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JH
Numerade Educator

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Problem 89 Hard Difficulty

The Cantor set, named after the German mathematician George Cantor (1845 - 1918), is constructed as follows. We start with the closed interval [0, 1] and remove the open interval $ \left( \frac {1}{3}, \frac {2}{3} \right). $ That leaves the two intervals $ \left[ 0, \frac {1}{3} \right] $ and $ \left[ \frac {2}{3}, 1 \right] $ and we remove the open middle third of each. Four intervals remain and given we remove the open middle third of each of them. We continue this procedure indefinitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of the numbers that remain in [0, 1] after all those intervals have been removed.
(a) Show that the total length of all the intervals that are removed is 1. Despite that, the Cantor set contains infinitely many numbers. Give examples of some numbers in the Cantor set.
(b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It is constructed by removing the center one-ninth of a square of side 1, then removing the centers of the eight smaller remaining squares, and so on. (The figure shows the first three steps of the construction.) Show that the sum of the areas of the removed squares is 1. This implies that the Sierpinski carpet has area 0.

Answer

a. $\frac{1}{3}, \frac{2}{3}, \frac{1}{9}, \frac{2}{9}, \frac{7}{9},$ and $\frac{8}{9}$
b. $\frac{1 / 9}{1-8 / 9}=1$

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Video Transcript

Let's go to the next page here for some more room. So for part A, we have the canner said This is where you take the closed interval zero one and then the first thing you do is you just remove the middle one third here. So everything between one third well, not including one third and then here will pick up at two thirds and go backto one. So removing this open middle third interval and then the next step is to just do the same thing, but for the smaller intervals. So here I'll take out the middle ther so I'm still a zero. And then this time I stopped at a third. That's a ninth and then two nines all the way back to one over three. And then over here, this's to over nine. Or you could think of it a six over nine to over three. So then we go one more over two seven over nine, and then we take out between seven over nine in the eight over nine and then you pick back up that ate over nine and go backto one. So the same thing we did from the first row to the second. We also didn't from seconds of third and I would continue in this fashion, and it's going to be very hard to draw the next one, so I won't be able to label it. But we take out the middle one third of each of these four lines. And basically I'LL stop writing there because it's just becoming too hard to draw. Now. I'd like to find this sum and removed length each that we removed the side length here. We removed the third and the first step from step zero two, Step one. We took out a third. Then, from the steps of the third row, we didn't take on the third, we took out a third of a third, so we really took out a knife and then we did it twice to let to over nine. Or you could think of that is to over three square to make this look like a geometric PSA and then going from the next line, we would take O a third of this ninth, but that would be won over twenty seven. And then there's a total of four of them. There's one there, there, there and here is well, so that will be for over twenty seven. And we could write that as too square over three, cute and so on. So we see that this is a geometric series and we see that are are the things that were multiplying by each time is two thirds, so we can go ahead and we have a some formula for geometric series. You take the first term of the whole series and then you divine by one minus. R In this problem, the first term in the Siri's is this one over three, and then we divide by one minus two thirds. But that's just the third over a third, and that equals one. And that will. That's exactly what we intended to prove for part A. Now, let's go ahead to the next page for part B. This time we're looking at Sir Pinsky Carpet. So you start with the square of side length one and then we come inside the square here. So let's maybe look at this middle one ninth. We want to remove. Think of the square is being this big square. You could break it up into nine pieces. We like to remove this middle square here and then eventually when we like to continue this pattern after I removed the square, then I'll go inside of each of these eight remaining squares and that also go inside there and also remove the middle one ninth. So we're not. We have eight left. We had nine and we took out a square. And so we have nine little ones. So remove the area. We would like to add that up and show that it's equal to one well. From the first square they're removed. Area you just remove the one ninth of the square. So one ninth, the whole ear has the whole square. The large square has area one times one equals one. So one ninth of the square it has area one overnight. Then in the next step, that's going a different color. Here, let's go to blue Now, after we zoom in and I'm each of these each of these squares here Now we're looking at squares of this size. They all have a little window will go in that little ninth will go in there and share all those over removing all of these areas here and I notice that there's eight of them. So let's write that we have eight of them and then what's the area we'LL Each of these squares has area one over nine, and then we were looking at one ninth of the inside. So that's one over nine squared a ninth of a night. So this is a over nine squared. Let me write. That is too cute. No, actually, let me write. That is eight a over nine. Swear now the next step would be to rinse and repeat the same process. Now I go to one of these squares here I goto one of these eight squares and then I go inside of that square and then so let me do it over here on the left, with the blue. Then I break this up into nine pieces total. I've removed the middle piece, but then I still have. He's eight remaining squares that are inside so we can see where the AIDS coming from. We always have a little windows left over. So there I would get I have eight of these green windows, but each of them have eight of their own smaller windows on the inside. So That's two IDs. And then here. How many nines? Well, we started off by taking a knife. Then we took another ninth. So and then here were This is our third time. So this time we're actually doing three nines there, the ninth of a ninth of the ninth. So here, the third time we removed the area would have eight squared over, man. Cute. And we could just see the pattern. Now, this is geometric geometric with our equals. Eight over nine. This is what you're multiplying by east time. And we know the formula for a geometric Siri's. You take the first term of the series, and then you just divided by one minus R. In our case, the first term is just one overnight. We could see that up here, but that was the first step. And then we do one minus eight over nine. Because this is our value, are God and simplify that. You get one over nine over a one over nine that's equal to one. And that is precisely the sum that we wanted for A and it also for a party. So this is our final answer