0 and 1 are not in the union. So the union cannot be [0,1].

What if I take x such that 0 &lt; x &lt; 1? Is x in the union?