Closed Set Proof: Homework Statement

[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 $\forall$1≤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 $\forall$1≤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 $0 < \delta \leq \delta_n$ 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 $0 < \delta \leq \delta_n$ 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 :

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

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 :

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

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 $0 < x < 1$? Is $x$ in the union?

 

[STEMucator]

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 $0 < x < 1$? Is $x$ in the union?

Yes x would indeed be in the union then. So I'm guessing I had to be particular about the endpoints i.e (0,1)

 

[jbunniii]

Right, (0,1) is the answer. So think for a moment about why this question was asked as part (b), right after you showed that the union of a finite number of closed sets is closed.

 

[STEMucator]

Right, (0,1) is the answer. So think for a moment about why this question was asked as part (b), right after you showed that the union of a finite number of closed sets is closed.

So part (a) told me that the union of finitely many closed sets is closed. Hence finitely. Now we have an INFINITE amount of closed intervals, but their union is (0,1), a single open interval.

I'm thinking that I have to argue that this must be open because $\forall x \in (0,1), \exists δ>0 | (x-δ, x+δ) \subseteq (0,1)$

 

[jbunniii]

So part (a) told me that the union of finitely many closed sets is closed. Hence finitely. Now we have an INFINITE amount of closed intervals, but their union is (0,1), a single open interval.

I'm thinking that I have to argue that this must be open because $\forall x \in (0,1), \exists δ>0 | (x-δ, x+δ) \subseteq (0,1)$

Yes, that's right. All open intervals are open sets, for the same reason. You can always find room within the set to fit a neighborhood around any point.

 

[STEMucator]

Yes, that's right. All open intervals are open sets, for the same reason. You can always find room within the set to fit a neighborhood around any point.

Wow is that really it? Man I love the real number system... even though its a tad bit broken, but nonetheless part (a) was very convenient here.

 

[jbunniii]

I'm off for the evening. I think you have this problem wrapped up now, nice work! I'm sure this stuff will become a lot more intuitive as you get more practice.

 

[STEMucator]

I'm off for the evening. I think you have this problem wrapped up now, nice work! I'm sure this stuff will become a lot more intuitive as you get more practice.

Yeah man, you deserve mentor status for this one. A month or two and I'm sure ill have this down pat. Though I'm sure I'll be back with some interesting exercises from my prof tomorrow. Thanks for all your help.

 

[jbunniii]

Wow is that really it? Man I love the real number system... even though its a tad bit broken, but nonetheless part (a) was very convenient here.

The real number system is pretty amazing. You're just starting to scratch the surface - wait until you start playing around with the Cantor set and measure theory and Baire category and so forth. Really cool stuff, and it all builds on the basics that you're learning now.