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Why does Cantor's diagonal argument fail to show that the set of all finite sequences of O's and 1'$ is uncountable?There is no contradiction; becaus...

Question

Why does Cantor's diagonal argument fail to show that the set of all finite sequences of O's and 1'$ is uncountable?There is no contradiction; because the sequence produced by taking the diagonal of the table counting all sequences and flipping its bits is not finite:There is no contradiction; because for finite sets we usually use the letters a,b.There is no contradiction; because infinite sets are usually countable:The argument does work The set of sequences named in the questio

Why does Cantor's diagonal argument fail to show that the set of all finite sequences of O's and 1'$ is uncountable? There is no contradiction; because the sequence produced by taking the diagonal of the table counting all sequences and flipping its bits is not finite: There is no contradiction; because for finite sets we usually use the letters a,b. There is no contradiction; because infinite sets are usually countable: The argument does work The set of sequences named in the question is actually uncountable and Cantors diagonal argument shows it to be 50



Answers

Show that the set of all finite bit strings is countable.

Let's follow these construction we put be zero to be the empty set. Now from a minus V zero, we take some element in some way. Let's say randomly on element X E X one course, be zeros empty. So when we remove nothing from A which left away in a is an infinite set. So while there are infinitely many elements, let's just take one. Let's call it X one. Then we put the one to be the said That only contains thes element X one. Now from a minus B one again, we take in some way on Element X two again, eh? As infinitely many numbers, I mean elements and we're not considering the elements of the one which is that just x one so well from infant. Infinitely many, we just remove one. So there are still in Fridley many, So we can take our next to say and then we put be, too to be said that contains X one and X two, and we keep repeating this process from a minus B n. We can still take some element because bien is going to be fine at, and so I am honest B and is going to be infinite. So there is something. This means that this process never stops because every bien again it's financially contents. Exactly. The elements x one x two all the way up to accent and so an A minus bien. He's still infinite. So there must. There must be something and in a month's bien, which means that we can take another element. Now we have the sequence off sets B one B two B C, which essentially means we have a sequence x one x two x three all the way up to Infinity because the process it never stops. So formally. Write this that Big B as the union off these sets bien But it's just means they were taking the said that contains x one x two, x three and so forth, which is an infinite set because the process never stopped. And he's countable because, well, we haven't admiration. We have x one x two x three Salt Well, there's objection with that, plus where X and goes to end and saw this set. Big B is a subset of a discounted Lee infinite as we wanted to see

So we're given comfortably. Many sets a n which accountable, and we want to show that their union is still countable, so the countable union of comfortable sets is still accountable. So to do that, let's fry the diagram where we have in the first column the elements off the set a one. So let's call them a one. A two, a three and so forth, then another column. You write the elements off the set a two. Let's call them B one B two B three and so forth. Then another call on the sets of a three. C wants to see three and we completely the ground. The diagram In this way. Now here, let's assume that all the sets a narrow, infinitely comfortable and all the elements are different because that's the hardest case. Because if some of the sets a we're fine, it, then that's not a big deal. The union off finance sets you still finance, so that doesn't change anything. And if there are some other lap ings between the sets, so if one element is both in some a one and a two, well, when we're uniting them, the union is Moeller, then it will be if all the sets were different. So the set where all the kids were all the sets are infinitely comfortable And all these joint is the hardest possible case. So let's assume that So all these a wanted to be three and all these numbers are all different. So the union of all the sets a wanted to a three and so forth We decide that contains all these numbers They wanted to be one and so forth. So now we want a way to enumerates these numbers. But again here the trick is going by diagonal enumeration. So this first element here we call it X one. Then we proceed diagonally So these x two and we go diagonally This is X three and then we proceed by the diagonals disease called x four We going diagonals disease x five Keep going Diagonals, thes xx and so forth. So, by drawing these diagonals, we're leasing all the elements off this union. Now the union ofall the end is going to be the set that contains x one x two x tree and so forth. And because a while we've laid out this diagram all these ex Sonics. Two extra A Are all the elements off these union when out is an enumeration off the elements of the union, and so it is comfortable.

So we have a set that's called a just accountable I would take s which is a subset of a But that means that the car the night of s he's lower equal than they can. I d away because as he's a subset away and they can tell right away is just I live with zero because a is countable so that leaves two possibilities there Either s he is finite or as is countable


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